The Time Everyone “Corrected” the World’s Smartest Woman (2015) | Hacker News Reader

The Time Everyone “Corrected” the World’s Smartest Woman (2015)

priceonomics.com

211 pointsjasonhansel303 comments

There are a number of things I like about the monty hall problem. There's the history, the unintuitiveness, the subtle easy-to-screw-up nature of probability problems, the calculation, the sociology, and the overconfidence of wrong experts.

Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness. In fact, it would be much harder to be so widely wrong now that any statistician can so easily just code up a computer to run it a zillion times and get an empirical answer. But despite that, your intuition might still say otherwise.

I'm fascinated by the gap between intuition and logic particularly because it can be closed as soon as you find the right lens to take to a problem. I like the example in the article about 100 doors, or even a million doors, where it really helps drive it home in your gut.

That moment when you find the right intuition for a thing is a magical one that makes me love math.

The day this was published in Parade magazine, I was sure that she was wrong.

Even though I was a computer programmer then (1990), I didn't own a computer. I drove across town to my parent's house to use my father's computer to prover her wrong.

I didn't even have to run the program. Just the act of writing the program made me realize that she was right. When I was writing the part where the host picks which door to open, I had a revelation.

Because in 2/3 scenarios you had to make the host choose the door without a car?

The host can never choose the door with the car, because then the game is over. Ergo the host has to choose the door with goat, every time.

this constraint was unclear to me in the original problem statement, which is why I thought she was wrong, too. It wasn't until somebody wrote in a letter saying they wrote a computer program that I understand what the rules were, and then it was obvious.

I wish the problem was stated more clearly to say, the host has to choose the door with goat.

Right, and the host is only making a meaningful choice in the 2/3 scenarios where you didn't pick the car in the first place.

In this case, though, I don't believe it's just calculation vs. intuition. The reason the 100 or million door problem helped me understand was because it clarified an implicit rule of the game: that the door the host chooses to open is not random. If it were random, it would make for a very boring game ("Uhh, there's the prize, you win, I guess that's game over.") Once I understood that the host would only ever open "goat" doors, that's when I better understood the argument.

That was exactly what happened with me. I hadn't understood in the original proposal that the host's choice is not random. At some point during someone's offer of the '100 door' version, I realised they were operating with this extra piece of information I lacked.

Indeed, that is a crucial bit of information that is not made explicit in the original framing of the problem.

I think it's made pretty explicit. In this version: "Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats." The host has to open a door that doesn't show the car.

Yeah - I don't think that statement clarifies. The rule that the host will never reveal the prize should be made explicit.

You're quoting Priceonomic's 2021 description of the problem. This is NOT the 1990 Parade Magazine description of the problem that generated all the responses.

Marilyn had the same framing of the problem as everyone else.

Marilyn made her key assumptions explicit in the original explanation she gave.

92% of people still disagreed with her.

And most of them seemed fully aware that she had the guiness world record for the highest IQ in the world.

To paraphrase Paul Graham's recent article, If the smartest person in the world proposed an idea that sounded preposterous, I'd be very reluctant to say "You are wrong."

One of the sources of confusion with this was Monty Hall did not always behave the way the problem dictates. He sometimes would open the prize door right away, or he would reveal a goat that was not behind a door or etcetera.

One of the sources of confusion with this is Monty Hall did not always behave the way the problem dictates. He sometimes would open the prize door right away, or he would reveal a goat that was not behind a door or etcetera.

Can’t it be random if there are two goat doors the host has to choose from? But I see your point about it clarifying the problem.

* Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness.*

It puzzles me that so many smart people got it wrong. It’s so easy to check your working.

When I first heard of this problem, I got it wrong too. Then I was told the answer, and to prove it to myself, it’s trivial to list the scenarios and simulate on paper. It’s still uncomfortable to think about sometimes, but I know how to show it’s true.

But all these people never thought to just do the working? Crazy!

This seems the most interesting aspect of the problem:

"However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that the player has picked door 1 and the host has opened door 3." [1]

[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_o...

Probability of an outcome always depends on what pool you're sampling from.

People pretty much always understand what the options are, but still get the analysis wrong. That is what I find interesting. Not the idea that there's some secret mechanism at play that alters the odds of a specific case. Even in that wonky "open the door on the right when possible" world, a contestant that always switches will still win 2/3 of the time.

I think people didn't check because the answer of 50% is so strikingly obvious (despite being wrong) that it seems like there's no work to check.

It's right depending on your assumptions. If the host has rules of only picking the right most door the contestant doesn't pick, and it happens to be a goat then the answer is 50%. Basically if the host is just as willing to show a car as show a goat and it is a goat which is shown, the answer is 50%.

There is a confirmation bias that people expect people who are aware of the Monty Hall Problem expect others to be wrong about it and attribute it to people not understanding. I think it's perfectly acceptable for someone to read the vague "the host, who is well-aware of what’s going on behind the scenes" and not assume that means he chooses to pick a goat. They're probably right, and the people saying 50% are reasoning along the lines of "there are two doors so 50/50".

For rigor, I think the host's rules need to be more strict: "The host knows what is behind the doors and will never reveal the car".

Or a 50% answer is correct if you lay out the principles of how the host acts given his knowledge.

Intention doesn’t matter. The problem states that the host opens another door and reveals a goat.

It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed.

Obviously if the host revealed the car, that would also affect the probabilities!

Whether Monte Hall is counter intuitive is a function of how the question is phrased.

When you phrase it in a way that underlines the mechanical nature of the host's decision, people get it right. When you phrase it in a way that suggests the host's choice is itself random, people get it wrong.

I think the first formulation primes people to think of it from the perspective of the host, which is the right perspective for this problem.

> When you phrase it in a way that suggests the host's choice is itself random, people get it wrong.

In other words, they still get it right; they get it right for the separate question that that phrasing implies.

If the host's choice is random, so that when you initially picked wrong it's equally probable that the host open the door with the car and then say "sorry, looks like you lost" (which is what I assumed when I first heard this problem, not being familiar with the show), then even if the host happened to open the door with a goat and give you a chance to switch, it doesn't matter if you take it or not. People are correct that, for that question, the probabilities are one in two for both of the remaining doors.

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all.

It's your perspective (narrowing the choice down) that changes, not the position of the car. Like getting a run of red in roulette and thinking the next one has gotta be black.

When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick.

When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door.

To look at another way, imagine if the host didn't reveal the content of the door, but gave you the option to switch to BOTH of the other doors instead of your door. Your odds of winning clearly go up, as you now have two chances to win (and all are of equal probability). That's equivalent to what's happening here. By showing you the losing door of those two, he doesn't change anything - there's still twice the chance it was behind one of the doors you didn't pick compared to the one you did, and by switching you win if it was behind either of them.

I've seen this explained ten different ways but this is the way I'm going to explain it from now on. thank you.

I wrote out a hopefully helpful intution down below [0]. Here's a terse part of it:

> Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch?

You're right that it's absolutely the narrowing down of the choice that's the key. If you happen to have not picked the prize door on your first try (more likely than not), the narrowing down will be your door and the prize door.

[0] https://news.ycombinator.com/item?id=27092614

This is the best explanation I've read thus far.

Ok. But you picked a very very very very small number between 1 and infinity. That can’t be random.

That it isn't random is precisely the point, isn't it?

The only difference that the "very very very very small number" makes is that the odds aren't infinity to one but two to one. Still winning odds.

Tons of common distributions have infinite support, but the details don't matter for this bit of intuition. If you want to formalize it in a simpler way, just consider it as an arbitrarily large finite number of doors with a uniform distribution.

Look at it from actual door, first guess door pairs -- which are the random uncorrelated events:

1,1 - switch loses (host may have shown door #2 or #3)

1,2 - switch wins (host showed door #3)

1,3 - switch wins (host showed door #2)

2,1 - switch wins (host showed door #3)

2,2 - switch loses (host may have shown door #1 or #3)

2,3 - switch wins (host showed door #1)

3,1 - switch wins (host showed door #2)

3,2 - switch wins (host showed door #1)

3,3 - switch loses (host may have shown door #1 or #2)

The choice of which door to show you is not a random event and is highly dependent upon the events which preceded it, because the host cannot show you the door it is actually behind, or the door that you picked.

incredibly succinct, thanks.

Here's another thought experiment.

Forget the opening of the door.

Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree.

Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors.

If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors.

Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?

It might help to have a contrasting example to clarify the key details.

Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two.

I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem.

When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner.

You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.

Intuitively it makes sense it would be half, because there was a 33% chance of winning but now it's 50/50 because 3-1 = 2, but when you read it logically it makes sense.

Yours is the first explanation I 'got', thanks!

Glad it helped, I just came up with it and it made it click more for me.

The other door is not irrelevant.

When you first picked a door, there was a 33% chance it was right, and a 67% chance one of the other two doors was right.

Once the other door got opened, it is still a 67% chance that the other two doors is right, but you now know which one of those two it would be - the one which wasn’t opened.

Thats what I'm stuck on, I suppose.

Imagine the entire situation is reversed. The Reverse Monty Hall problem.

I'm given a choice between two doors, once of which contains a car and one contains a goat. I have to choose, 1 or 2.

Then the game show host reveals that there was also a third door, which contained a goat, which is no longer relevant and never was relevant. I'm then also asked to choose a door (which is also irrelevant since the problem is backwards and the supposed aim is to get the car).

Even if that last step repeats 1000 times with 1000 doors and 1 car, each removing a goat-door, the only relevant choice is still the first one as the host appears to be adding new information, but it's always irrelevant information as a new choice is always made at the end.

In your reverse problem, I don't see how the new door can ever be relevant, as you know the car is behind 1 or 2, so you're only ever interested in picking between those 2.

In the original problem, its presence _is_ relevant, as the car could be behind doors 1, 2, or 3. Say you pick door 1 - there's a 1/3 chance you are right.

The host is then left with doors 2 and 3. We know there is a 2/3 chance the car is behind _one_ of these doors. When the presenter reveals a goat (say in door 2), he is reveling information about this set of doors - there is still a 2/3 chance that the car is behind one of the doors in this set, but there's only one door left we don't know anything about (3). There is therefore a 2/3 chance that it is behind _this_ door.

I think it boils down to the fact that time isn’t reversible, so the “reverse problem” isn’t like the “forward” version.

Let’s take some new problems. Suppose you have 100 doors and no switching. Your probability is 1/100, even if the host later opens a goat door, so in that sense the new information is irrelevant. But if we “reverse it” and the host opens the door first and you guess second, your probability improves to 1/99. So now suddenly the same information is relevant. Two things to observe here, one is that the forward and reverse problems are different, the other is that the relevance or irrelevance of the information depends on the direction of time. If you learn the information before you act it is relevant, afterward it is irrelevant.

One way to think about Monty Hall is you’re deciding which of these games to play. If you will stick with your first decision, you are sorta turning it into the toy problem above, where you decide the door first and then the goat information is irrelevant. Vs if you will switch, the goat door is opened before you decide, which is relevant.

Another way to think about it is with two contestants. Let’s say I pick the door initially, then someone opens the goat door, and finally you decide whether to switch. In this scenario, you don’t have self-preference bias to stick with “my” original door. In fact, my decision was the irrelevant information. It doesn’t matter at all what door I picked, what matters is whether you pick the right door, and involving me at all is a kind of misdirection to anchor you to the 1/3 probability.

Think from the perspective of your first choice. You had a 66.66% chance of choosing the wrong door.

So now I just eliminate one of the doors you didn't choose. There are two left, including the one you chose first, which only had a 33.33% chance of being correct. Nothing else has changed about the problem.

What if we started with a thousand doors? You chose 1, with a 999/1000 chance of being wrong.

Of the remaining doors, I open 499 doors. The prize is behind one of the remaining 500 that you did not choose the first time. Do you want to have a chance to pick from one of the 500?

I like expanding it out to 1000 doors. Even though the logic is the same between 3 and 1000 doors, the chances are more intuitive with more doors.

I think the part that really throws the brain off with the 3 door version is that the host opening a door seems like another random event but it's actually dependent on the state of the game at that point. Our natural intuition of the chances is for a version of the game where you pick a door, then the host randomly picks (and does not open as you also haven't) a different door, and then you have to choose if you want to switch to the door that neither of you picked.

> The other door is no longer relevant and doesn't affect the new state at all.

There is no new state, that's the thing. By picking one door you split these in two groups: one group with 1/3 to win, the other with 2/3 to win. When the host reveals the door with a goat behind it from the group that is 2/3 to win, it doesn't change the fact that that group had 2/3 to win to begin with.

I mean: that's how I see it.

The trick is that the host does not pick a door to open at random... If they did there would be a 1/3 probability they opened your door. If they didn't open yours under that circumstance, there would be a 50/50 chance you already picked the good door.

But it's not random... Monty will never open the door you chose. That's what messes up the intuitive probabilities.

Not only will he not open the door you choose, he won't open the door in front of the car either. I think that's the key point that confuses people. He knows where the car is and he always exposes a goat instead.

It's the presenter sharing knowledge with you, by opening a door that isn't the one you choose nor one that has the prize that changes the odds. And it doesn't help you if you stay with the same door because he never opens your door anyway, so it only adds knowledge about the other doors.

Maybe someone can help me with this -- what if for round two, you flipped a coin: heads you pick the door you picked previously, tails you pick the other door. Does that change things at all, since you're "switching" each time?

Edited -- Actually @haberman's comment that "By opening a door, Monty lets you cover 100% of the "everything else" makes the most sense to me as I think about this. Imagine there is no Monty hall, just three doors, and I say "you can choose one door and you win if there's a car behind it, or choose two doors and see if there's a car behind it." Clearly you're better off choosing two doors. And that's in fact, functionally what happens by switching after a door is eliminated.

This shows how much of our intelligence is crystalized (external). We look intelligent because we're reusing the solutions found by others. As soon as we can't do that it becomes apparent that the intelligence of a single person is not so great. It took us millennia to get where we are.

I was trying to think this through and realized it would be faster and more accurate to just write up a little script. Pretty sure this is accurate:

https://replit.com/@NateWildermuth/Monty-Hall-Problem

Thinking about this was a lot easier once I could look at the code. It seems like what's happening is that when changing doors, you are actually changing to two doors rather than one, giving you 2/3 odds?

I don’t think you need the 100 or 1,000,000 doors thing.

If you decide before not to switch then your goal is to pick the car which you have a 1/3 chance to do.

If you decide before to switch then your goal is to pick a goat which you have a 2/3 chance to do.

My favorite thing is when somebody says, “what if I want to win the goat?” Because you instantly know they’re a contrarian for its own sake.

The thing that bothers me about this problem is that (as far as I recall) it's not asserted that the host always gives the player a chance to switch. If the host sometimes gives people a choice and sometimes doesn't, based on unknown criteria, then all bets are off.

In the probability puzzle, the chance to switch is part of the basis of the puzzle. It is always offered.

In the actual game show, the rules were more flexible. The OP article mentions this. For example, the host could "offer the contestant cash NOT to switch." That introduces a more psychological poker-like aspect, but that example still allows for switching and doesn't change the underlying probabilities.

Yeah, for instance if the host only offers the switch when you've picked the car, obviously you shouldn't switch. That said, this does not seem to be the thread the respondents were pulling at...

So this is at least in part about the "Monty Hall Problem" and why it's solution not intuitive. The article missed an important angle: when the host opens a door, he's giving you more information, which explains why it's better to switch. If you're the host, you need to know which door the car is behind to do your job 2/3 of the time, to avoid revealing it.

It's this quality of unexpected information exchange that I find most fascinating about this particular puzzle!

When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else".

By opening a door, Monty lets you cover 100% of the "everything else" partition using only one guess.

So now you get to choose between partition 1, which covers 1/3 of the board, and partition 2, which covers 2/3 of the board.

I think the confusing part for many people is the unstated fact that the host takes all prizes into account when revealing a door, and that the host is forbidden from revealing the car. If the host chooses randomly then his pick adds no information (other than the extra revealed prize)

I don't think this actually matters. Suppose Monty doesn't actually know which door has the prize, and just picks randomly between the two. Now the only difference is that Monty might accidentally pick a prize. Suppose he does. What are the options now?

1. The game goes on, and you can just switch your guess to the door he picked. (WIN)

2. The game resets due to Monty's error and you play again. (REDO)

3. Monty just decides that you lose since he picked the prize. (LOSE, NO CHOICE)

Options 1 and 2 don't deprive you of the car. But option 3 doesn't give you a choice, so there is no dilemma.

So conditional on having a game where no prize has been revealed yet and you get a choice to switch, I think it still pays to always switch.

Another way of putting it: you get to choose between partition 1 covering 1/3 of the board where you get 1 guess, or partition 2 covering 2/3 of the board where you get two guesses. It works even if all of the guesses are random.

>Now the only difference is that Monty might accidentally pick a prize.

This a mistaken conclusion, as actually there is a difference if a goat is revealed by devious Monty (who makes sure he doesn't reveal the car) and dumb Monty (who picks at random), because the rules of the game really are different. The apparent outcome (you pick door, Monty reveals a goat) is the same. But the process is different. In the latter version dumb Monty has all the same information as you, in the former version devious Monty gets more information that you do and changes his behavior based on that information.

In programming terms, devious Monty has an "if" statement in there. (Something like, `if (random_choice == prize_door) { swap_choice(); open_door(); } else { open_door(); }`.)

TL;DR: I was wrong (but I did understand the distinction you are drawing)

I'm aware that there is a difference between devious Monty and dumb Monty. The question I was trying to answer is whether this difference affects the actual outcome of the game, or the choice you should make.

I ran a Monte Carlo simulation to test my analysis, and this proved that my previous analysis was in error. If Monty doesn't know to avoid the door with the car, then it makes no difference statistically speaking whether the player switches or not after Monty reveals a goat.

If the rules are such that you win when (dumb) Monty picks the car, then your overall chances are now 2/3 instead of 1/3, since you effectively get two guesses (your guess and Monty's guess). But it makes no difference whether you switch after Monty picks a goat.

On the other hand, if (dumb) Monty choosing the car results in either (1) you losing or (2) a reshuffle and redo, then your overall chances are only 1/3. In this case, it also makes no difference whether you switch after Monty's guess.

if monty picks a prize the game is over

I don't think people are confused by this in their initial solutions, but rather this after-the-fact problem modification happens to match the incorrect intuition people have.

The instinct is strong that "I am right" so "I must have just solved the wrong problem", thus "the problem was confusing".

Rather, I think you will find that most people don't expect the host will ever pick a door with a car until after you tell them that switching is better, and rationalization begins.

Additionally, it's on the solver to realize that a problem is under-constrained, and to ask for more constraints. That means that even if you were earnestly confused by the question, you presented a solution to an under-constrained problem. This is like answering "what is the square-root of four" with "it's negative two".

yeah, it’s reasonable to attribute those defects to the solver and not their situation and circumstances

I have never heard this formulation, and I like it quite a lot. Thanks!

It is Bayes’ Rule essentially written out (sample spaces)

Interestingly, if the host picks randomly (and if he reveals the car, you... start over, or you get the car, or you get nothing, or ... it doesn't matter because it happens not to have happened in the time we're considering) then you are faced with a 50/50 chance.

Yeah - and at least for me this was part of the confusion. The problem is often stated incorrectly.

Critically the host always removes a goat.

If the person presenting the problem just says “the host opens one of the doors”, but doesn’t specify he’s always revealing a bad door then it’s not clear why that matters.

The many door example makes it easy to intuit as well.

I think when stating the problem, some people assume the listener has actually watched Lets Make A Deal. You can under-specify the problem for people who have watched it, as the fact that Monty shows the worthless prize is already known.

That’s obvious. If he reveals the car, he isn’t going to ask you if you want to switch. A goat is a goat. It doesn’t matter to you which goat you get.

It’s not obvious if you think revealing the car ends the game (and you lose).

I think when I was first told it the person said “takes a door away” which is even less clear.

Years ago I challenged you on this and you suggested I write a simulation to check. I did just that and my understanding of the problem improved dramatically. That exchange made a pretty big impact on me and is something I think about fairly often, so thanks. :)

https://news.ycombinator.com/item?id=9079260

I also wrote a simulation, in PHP, when I first encountered this problem.

  <?php
  $wins = 0;
  for ($i = 0; $i < 1000; $i ++)
   {
   $car = mt_rand(0, 2);
   $choose = mt_rand(0, 2);
   if ($choose == $car) $wins ++;
   }
  echo "stay: " . ($wins / 1000 * 100) . "%<br />";
  // switch
  $wins = 0;
  for ($i = 0; $i < 1000; $i ++)
   {
   $car = mt_rand(0, 2);
   $choose = mt_rand(0, 2);
   if ($choose != $car) $wins ++;
   }
  echo "switch: " . ($wins / 1000 * 100) . "%";

This made my week :)

I don't understand how this could be. If you start over when the host picks the car, then isn't that the same as the host picking the goat every time, i.e. the same as the host knowing.

This is a small enough situation that we can do complete enumeration of the states.

Let's say you pick door 1. Let's go through all the possibilities: the states of the 3 doors, which one monty reveals, and what do we do, and what is the result?

  1 | 2 | 3 || Monty Reveals | You switch | Result
  ---------------------------------------------------
  G | G | C ||      2        | Yes        |  Win
  G | G | C ||      2        | No         |  Lose
  G | G | C ||      3        | N/A        |  Start Over
  
  G | C | G ||      2        | N/A        |  Start Over
  G | C | G ||      3        | Yes        |  Win
  G | C | G ||      3        | No         |  Lose
  
  C | G | G ||      2        | Yes        |  Lose
  C | G | G ||      2        | No         |  Win
  C | G | G ||      3        | Yes        |  Lose
  C | G | G ||      3        | No         |  Win

Count 'em up: When you switch, 2 wins and 2 losses. When you don't, 2 wins and 2 losses. Of course, the situation is symmetrical for any starting guess you make.

You are not enumerating the options correctly. You have 3 rows for the GGC and GCG options but 4 for the CGG option, which is assigning the CGG option higher probability. In reality, they have equal weight.

Agh, you are right. There should be two “start over” rows in each of GGC and GCG. I can’t write “N/A;” we must say “yes/no” in those.

Doesn’t change the result though.

What would this table look like if Monty always revealed a goat? Would any win or lose row be added or removed?

I know the result is right, but I don't think this table illustrates the reason.

Fair warning: I haven't rigorously verified the math on this, and tfa is all about the importance of vigorously verifying the math.

In the case where the host revealing the car means you restart, there are three equally likely situations after the host opens a door: - You picked the car, and the host revealed a goat - You picked a goat, and the host revealed the other goat - You picked a goat, and the host reveals the car (restart)

Of the two terminal cases, one gets you the car if you switch, and one gets you the car if you don't switch. In the non-terminal case, it doesn't matter what policy you have because you don't even get a chance to apply it.

One way to think about it is to imagine Monty picking randomly in either case, but then (internally and invisibly) correcting his pick.

Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):

    A) Car 1, Monty 2
    B) Car 1, Monty 3
    C) Car 2, Monty 2
    D) Car 2, Monty 3
    E) Car 3, Monty 2
    F) Car 3, Monty 3

If Monty doesn't correct in cases C and F, then when he shows us a goat behind (say) 2 then we learn we are in either A or E - it's 50/50. If Monty does correct himself, then we might have been in A or E or F.

Ah, yep, that’s right. Another way to see it is that we’re interested in the probability that your door has a goat behind it, given that you didn’t need to start over:

P(you chose goat | host didn’t choose car) = P(you chose goat, host didn’t choose car) / P(host didn’t choose car).

The numerator is 2/3 * 1/2, and the denominator is 2/3, so the ratio is indeed 1/2.

(A rejection sampling loop, where you repeatedly simulate a process until a condition holds, has the same distribution over final outcomes as the conditional distribution—so repeatedly restarting the game if the host chooses the car induces the same distribution on final results as simply conditioning on the host not choosing the car.)

At first I agreed with you, but then I thought about the variation with 100 doors (this is what originally helped me grasp the correct answer): you choose a door out of 100 (with very slim chance of choosing the correct one), the host opens 98 of the remaining 99 doors and there are goats behind all of them. Now it's obvious that the car is in the remaining door.

If the host is opening random doors, then: * If you chose a door with a goat, in 98/99 cases we would start the game over because the host opened the door with the car. * If you chose the door with the car, the game cannot start over because the host can only open doors with goats.

This means that when you choose a goat, the randomness gives you another chance to choose the correct door.

"Now it's obvious that the car is in the remaining door."

No. It could be behind the one you chose, surely.

I understand the explanations, or at least some of them; but I still don't get it. I thought I was intelligent and numerate, and this is making me sad.

I don't see why, after Monty reveals a goat and invites you to switch, you can't re-assess the probabilities from scratch. There are two closed doors, car behind one, goat behind the other, and you have no evidence which is which; therefore 50:50.

Well, I don't know. Advantage moves to the contestant: they close the doors, but everything's gonna remain in the same place. Steathily repositioning an automobile on a soundstage might be possible (albeit difficult), but just you try moving the damn goat

No, because the probability of a do-over depends on whether the contestant initially picked the door with a car.

This is exactly why the Monty Hall problem doesn’t apply to Deal or No Deal. Intuitively, it disappears when doors are opened randomly because, by Bayes, the likelihood you picked the good prize goes up every time you reveal a bad prize.

This is always how I thought about it. The host is giving you more information about the 2/3 doors you did not choose, while the amount of information stays the same about the 1/3 door you did choose. It's this asymmetry that makes switching make sense.

I'll suggest we encountered the same class of inadvertent information sharing via a recent post to this forum, The Case of the 50ms request:

https://news.ycombinator.com/item?id=27053941

Having recognised (from crystallized experience) the nature of the problem from the very first hint, I went straight for the jugular at the first opportunity; any residual uncertainty about the diagnosis was extinguished by the wording of the answer prompt, which to my mild disappointment took all the drama out of the reveal.

Sadly, there was no automobile prize on offer. No, not even a goat.

The explanation that best helped me 'intuitively' understand things is to rephrase the game as being two separate games: choosing one of three doors, and after that having the option to go for a coin toss instead. Obviously you'd go for the coin toss.

Now I understand that the reason why that worked for me is precisely because it becomes much more clear that it's really about new information, and what tripped me up was trying to maintain a kind of 'narrative', I suppose?

I still don't think I really "get" it, but this explanation makes the most sense to me. Thanks!

Like of course I would switch and go for the coin toss, but I don't know how to convert that certainty into actually probabilities, tho I'm sure that the odds of my choices are not 1/3 versus 1/2, because there's probably some interplay.

If I think about it more I start to get confused again, along the lines of: does the first choice even matter then? and what if I reverse my first and second choices (as in I pick 1, then switch to 3, but what if I picked 3 then switched to 1), how can I still have better odds by switching, since there's only 1 right answer? But sticking with the coin toss, I feel, "OK this makes sense."

the first choice matters insofar that you can choose to stick with it, so while the coin toss in isolation is 50/50, for the game as a whole you need to consider the option of sticking with that initial 1 in 3 chance along side the 1 in 2 chance.

I like to think about this slightly differently: at first, you bet based on no information, and you know you’re probably wrong. In advance of the host helping out, though, you have no way to put money on your initial bet being wrong. (It’s like playing roulette — the casino is not going to give you even odds betting on two dozens, which would have almost 2/3 chance of being right.)

After the host helpfully takes a door out of play, though, you can bet that you were wrong, and the payout doesn’t change. You already knew you were probably wrong, and now you can bet on this. Of course it’s a good deal. A player can probably avoid the cognitive dissonance by thinking of their initial bet as being their best guess at finding a door that doesn’t have a car behind it.

This was one of the explanations that I find elucidating as well, but just now I realize that the change from 1/3 as an estimate to 1/2 is exactly the result of absorbing new information, which makes this explanation unsatisfactory.

Isn't he also removing relevant information as well? The third door is no longer part of the game so the information he gives you ceases to be relevant.

I don't think you can "remove" information from A probability problem in this way. And even if you could there's no reason to assume it would zero-sum with the later information gain.

Happy to be wrong though, been a long time since i dabbled in probably.

An intuitive way to think about the problem is from the perspective of the host.

Sometimes the host has 2 goats he can pick from to show, and sometimes he only has 1 goat he can show.

66% of the time the contestant chose a goat, so the host has to reveal the only other goat 66% of the time. So switch.

The way I think of it, if you pick a goat on the first try (2/3 chance), you get the car by switching

The 2 goats didn't make sense to me but the host perspective does I think, like:

after player's first choice, there's two cases.

Either player picked the car, or they didn't.

If player picked the car, then host has two choices to pick from, if player didn't pick the car, the host has 1 choice to pick from. (At this point you can sort of imagine a binary tree with 3 leaves and 5 nodes)

So either player picked the car the first time with a 1/3 chance (so, less likely), or didn't (more likely).

So it's more likely for player to miss the first time (unless you're a precog psychic like me), which, "paradoxically (but not really)", by missing (by failing), player actually improves odds for second guess. So, on average the second guess is more likely to hit, because, on average, the first guess is more likely to miss.

Now finally it makes sense to me. Because it's more likely to miss on first guess, it's also more likely to hit on second guess, so it's more profitable to switch, on average.

Do I get it now? I think I do! Woot! :P :) xx

This is the most intuitive formulation for me.

From reading the problem as posed in the article, the confusion seems similar to the "plane on a treadmill" thing, where people are interpreting the premise of the problem differently. (In plane on a treadmill: "the treadmill moves backwards @ -1 * plane airspeed" vs "the treadmill moves backwards to keep plane airspeed at 0". In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

> vs "the treadmill moves backwards to keep plane airspeed at 0".

You might be correct in thinking people misunderstand it as that, but that's physically impossible, so I don't think you can argue that people have some sort of alternate understanding under which they are actually correct. That's just one type of wrong reasoning people might apply to the problem.

Add to that, in your hypothetical understanding of Monty Hall, that actually still doesn't change anything. Because even if the host picks a door completely at random, you still should always switch, it would just sometimes be the case that the host goes "ohh, that's to bad, I revealed the car and you can't win it now", but obviously that doesn't do anything to change your odds, because you're always better off or the same by switching, it's just in the unfortunate cases you'r switching from 0% change to 0% chance. But that doesn't do anything to change the fact that if he didn't reveal the car you're still going from 33% to 66%.

It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50.

Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This actually cancels out the effects of the normal Monty Hall problem, and we are left with a 50-50 chance.

> imagine every time the hosts picks the car the universe resets

Do you believe that we live in a universe that is reseting, every time Mr Monty hall opens a car?

As in literally Mr Monty Hall. From the game show. In real life.

Because the purpose of the question is to mimic the show.

You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car, and the result is the same.

> You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car

In the actual show, do you believe they are sending in a new contestant every time Monty hall picks are car?

Do you think that is what is happening in the actual game show?

It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person.

As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic.

In the regular situation, the host always remove a goat. because otherwise the quiz show is kinda boring. But in the modified version we are considering where he just picks a door at random, then in 2 of the 6 possible outcomes for the door the host picks, he'll be removing the car. Now since the problem has the host showing the doors content, that should be the end of the game show. Who want's to see someone pondering if they should switch between a goat and another goat right? But that doesn't change the logic at all, because the conclusion of "you should always switch" isn't impacted in any way. 0% to 0% is just no change. And in the rest of the cases your'll go from 33% to 66%.

Now of cause what would happen in your odd example where the host has a universe resetting button is that you don't have any choice at all, because no matter what the host will reset the universe until they maximize ratings, which might have you get the car or might have you not get the car, but there's no choice to be taken and the outcome that has maximum rating will always happen with 100%. That's why most stats problems avoid introducing "then the host resets the universe" in their problem description, it kind of ruins the whole point of calculating proabilities.

Think of it this way. In the original problem, there was a 1/3 chance of your original choice being the car, and a 2/3 chance of the other two. The host knowing which door to open gives you information without changing the odds, so there's still a 2/3 chance you picked wrong.

When the host does not know which door to pick, there's a 1/3 chance that you picked right, a 1/3 chance that the host opens the door with the car, and a 1/3 chance that you should switch. The host opening the door with the car ends that branch, so you're left with 2/3 of the original probability, and your odds for either branch remaining are 50:50.

I agree with you that "the universe resetting" is a dumb way to visualize or explain the problem. But if Mr. Hall opens the first door randomly (essential that it was a random choice), and it revealed a goat, then no, switching does not provide an advantage. You do not go from 33 to 67%. Instead, you are left with 2 options, both of which originally had a probability of 33% of containing a car, and which, now that the third option has been eliminated, are both currently 50%.

It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.

I didn't state that it doesn't change your odds. Obviously going from 33% to 0% because the car isn't possible to win anymore will be a change of your odds. I stated it doesn't change anything, because in one situation you go from 33% to 66% by changing and in the other you go from 0% to 0% by changing. So there is no difference to the logic of always switching because it either improves your odds or keeps them the same.

If the host reveals a goat at random rather than by special knowledge, your odds don't go from 33% to 66% by switching. The case where the host reveals a goat is only two out of three, not three out of three because one third of the time, the host will reveal the car. In one of those two cases, you picked a goat and in the other you picked a car. So you're at 50% odds whether you switch or not, if the host doesn't know where the car is.

If the host does know, then there are three out of three cases where the host reveals a goat. In one out of three cases you picked the car but in the other two cases you picked the goat. So that's why your odds go up if you switch.

The problem is that there are two interpretations of the problem, and is is not obvious that the incorrect one is wrong unless you know that Monty knows what's behind all the doors, and chooses never to open a door with a car behind it. The question isn't "does Monty showing a goat change the odds that it's behind the door that neither you nor Monty picked", it's "does it change that probability from 33% to 50% or from 33% to 66%".

Scenario 1: The first, incorrect, interpretation of the problem is "You choose a door, which has either a goat or car behind it. Monty then chooses one of the other two doors, which will also have either a car or goat on it, and opens that door. You then have the choice of whether to switch doors or stay with your original choice".

Scenario 2: The second, correct interpretation of the problem is "You choose a door, which has either a goat or a car behind it. Monty then looks behind the other two doors, and chooses the one that has a goat behind it. If both have goats behind them, Monty chooses randomly. Monty opens his chosen door. You then have the choice of whether to switch doors or stay with your original choice".

    +-----+------+--------+---------+--------+--------+------------+------------+
    | Row | Car  |  Your  | Monty's | Result | Result | Frequency  | Frequency  |
    |     | Door | Choice | Choice  | Stay   | Switch | Scenario 1 | Scenario 2 |
    +-----+------+--------+---------+--------+--------+------------+------------+
    |   1 | #1   |     #1 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    |   2 | #1   |     #1 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |   3 | #1   |     #2 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   4 | #1   |     #2 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   5 | #1   |     #3 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   6 | #1   |     #3 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |   7 | #2   |     #1 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |   8 | #2   |     #1 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   9 | #2   |     #2 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  10 | #2   |     #2 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |  11 | #2   |     #3 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  12 | #2   |     #3 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |  13 | #3   |     #1 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |  14 | #3   |     #1 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  15 | #3   |     #2 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  16 | #3   |     #2 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  17 | #3   |     #3 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  18 | #3   |     #3 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    +-----+------+--------+---------+--------+--------+------------+------------+

In scenario 1, before any door is opened, you chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. Monty then opened a door which happened to have a goat behind it, which eliminates rows 3, 5, 7, 12, 14, and 16. Now you have a 6/12 chance of winning the car if you stay, and a 6/12 chance of winning the car if you switch, and this is how you come to the conclusion that there is no advantage in switching.

In scenario 2, you still chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. However, this time Monty's door-opening doesn't eliminate any rows with nonzero probability, since rows 3, 5, 7, 12, 14, and 16 have zero probability to start with. As such, you still have a 6/18 chance of winning the car if you stay, and a 12/18 chance of winning if you switch, so you should switch.

The presentation of the problem included the phrase "the host, who knows what's behind the doors, opens another door", so I think in that case the interpretation that Monty deliberately reveals a goat is more appropriate (in addition, obviously, to being what was intended), although I agree there is wiggle room enough to admit both possibilities and clarity about which case you are discussing is important as it changes the answer.

It does, but it doesn't emphasize it enough to allay confusion, particularly for those unfamiliar with the "gotcha" in the problem. It's easy to overlook in that description unless you know that it is the important clue to understanding.

Kind of like the classic riddle "John's mother has four children. They are named March, April, and May. What's the last child's name?" which uses underemphasis for almost comedic effect.

> as it changes the answer.

It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.

I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise.

I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.

I owe you an apology, you're correct, I was wrong. I can't pinpoint where my reasoning failed me exactly, but my confidence certainly was misplaced, and that's from someone who did do the simulation originally to understand the outcomes, but not to understand the distinction between the host deliberately or randomly revealing the door. I'm sorry, but thank you for you persistence in correcting me. It was a fruitful learning experience to dig into this problem again.

> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.

Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter.

There's a difference here, that our language obscures, between procedure and hypothetical.

The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door.

If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.

(Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.)

The intentions of the host do matter.

Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door.

I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter.

Breaking it down into equally likely cases, assuming the contestant picks door 3:

    A) The car is behind door 1, Monty picks door 1, Monty corrects.
    B) The car is behind door 1, Monty picks door 2
    C) The car is behind door 2, Monty picks door 1
    D) The car is behind door 2, Monty picks door 2, Monty corrects
    E) The car is behind door 3, Monty picks door 1
    F) The car is behind door 3, Monty picks door 2

When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B.

If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely.

If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.

So, thinking about it really hard and reading about it online:

My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter.

But I think your example here doesn't show that? Are you trying to illustrate the Monty Fall variation?

The hubris of some of the her critics - especially ones with titles - is remarkable.

When someone famous with a reputation of always being right is suddenly wrong, there's a scramble to be the one who corrects them. It's a point of pride and fame.

The Monty Hall problem is particularly troublesome because the statistical connection is for some reason counterintuitive and difficult to grasp unaided, even for experts. Being that rare person who just gets it right away means that you'll have a lot of people thinking it's you who have misunderstood.

So in this case I'd say that this kind of reaction was rather normal (albeit with a sexist bent in a number of the responses).

It won't be long before we see some here I predict.

Hubris of title and its correlation to being dead-ass wrong is a time-honored tradition.

I'm willing to forgive pretty much anyone who's fooled by something that also fooled Paul Erdos: https://www.wired.com/2014/11/monty-hall-erdos-limited-minds...

Being wrong isn’t a crime. Harshly lambasting a woman for, largely, being a woman and disagreeing with you is the problem.

I don't see the problem. We are still doing this today. Pick a hot button topic that has some "consensus" then watch how outliers get treated. For example, COVID and climate change. I'm not saying people going completely against the narrative but professionals publicizing concerns or data that just doesn't fit the current narrative.

This is more of meritocracy run amok. I've got the fancy credentials which shows I'm brilliant, and I know it.

World renowned Harvard Philosopher Michael Sandel on The Tyranny of Merit: https://youtu.be/Qewckuxa9hw

Jumping to sexist assumptions is also a problem.

So is not reading the article:

> Maybe women look at math problems differently than men.

> “I still think you’re wrong,” wrote one man, nearly a year later. “There is such a thing as female logic.”

The epilogue: "A few days after Erdos left, he telephoned to say that Ron Graham of AT&T explained to him the reasoning behind the answer and that now he understood. He proceeded to tell me the reasoning but I couldn’t fathom his explanation." -- Vazsonyi [1]

[1] https://web.archive.org/web/20140413131827/http://www.decisi...

Something tells me Erdős didn't resort immediately to personal insults when faced with the problem, though.

Something tells me Erdős wasn't immune to sexism and would've probably not been immune to being a bit sexist here too.

Something tells me he probably didn't care who was explaining it to him, as long as he understood.

I'm not sure about that since Erdos had the typical pre-GenZ-wokeness vocabulary of a nerd, where everything they said was worded as an insult to people less smart or with different hobbies than him. To the point where it's in his Wikipedia article:

> Women were "bosses" who "captured" men as "slaves" by marrying them. Divorced men were "liberated".

> People who stopped doing mathematics had "died", while people who died had "left".

> Music (except classical music) was "noise".

(Did boomers ever notice all their wife-based humor was about how much they hated theirs?)

> Did boomers ever notice all their wife-based humor was about how much they hated theirs?

Probably, because the husband-based humor was the same.

> Whereas only 8% of readers had previously believed her logic to be true, this number had risen to 56% by the end of 1992, writes vos Savant; among academics, 35% initial support rose to 71%.

that 71% leaves 29% of _academics_ not getting the elementary math of the problem. I'm baffled this is _that_ hard?

"Academics" is a very loose definition. I imagine a very low percentage of PhDs out there have studied math seriously beyond a middle school level.

I think the lesson is that the are a lot of bad academics out there, who either repeat what they've been told without understanding it themselves, or whose confidence exceeds their ability, Dunning-Kruger style.

That 29% is probably made up of men that honestly think she should be pumping out babies instead of doing men's work. I wish I was being sarcastic or otherwise kidding.

I’m disappointed that this entire thread is focused only on discussing the Monty Hall problem and conveniently pretends her gender, and how she was treated, is unworthy of discussion. Come on!

If we can’t even mention how STEM women are often treated as default-incompetent in this, _very_ egregious case then we have no hope of being able to have the discussion at all.

You pick the car with 1/3 probability and a goat with 2/3. If you picked the car and switch, you lost. If you picked a goat and switch, you win because the only other door left has the car.

The problem is simple to reason through. The hard part is convincing yourself that you need to think through it given what seems "obviously correct".

This is the same explanation I arrived at too. Crucial to answering the problem correctly is understanding that the host doesn't reveal a random door, but a door that you have not chosen, and also is wrong. And then the reasoning follows naturally to the correct answer. It's amazing to be so prejudiced that someone would be adamant on the incorrect answer, when it can be explained in 3 simple lemmas.

The simplest explanation I've heard that has helped it click for people is "You can choose Door A, or both Doors B and C". Clearly choosing both B and C is a 2/3 probability. Then you guide them to understand how the host revealing one of B or C doesn't detract from the fact that you are still choosing both B and C to begin with. And finally establishing that switching is equivalent to choosing B and C instead of A.

I'm familiar with the problem and my intuition is still wrong.

I wonder if you couldn't design a gambling machine that has a variant of the Monty Hall problem built-in favoring the house (naturally).

Here's the intuition that helps me.

Imagine you're lost somewhere in the entire universe, trying to find your way home. Friendly (but tricksy) aliens offer you a deal. First, you pick some coordinates.

(Possibility 1) If the coordinates you pick aren't on earth, they'll take you to earth.

(Possibility 2) If the coordinates you pick happen to be on earth, they'll take you god knows where.

You also have the option to be taken to the coordinates you picked.

As a lost spaceman, perhaps going by the name Arthur Dent, what's your most likely path home? Odds are, if you pick a random point in the universe, it's not on earth. The odds are literally astronomical. So you're almost certainly going to find yourself in possibility 1.

Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch?

The best intuitive explanation I know modifies the problem to have 100 doors, and after you choose one, the host opens 98 doors that don’t have the prize.

I think stordoff's answer would be helpful:

When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick.

https://news.ycombinator.com/item?id=27093045

From what I understand, that would just be overcomplicating the matter, because most modern video gambling machines just use a set number of wins per time interval.

What's more, laws tend to limit how much the machines pay out, not how little. (Other than making sure the game is 'fair'.) Someone who loses a lot tends to limit how much they have to play. By contrast, the thrill of frequent winning gets people to lose more, in total, then they otherwise would haver.

Actually they use a RNG that in the case of slots, has a kind of virtual reel with multiple assignment.

If you like, add a bit of skill that gives the primates the feeling of control (video poker) or when you let loose of the handle or press a button (which simply grabs a number).

Stir in a bit of marketing pizzazz (off-by-one on slots, enormous number of simultaneous smaller bets, low payoff better odds vs. high payoff worse odds) and you can get the mark to stand there for a bit longer.

Networking, recognition and databases on players, etc. add more wrinkles.

In any case, I don't doubt that a Monty Hall bet could be wrappered into the game somehow to give the appearance of better odds than really exist.

(hopefully I got this somewhat right, I've never designed any gaming stuff. I'll bet that a gig at Bally or IGT would be a real eye-opener.)

This is absolutely false in virtually all commercial regulated gambling casinos in the United States.

That is incorrect, and most likely illegal anywhere gambling is legalized.

The machines are random, with no enforcement on wins per time interval.

Source: I work in the industry.

I have a Ph.D. in slot machines and you blew it big time. In Nevada they are very much controlled, just not by time interval.

Oh, and fwiw, I know several people who always win when they come to Las Vegas. I know because they've told me so. This proves that you don't actually work in the industry.

(I really hate that I have to disclose that I don't really have a Ph.D. is slot machines.)

Imagine there were 1000 doors, one of which had a prize, and you picked #123. The host then opened all the other ones except #456, and they all had goats. Should you switch?

I don't think you could. In most jurisdictions, gambling machines are tightly regulated to be random, and definitely not skill based.

Looks close to the one armed bandit problem which is fundamental in Reinforcement Learning.

Another statistically unintuitive problem (which I've witnessed a lecture hall enter a state of uproar over):

There are 2 red and 2 blue balls in a box. One ball is removed at random, what are the odds that the ball is blue?

Now we repeat the problem, but before examining the ball, we remove a second ball. We observe that the second ball is blue. In this case, what are the odds that the first ball is blue?

1/3 - If the first ball was blue, the probability to pick the second one is 1/3, while if it was red, the probability to pick a blue ball as second ball is 2/3. The prior probability for each scenario (first blue ball vs first red ball) is 1/2, so the posterior is 1/3 that the first ball was blue.

A neat trick to reason about those cases is to use odds. The prior odds for the first ball being blue vs red is 1:1. The odds for the first ball being blue vs red, given the second ball is blue is 1:2. We can just multiple the odds to get (1*1):(1*2) = 1:2 as the posterior odds.

This doesn't seem impressive because the prior is 1:1, but using this method you can easily calculate the odds in the scenario where there are 4 red and 2 blue balls. The prior odds is 2:4=1:2, the conditional odds is (1/5):(2/5)=1:2, 1:2 * 1:2 = 1:4, i.e. 1/5 chance that the first ball was blue.

I might be wrong here, but this is my understanding.

You'll always be able to show a blue ball as the second ball after the first is drawn. So arguing that it tells you something about the statistics of the state of the system after the first has been drawn is wrong. The chance that the first draw is blue is 50/50. The arguments that lead to 1/3 are trying to use the second event as a statistic for the state, however for that to be appropriate the problem would have to be phrased as:

You draw one ball and set it aside, then draw another, if the second ball is red, you start the entire thing over, if it's blue, then you continue the experiment.

Because otherwise the assumption that the second drawing can tell you the statistics of the underlying state is wrong.

The question of the odds really boils down to, did we randomly pick a ball and observe it? And if so, what would have happened if we didn't observe what we specifically stated for this instance.

Love it. The answer is 1/2. The first event occurs with a population of 2,2 and the second event has no causal effect. The trap is that the observation doesn’t change the sample population of the first event, exactly like the Monty Hall case: opening doors itself does not change the probability to 1/2 as people think it does (it remains 1/3) — it’s the switch that changes the odds.

Similar question with a similar trap: there are new neighbors moving in next door, and you know they have two kids. You see a boy in their yard, so you know they have at least one boy. What’s the probability they have two boys?

Take a look at the other replies. I like paxys' for the simplicity.

The second ball being selected doesn't change the first event, but it does change our understanding of it.

An extreme version: There's a bowl with 3 red and 1 blue balls. We remove two balls again, and the second one is blue. What are the odds that the first one is blue?

Your two kids problem is actually pretty complex, in the form you phrased it. Wiki has a decent explanation (I contend that your question is equivalent to the second question in the wiki article): https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

I’m aware that it’s complicated. I intentionally phrased it in a way that makes it clear that the family moving in is not somehow selected from the set of families with at least one boy, but rather the observation is unrelated.

This is the distinction. I believe that the 1/3 analysis may also be incorrect for the way you phrased your question. If you had said: “we select a second ball, and only observe the first ball if the second is blue — what are the chances of it being blue?”, then the second observation controls the population. Otherwise it’s semantics over exactly what probability we are trying to define?

I think the issue with these questions is that we are asking about “probabilities” which only make sense with repeated iterations. So you always have ambiguity in the construction of the question and interpretation when asking about things that are a “one time” event like both these examples.

I don't think this is correct. It matters whether the person removing the second ball can see the colors and choose accordingly. If you can, and you willingly remove a single blue ball, you have the Monty hall problem where you stick to your choice. It does not influence the chances, it is still 1/2. But the way I read the problem, you choose a ball at random, look at its color, it happens to be blue. Now this gives you information on the composition of the remaining balls. The chance is 1/3. To drive home the point that removing a ball "can influence" the odds of the first ball: what if you took out an extra ball that turns out to be blue? You now have two blue balls, leaving the chance the first ball was blue zero.

This is turns out to be rather straightforward once you realize that this is perfectly symmetric: it's just a matter of in which order you are revealing the colors. So this problem is exactly the same as looking at the first (second) ball, seeing that it's blue, then drawing a (revealing the already drawn) second ball. So there are two reds and one blue 'left'. So 1/3. (I am 95% sure...)

Probability of draws:

B then R -> 1/3

B then B -> 1/6

R then B -> 1/3

R then R -> 1/6

Removing the cases where the second draw was R:

B then B -> 1/3

R then B -> 2/3

Therefore the first ball being blue has a 1/3 chance.

Yeah, 1/3. Once I labeled them as R1 R2 B1 B2 and just listed all possibilities, it became clear where my error was.

I guess it shows for those who wrote her about female logic that it was, in fact, they who were overly swayed by male logic.

> In Joseph Bertrand’s box paradox (1889), three boxes are presented -- one containing two gold coins, one containing two silver coins, and the final containing one of each. Assuming the participant draws one gold coin from a box, the problem then asks what the probability is that the other coin in that box is gold. Bertrand, who concluded that the probability was ⅔, was lauded for his ability to look beyond the obvious.

The initial distribution is:

    [gold, gold] [gold, silver] [silver, silver]

With six possibilities for drawing a coin:

    1. [----, gold] [gold, silver] [silver, silver]
    2. [gold, ----] [gold, silver] [silver, silver]
    3. [gold, gold] [----, silver] [silver, silver]
    4. [gold, gold] [gold, ------] [silver, silver]
    5. [gold, gold] [gold, silver] [------, silver]
    6. [gold, gold] [gold, silver] [silver, ------]

After drawing gold, you are left with:

    1. [----, gold] [gold, silver] [silver, silver]
    2. [gold, ----] [gold, silver] [silver, silver]
    3. [gold, gold] [----, silver] [silver, silver]

In 2/3 of the scenarios, you drew from box 1 and the remaining coin is gold. In 1/3 of the scenarios, you drew from box 2 and the remaining coin is silver.

Interesting that they chose to identify Marilyn vos Savant in the headline as "Woman" instead of by her name. Also interesting that in doing so, they qualified her then believed to be world's highest IQ, male or female. Looks like they were attempting to be woke but ended up being misogynistic instead. It's also worth noting that many people did agree with her so it wasn't "everyone" correcting her.

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

If you are talking about a simple lottery with N tickets and a fixed probability of any one ticket being the winner = 1/N, then the screamers were correct. To see this, imagine that there are a total of two tickets (N = 2). You buy one ticket. The probability that it is the winner = 1/2. After buying the second ticket, is your probability of winning somehow > 1? Or has it exactly doubled?

But you are wrong: it's exactly double the odds.

If you bought all the tickets, do you think you'd have over 100% chance of winning?

Edit: looks like leephillips beat me to it

Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence.

P(A) = first ticket wins

P(B) = second ticket wins

P(A|B) + P(A|^B) + P(^A|B) = 1 - P(^A|^B) = 1 - (1 - 1/N) * (1 - 1/(N-1)) = 2 / N

It is slightly > 2 / N if the ticket is revealed before the next one is chosen because then it does throw away that possibility.

I think if you buy a first ticket, and somehow found out (but lotteries usually don't allow this) that it wasn't a winner, and only _then_ buying a second one, your reasoning would be correct.

There are still N possible outcomes, and you have 2 of them instead of 1. Your chance of winning is 2/N instead of 1/N and has exactly doubled.

And most lottery games allow numbers to be re-used, so you haven't "eliminated" anything. If you exhaustively bought all N number combinations, you have a 100% chance of winning, but you also have a decent chance to split the pot with someone else who also bought the winning numbers.

Next time your family screams and shouts at you that you're wrong, maybe you should actually listen to what they're saying?

Piling-on and screaming isn't a rational argument for anything. No listening can occur. None of them had above a high-school education.

Your piling on isn't helpful either, so can it.

I may misunderstand you, but replace the lottery with a coin toss. Buying a ticket gets you a 50% chance of winning. This reduces the pool of possible outcomes by 1 for the next choice, so there is precisely one choice left. However buying a second (different) ticket does not give you a 150% chance of winning.

If you buy heads and tails, then you have a 100% chance of winning. :) Unfortunately, each ticket cost 5 cents.

(See above.)

I may be missing something (not a mathematician, probabilist) but doesn’t the fact that the host would never reveal the car change the calculation? In the table shown in the article game 3 & 6 are nonsensical so with the remaining 4 games the odds are even between switch & stay.

The fact that the host never shows the car is critical to the solution. You choose a door. There's a 2/3rd chance that the car is behind one of the other doors. Monty is kindly going to show which of the other two doors not to choose. ... The other way to think of it is, you choose a door, and then you're asked "would you like to keep what's behind your door, or would you like to have the sum of the other two doors?"

Not at all. Once the door is opened, it doesn't matter if the host knew there was a goat behind it or not.

Edit: The host's knowledge only matter if the host can choose not to open a door.

If the host opens the door with car, and asks you if you'd prefer to switch to the unopened door, then you have a 100% chance of being wrong, as the door with the car isn't one of the choices. But, sure, my second point still stands, that it's effectively asking if you want your door or the sum of the other two doors.

But, to be clear, the question states that Monty opens the door with the goat. If you were to write a monte Carlo to test this empirically, you'd have to have your "Monty" choose the door with the goat.

You appear to have completely edited the comment I was replying to. My "not at all" does not apply to your comment as it currently reads, which is correct.

On HN it is considered polite to append corrections rather than replace content that has become part of the discussion.

No, it's correct. The easiest explanation, IMHO, is:

1. If you already selected the correct door, you should not switch, obviously. Equally obviously, the odds of being in this state are 1/3, because your initial choice was random.

2. If you did not select the right door, then the host clearly revealed the only one of the two remaining doors without a prize. In this state, you clearly should switch to the only door remaining, which contains the prize.

That is, the likelihood that switching will get you the prize is 2/3, because it corresponds to the states where you initially guessed wrong.

This is a very good explanation. I hope you won't mind if I say the exact same thing you did in just a slightly different way that might help it click for someone still not getting it.

Your initial chances of choosing correctly are 1 in 3. So the car is behind the door you chose 1/3 of the time. After the host reveals which of the 3 doors definitely does not have the car it's still the case that your initial choice will have the car 1/3 of the time [0]. That means the other door must contain the car 2/3 of the time. So you should switch.

[0] The haters are correct that the host hasn't changed the odds that your initial choice is correct. They just aren't tracking the implications of that fact.

The fact that Montey only reveals a goat is key to the probability calculation. When the player selects a door, they have a 1/3 probability that they have chosen the correct door, and Montey has a 2/3 probability of having the car. When he reveals the goat (which he always had behind one of his doors) that 2/3 probability shifts to the remaining door. There is no chance in Montey's selections, he knows where everything is to begin with.

Thanks for the helpful examples, makes sense. I had assumed it was accurate and I just wasn’t seeing it so good to understand why (also sleeping 4 hours a night with a new puppy so that doesn’t help).

The table is for the host revealing one door that you didn't pick and that has a goat behind it, not for the host revealing door #3.

"You made a mistake, but look at the positive side. If all those Ph.D.’s were wrong, the country would be in some very serious trouble."

And that, boys and girls, is how we got adtech.

This is why I no longer call people Dr.

I have a relative with a PhD in Economics. His personality leads me to believe that he would totally be one of the people to write in to Marilyn. He does sign things with Dr. Drives me up the wall.

Lol. I worked at a biomedical informatics shop with IITians and other folks who were MD + PhD CS. No one referred to anyone as doctor and no one there took a MENSA test. I think it's safe to say the only "doctors" are clinical MDs, podunk colleges to seem elitist, and Spies Like Us during the surgery scene.

Spies Like Us - Doctor, Doctor...

https://youtu.be/9UZWfMgjqAs

Only losers take Mensa tests and talk about their IQs. But plenty of PhDs use the Dr title in the right context. It's just the height of jerkdom to insist on people addressing you with it.

I don't know about "losers," but maybe insecure- or envious-behaving people.

I was never told what my IQ tests were, but the educational system kept pulling me out of class to arrange triangles for years. Later, I was lazy and took the SAT without coaching, prep, or studying. Missed one on the math section, but let's not talk about the English section. ;@)

I don't even believe in IQ tests as a valid, quantitative, reductionist, relative-performance ranking for anything other than taking IQ tests. Intelligence is multidimensional, multi-domain, difficult to linearize to orthogonal properties, and inferred based on a particular performance, not quantified directly by opening-up someone's head.

Can't people just be people better in some areas? I can't draw for beans, my handwriting looks like I have advanced Parkinson's, and the neighborhood animals tip me to not sing in the shower.

What's the right context? Why are PhDs (and MDs) the only people that deserve a special prefix in our society? I know loads of people that have worked harder and smarter than people with PhDs, but they don't get a prefix.

So do I, but titles don’t mean that you are more worthy, they are just a convention that indicates a particular thing. For example, the titles Mr., Ms., Prof., and so on. Different countries and cultures use titles differently; in some societies, such as in the US, we use the title “Dr.” for someone who has received one of several (earned, not honorary) degrees from an accredited institution. These are the MD, the PhD, and some other degrees that have a D in them, such as the EdD (Biden) and the DrPH (doc. public health)¹. This is so even though these degrees involve little more than vocational programs with two years of coursework.

The right context are things like business cards, listing people’s names on the program of a conference, or any place where you want to indicate someone as a degree holder, to lend a little formality, indicate respect, or try to impress the innocent. It is no more incumbent on you to use the title as it is to use someone’s preferred pronouns. It’s up to you. And to make an issue of it is pitiful and boorish.¹

[1] https://www.thelily.com/a-white-city-official-refused-to-add...

> the height of jerkdom to insist on people addressing you with it

Tell that to Dr. Jill Biden. Who doesn't even have a Ph.D.

Reading "Ask Marilyn" in the paper was one of those Sunday rituals that I miss from my childhood. Along with the puzzler from Will Shortz on NPR on the way to church and reading the Sunday funnies.

I kinda miss the newspaper, the way it was back then.

The problem with the Monty Hall problem is that it's based on implicit rules. Every explanation I've ever hear never says that no matter what they will never remove the winning choice.

There is a written rule: it's illegal. The Communications Act was amended in 1960 making it a federal offense for game shows to rig the outcome. Surreptitiously and intentionally removing the main prize on offer easily qualifies as just such a deceptive practice.

https://www.visitthecapitol.gov/exhibitions/artifact/communi... if you need that in ink & vellum.

Unlawful behavior can be reasonably excluded from both the construction and the answers to logic puzzles, except when noted otherwise (or otherwise reasonable in context), which is why taking Monty hostage, busting through all three doors, jacking the car, snatching the goat, and driving off into the sunset, is not a reasonable response to the problem either, no matter how great TV it might make for.

They show all the doors. Why would they ask you to switch doors, if they already shown you the car?? Think about it.

Ahh yeah that does make sense. I didn't realize they actually show you what the prize would have been.

If I were her I would have used the opportunity to get rich.

"Note to the 10,000 people who sent me letters saying I'm wrong: I'm willing to put my money where my mouth is, are you? I will pay anyone that can experimentally prove I am wrong $1000. If you can't do it, you owe me $1000. You have to put up the money before attempting any experiments. Put up or shut up."

The problem stated as stated by vos Savant hinges on the words "Monty (the host) knows". It implied that Monty intended to only open a door with a goat, so using his behind-the-scenes knowledge he opened door #2 (goat). However, such knowledge does not imply intention a priori. It could be that Monty doesn't give a f---, and he made the choice completely at random, for example by flipping a coin. Under this interpretation, he could have chosen door #3 (car) with equal likelihood, despite knowing that it would end the game early. In this case, the 50-50 answer would be correct.

I think this is what trips many readers up, given that we tend to associate impartiality with game show hosts. A better statement of the problem might be something like "The producers have instructed Monty in advance that after the initial selection, he must open one of the unselected doors to reveal where a goat is (under such instruction he will never open the door with the car.)"

To me the best explanation of the Monty Hall problem is to imagine there are 1000 doors. You pick a door and the host opens 998 other doors. Since you almost certainly did not choose the correct door, the host has almost certainly identify the correct door, so you should change your guess.

Then apply the same logic for 100 doors, then 10 doors, then 3 doors.

I still don't get it.

1. Three doors.

2. Two doors have a goat, and one has a car.

3. The contestant has a 1/3 chance of winning the car?

4. The contestant loses.

5. New game, and odds?

6. There's a 50/50 chance of winning?

(I'm assuming Monte Hall has no clue to where the car is. He is just opening doors.)

7. Could someone explain it to me, and thanks in advance.

(Off topic but a fawn had two babes in my back yard. It was pretty amazing. At first, it looked like the mother abandoned her babies, but she didn't. The babies stayed in the same spot for two day. By the third day, I thought she abandoned them. I looked into care. I was informed to leave them alone. They said, if you knew for sure the mom was killed, you could bottle feed them Goat's milk. Cow's milk is not good for baby deer. I didn't overreact, and mom was there all along. She now just visits with her healthy kids. This has been my best spring in memory. Happy Mother's day to the moms out there.)

Monte Hall DOES have a clue where the car is, and is NOT "just opening doors"

With the full knowledge of where the car is, Mr Hall opens a door that is NOT the door the contestant chose, and is also NOT the door containing the car.

This is the crux of it!

You have three doors. As you note, you have a 1/3 chance of winning.

Monty now reveals a losing door. At this point, your door has a 1/3 chance of still winning - the probability of that choice can’t change. However, as we now know one door has a 0/3 chance of winning (it’s been revealed) the remaining door must have (1-1/3) chance of winning. Thus, the remaining door has a 2/3 chance of winning.

I'm going to disagree with your framing here. Probabilities DO change as more information becomes available. For example, prior to rolling a pair of dice, you have a 1/36 chance of rolling a 12. However as sooner as you see the value of the first die, the probability changes to either 0 (most likely) or 1/6. In your description, if Monte opens a door randomly -- which you don't make clear whether he does or doesn't -- then the presence of the goat DOES change your P(winning a car).

He never chooses randomly in the Monte Hall problem. He’s always going to reveal a goat. That’s why the maths are so straightforward - and yet so counterintuitive.

> He never chooses randomly in the Monte Hall problem.

But you never state that, which is why I took exception with your explanation. You also state that probabilities never change, which is also untrue as more information is revealed.

You’re ignoring the fact that you already have information when it switches to two doors.

Any article that is secretly about the Monty Hall problem should have a very prominent trigger warning at the top

My intuition keeps betraying me, but a truth table finally convinced me:

  | Car | Goat | Goat | Switch? | Win? |

  |-----+------+------+---------+------|

  | T   | F    | F    | F       | T    |

  | T   | F    | F    | T       | F    |

  | F   | T    | F    | F       | F    |

  | F   | T    | F    | T       | T    |

  | F   | F    | T    | F       | F    |

  | F   | F    | T    | T       | T    |

Number of wins when not switching: 1

Number of wins when switching: 2

EDIT: Woops, hackernews does not like an org-mode table. Found out you can indent by 2 spaces to preserve format.

just the fact that we have SO many comments here (and always when it comes to this topic) about how this problem can be misunderstood (e.g. knowing that the showmaster will never open the winning door before yours and that he will ALWAYS show you a blank door), means that we don't quite know how bad people are in intuitively understanding the probabilites because the real problem here (and elsewhere) was the meaning of the question, not (only) finding the answer. Nonetheless, we have a lot of studies that show how bad we are with probabilites (e.g. that the probability of two defined rare things happening at the same time is not only rare but extremely rare - unless they are interconnected of course) so there is certainly truth in here, but not quite as much as it appears at first. Nonetheless, I really like how much more intuitively clear the problem gets once you're talking about 100 doors instead of 3: So if the question were: "You're taking 1 out of 100 doors, say number 27, and then the rules of the game say that the showmaster must open 98 of the other doors that all hold a goat, meaning only your door and another are remaining. In one of those is certainly the price. Will it be rather in yours or in the other? Remember that you chose randomly out of 100 doors while the shomaster had to take care not to take out the winning door when he took out the other 98 doors (unless it was already your number 27, then he could have chosen freely). So where is probably the price?"

I wanted to prove this to myself programmatically so I made this Python script: https://gist.github.com/snitzr/7ecf0abaacce610893e9364b13bce...

It’s better to switch.

One way to think about it is in an exaggerated way. If there was a googolplex number of choices and you chose one and then all but your choice and another door were taken away, you’d have almost a 100% chance of being correct if you switch.

Hall clarified that things worked a bit differently than the scenario presented by the Parade reader in vos Savant’s column.

This is partly why I think it's hard for some to comprehend "the Monty Hall problem". If you've ever seen Let's Make a Deal, you'll notice the big deal is set up very close to, but still materially different than how "the Monty Hall" problem is set up.

I wonder if people who have never seen the show can understand the math quicker...

I find that the phrasing of the Monty Hall problem makes the solution less certain than stated.

> "You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat."

That's what the gameshow host did in this one trial, but why do we assume that the host would do the same thing every time? Perhaps the host only tries this other-door-goat diversion tactic when the contestant first selected the door with the car!

“The winning odds of 1/3 on the first choice can’t go up to 1/2 just because the host opens a losing door,”

Wow. I've never seen it phrased so simply. _of course_ it doesn't change–if you picked the door with a goat, the host has no choice but to open the door he does. If you pick the door with the prize, whatever door he picks simply doesn't matter. There's no new information to learn about your own pick from that.

I guess Monte Carlo simulations weren't well-known or easily accessible back then? It would have been the perfect way to shut up her critics.

i was about 15 when this went down and immediately wrote one that confirmed she was right. formative experience in my life.

You don't need Monte Carlo, you just need a table with three rows. GGC, GCG, CGG. Pick the first door, cross out the other goat, and switch. you will lose once and win twice. if for some reason you think the door you pick at first matter somehow, make nine rows, and you will lose three times in win 6. Easy peasy.

You can't fight fire with fire, but you can fight Monty with Monte. :-)

You often can and should fight fire with fire. For a literal instance, https://en.wikipedia.org/wiki/Controlled_burn

This particular one isn’t that hard to just do by hand. Which I did at the time, was surprised at the result, and stared at the column long enough to convince myself why she was right.

Using Bayes theorem? But if you know to use Bayes theorem, then you're already formulating the problem correctly, which is the problem all the critics are running into.

Edit: Someone mentioned you could manually "simulate" it with a pen and paper, which I think is what you meant. If so, very good point.

I was unable to convince my father of this, so he and I just put together a little model with paper plates and coins and carried out the simulation manually until we had twenty or thirty trials and he became convinced by seeing it in action.

You can brute force this without knowing Bayes’ theorem (see the grids in the article as an example of iterating through every possible solution then counting to arrive at the right probability).

Ah! I see, I thought the commentator was saying to solve it mathematically. Manually "simulating" it makes sense.

A simple probability table works just as well

Not really because you can create a valid Monte Carlo model that does "prove" the odds are 50-50 if you don't incorporate the essential rule that throws most people off - that the host's choice of which door to reveal is not random.

they were. the critics didn't use them. she told them to "try it" in one or two columns before she had to explain it very slowly and methodically in a third, as I recall. I'm old. I read each of these as they arrived in the paper.

> I'm old. I read each of these as they arrived in the paper.

So back in the day, Parade magazine was a thing that people actually read, and not just ignored or recycled immediately? The past really is a foreign country.

You could do it in Excel in probably 20 minutes.

You could do it with a pen and paper in less than 5.

The point of doing it in Excel would be to convince someone who won't be convinced by the pen and paper demonstration.

Ok, so this is about the Monty Hall problem. A few remarks:

1. It is crucial exactly how the problem is framed. The "correct" framing is:

a) Contestant choses a door, b) Monty Hall will pick a door (different from the one the contestant chose) SUCH THAT a goat is behind it (if there are two, pick one randomly), c) contestant is offered a choice to switch. In that case, the correct answer is to switch, increasing the chance for a car from 1/3 to 2/3.

(The alternative framing is this: a) Contestant choses a door, b) Monty Hall picks a door randomly (different from the one the contestant chose), and if a car is behind that door, then the game ends, otherwise, c) contestant is offered a choice to switch. In this case, switching doesn't matter, the chance of getting the car is indeed 50/50, switching or not.)

2. Not everyone "corrected" her, some people did, and some people supported her. The confusion arose, arguably, because of the unclear framing. (One could argue that the phrase "the host, who knows what’s behind the doors" hints that the first framing is intended, but why not make it explicit? The host could know what is where and still choose randomly, in which case we're at the alternative framing.)

What does it feel like when you're wrong about something? It doesn't... it only feels bad after you figure out you've been wrong.

I love sharing this puzzle/problem, and hope to once again share it with new friends once the plague is over.

I'm no way an expert, so even if Ms. Savant beautifully explained it; my puny brain thinks... "how is it different from starting the game with 2 doors?". If you choose again, how is the probability not 1/2

Because you aren't starting from scratch - that you had a 1/3 chance to begin with is important. Say there are three doors, A B and C. You pick A:

P(A wins) = 1/3

P(A loses) = P(B _or_ C wins) = 2/3

The host then reveals that there was a goat behind door B. This doesn't change the state of anything (there is always a losing door in the two you didn't pick, and he is always choosing to show that one, and none of the items move). This means the probabilities remain as they were above. However, we know that B didn't win, so we can simplify it to:

P(A wins) = 1/3

P(A loses) = P(C wins) = 2/3

Therefore, if you switch to door C, you have a 2/3 chance of winning, rather than 1/3.

The only way the probabilities would go back to 1/2 for the second choice is if the prize and the goat were shuffled after B is revealed. However, they are not, so the chance that you picked the right door initially is fixed when you picked it.

To think about it another way, your initial choice of A means there's a 1/3 chance it is in A, and a 2/3 chance it is not and is behind one of the other doors. By ruling out B, we don't change the 1/3 chance that it was initially behind A. That means when we are asked again, there is still a 1/3 chance it was behind A, and a 2/3 chance that it was not. However, there's now only one thing that is not A, so there is a 2/3 chance it is behind door C.

Because you are throwing away information that you already have.

I read about this years ago in a very good book (for the layman) on the subject of probability: "The Drunkard's Walk." I highly recommend the book.

Excuse my GIMP skills. https://i.imgur.com/MW90pVZ.jpg

One appealing thing to me about math is how egalitarian it is. It doesn't matter where ideas come from, they are probably right or wrong. It is (maybe unhealthily) rewarding to see critics acting like its another discipline where the senior or otherwise higher status person's opinion has more weight, only to make themselves look ridiculous. Contrast this with so many workplace things where senior people say dumb or incorrect stuff all the time but there is not the same objective way to call them out on it.

> One appealing thing to me about math is how egalitarian it is.

I simply cannot believe anyone could read this article and come to this conclusion. Do you think the same flood of unduly harsh criticism would have resulted from the same column written by Maury von Savant instead of Mary?

good illustration https://upload.wikimedia.org/wikipedia/en/thumb/c/cd/MontyIn...

Which public intellectual introduced Monty Hall to the public when awareness of the problem was at a similar low, and received the fewest incorrect responses?

For the glib answer: probably the public intellectual who received the fewest responses, period.

That's probably true, since satisfaction with an explanation tends not to motivate letter-writing in any situation.

2016

There’s a bit of historical revisionism at play in the article.

Here’s an accurate representation by Aaron Brown of what transpired: https://www.quora.com/Why-do-some-PhDs-argue-against-Marilyn...

Quoting for those who don’t want to go off-site:

“You have to have lived through this ancient travesty to care about it. It’s probably best forgotten.

Marilyn Vos Savant published an incorrect answer to the Monte Hall problem. She got the correct answer from a lot of readers, including me. She then published a second incorrect answer claiming that all the letters she got—including from math PhDs were wrong. Finally she got the answer right—the one she undoubtedly got from any math PhD or intelligent person who wrote her—but continued to claim (a) that her previous answers were correct and (b) that all the letters were wrong.

As a result, the public is mostly misinformed about this simple problem.

The set-up is simple. There are three doors, one of which has a valuable prize behind it, and two of which have worthless prizes. You select one door. The host opens one of the other two doors to show you a worthless prize, and offers you the opportunity to switch your choice.

The key to this problem, which Marilyn finally saw in her third solution, is the knowledge and intentions of the host. If the host doesn’t know which door has the prize, there’s no reason to switch or not switch, it’s the same either way. If the host knows the door and is trying to hurt you, don’t switch. If the host knows the door and is trying to help you, switch.

Marilyn’s first answer was to always switch, regardless of the host’s intentions. This is also the answer that most people now believe. It’s a common error that people make all the time. That’s why people wrote in to correct her.

In order to justify her incorrect first answer, she claimed it was obvious that the host used the strategy of always opening a non-prize door and offering you the chance to switch. This does indeed justify switching. But that assumption was missing from her first and second answers, along with any discussion of the host’s knowledge and intentions mattering.

Her justification that the assumptions were too obvious to need stating was the actual television program Let’s Make a Deal, hosted by Monte Hall, which eventually gave the name to the problem. But anyone who watched the show knew that was not Monte’s strategy. Sometimes he opened another door and gave a chance to switch, sometimes he didn’t. He was well aware of the problem if he followed Marilyn’s assumed strategy, and was scrupulous about not giving any advantage to the contestant. You can tell this from statistics on the show—on average it didn’t pay to switch—and his published statements.

Clearly there are much bigger problems in the world, and it doesn’t pay to get upset about misinformation put out by popular vain people. If was very frustrating at the time, but it seems quaint that such things upset us in the 90s when there is so much worse information put out today by even more popular and more vain people.“

That quora answer does not quote the actual question she received. This archive claims to have the actual question: https://web.archive.org/web/20130121183432/http://marilynvos...

> the host, who knows what’s behind the doors, opens another door, say #3, which has a goat

The question, if that was indeed the formulation, very explicitly does state that the host both knows what's behind the doors, and uses that knowledge to show a goat.

Unless someone provides evidence of more ambiguity in the original question, I'm going to have to trust Marilyn's website to be quoting the question correctly... And the question seems to make the assumption very clear.

No, the experts didn't get it wrong because they got hung up on a (fairly ridiculous) misinterpretation of the rules. They got it wrong because they didn't bother doing the math.

And yes, this is deservedly a strong cultural memory, because it's so easy to fall into that error without realizing it. There are some PHDs who got it wrong, and are _still_ making frankly ridiculous arguments about how the rules were vaguely defined, rather than simply admitting that they made basic mathematical errors.

Be careful that you're actually fighting against revisionism, and not falling for revisionism.

Wait, if she did get the original wrong and only presented the correct answer on her third solution, where's the proof? This whole thing would be a waste of time if that's true.

Why would he open a door with the car?? It makes no sense to offer someone to swap if they’ve already been shown the car.

Is "Vos Savant" her real name? It's kind of a giveaway.

It actually is the family name on her mother's side, though she changed it to that back from her mother's married name (Mach).

> Mach

Amusing that her alternate name still exudes intellect.

https://en.wikipedia.org/wiki/Ernst_Mach

Or that she's very good at IPC.

She's a savant who's named savant.

Am I the only person who finds this funny?

---

Mommy, I want to be a savant when I grow up.

But you are already a savant.

You know what I mean.

Vos Savant is her mother's maiden name, which she chose to use over her father's.

Her mother was Marina vos Savant.

Smart lady but IQ tests are definitely not useless, if you disagree try this test: https://test.mensa.no

Why would talking that test convince me of the usefulness of IQ tests?

How they're useful is debatable.

But if you had a hundred people take an IQ test, and divided them into quartiles, you would notice some clear and strong differences between the groups, in a lot of different aspects.

What to do about the differences is an exercise left for the reader.

IQ tests apparently have a very wide standard deviation, and regression to the mean would also be an issue (if you select for high IQ, you are also selecting for medium IQ with large positive error).

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