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Comment by bhc3 | Hacker News Reader

My crack at it...

1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it.

But there's a 2/3 chance it's behind one of the other doors.

So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose).

You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B.

So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance

Make the switch every time.

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Dammit you beat me to it! Although maybe I can contribute because I was going to state it thusly...

Imagine a related Monty Hall problem, where you select a door, and then Monty immediately asks (without revealing anything), "Do you want to keep that door, or would you like to pick the other two doors?" Clearly you'd pick two doors instead of one.

When monty opens a door and gives you the "choice to switch" he is making noise designed to make picking two doors look like picking one door.

Thank you. When you put it that way, it makes so much more sense to me. Opening the door does not change the odds. Why is that so counter intuitive?

Glad it helped! Why it's counter-intuitive-ness is a really, really good question because like 99.9% of people I found this one hard to accept, too.

All I can say is that it's an incredibly effective verbal/logical obfuscation that relies on synergistic choice of number of doors and the (in my view) totally bogus part about Monty opening one of the other doors.

In fact, some people (in this very thread!) are still analyzing the problem as if the opened door represents new, 'very valuable' information! It is truly just slight of hand to make it seem like you're only picking one door when you are picking two doors.

My theory on why it's such a difficult problem to understand is that it's too easy to get caught up in the real world problem, when this isn't actually a real world problem.

In the "real world" problem there are additional variables to consider, such as whether or not the host would open a door containing a car, whether that would result in a win or a loss, whether the host would offer a switch in all cases, whether if so, there is any additional incentive for switching (which wouldn't adjust the probability of winning, so is a red herring, but "intuitively" it seems that if the host is trying to make you lose by trying to make you switch, probability should go down).

When you try to conflate the real world problem, which actually has multiple different probabilities depending on exact scenario, into an answer to what's generally accepted as the "Monty Hall Problem", some of this gets in the way.

In addition, there are two probabilities in the "Monty Hall Problem" - the first is whether your switch results in a win. The second is the probability that, if you switch, it was the correct choice.

Well except that you don't actually get to pick two doors. it wouldn't be much of a problem if it was posed as "You can pick 1 door or 2 doors, what do you want to do?" Since you only get to pick one door then Monty opening a door to reveal a goat does provide valuable information.

I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.

I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.

One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.

Note that this depends on whether Monty's choice of door was informed by the location of the prize. If he picks randomly and just, this particular time, happens not to have revealed the car then it doesn't matter if you switch or not. Of course, might as well switch anyway in that case just in case you misunderstood...

I don't think that's correct.

The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.

Years ago, I agreed with you. I wrote a simulation to demonstrate it. The simulation showed otherwise. I encourage the exercise - others have described the same experience.

You're right.

That's really interesting and counter-intuitive.

https://gist.github.com/ecdavis/da8f67258860e9f35620

EDIT: After thinking about it for a while it seems obvious and I feel fairly stupid.

1/3rd of the time your initial choice was correct. The random host always reveals a goat, you take the opportunity to switch and lose as a result.

1/3rd of the time the random host reveals a car, the game ends without an opportunity to switch.

1/3rd of the time the random host reveals a goat, you take the opportunity to switch and win as a result.

My understanding was that if the host picks randomly, then you are still better swapping if a non-winning door is revealed. The catch is you can't switch to the winning door if it is revealed. So, going into the game, your odds are not as changed. However, at the point of possibly swapping, you are down to chance that the remaining door is a winner. Roughly 1/2. Compared to your initial chance of 1/3 on the first pick.

I think to enumerate the possibilities you'd have to see that if you picked the winning door, there are two ways the host could leave doors for you to swap to and lose.

If you picked a losing door, there is only a single way for the host to reveal a losing door.

So, at the point you are looking at a losing door and making a swap, there are 4 ways you could have gotten there. You picked the winning door, and the host showed either of the two losers. Or you picked either of the two loser doors and the host showed you the other loser. Four possibilities, two of them you win if you swap.

Your understanding does not jive with my simulation. Write the simulation. It is certainly possible that my simulation is in error, but as I said it surprised me so I looked closely for error, and I have several times heard people report independently reaching the same conclusion on writing their own simulations and never (yet?) the other.

Thought even more on this. Did not get a chance to run a simulation, but thought of it while driving around.

It finally jives with me that your odds at swap time are only 50%. Seems kind of obvious when you think of it as random events and you are at the end with it definitely behind one of two doors. Either door is clearly as likely. Now, your odds of getting to this point are vanishingly slim, the more doors there are. Which makes sense.

I think my intuitive block comes in in that your odds of winning the game are not increased in this scenario at all. Which, I knew. I think I even stated it at some point. Still a hard block to get around.

I'll try to write up a simulation this weekend. I'm not entirely sure where we are disagreeing, though. I'm not saying that you have 1/2 chance of winning the game. Only if you have the chance to swap, you win half of the time if you do. That seems to be what you said in a sibling thread.

I agree with the parent poster that the participant's awareness of whether the host acted randomly or made an informed decision is critical for the participant to decide whether set B's 2/3 probability shifted/concentrated into the one unopened door in set B or whether set B's overall probability got reduced.

I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.

This. However, the alternative scenarios aren't the "Monty Hall Problem" - - the host will never open/blast a car door.

I usually find this problem annoying, not because it's all that difficult, in fact it's quite intuitive - when you're told the exact parameters defining the Monty Hall Problem and systematically work through them.

In my experience though, it's used more often as an exercise in diminution, a sick wet dream of probability teachers, where the learning party isn't aware of the problem, and usually either hasn't been explained, or doesn't quite grasp, the exact circumstances around whether the host's choice is random or decided.

There are lots of "it depends" moments that can be applied to incomplete descriptions of the problem, including (amazingly) whether the host offers a choice at all - this is the one that seems to trip up most people, as they might start to question the "motives" of the host (which are irrelevant in the actual statistical problem).

See my response in a sibling. The host revealing does matter, in that otherwise you don't get to act on the reveal. If the host just blasts it away as you said, then the probabilities don't change. If the host reveals, then there is a chance you don't even get to the swap before you lose. But if you do get to the swap, you have better odds of winning.

This is wrong. Whether "host reveals car" leads to win, loss, or replay, once you are faced with the choice the odds are 50/50 if the host picked randomly.

Isn't that what I said? Your overall odds of winning the game are not 50/50, but once you are at the choice, if you are choosing between two doors at the end, you are then at a 50/50 chance if you swap.

Hmm. You said, "if you do get to swap, you have a better chance of winning." Rereading, it seems you might have meant "better than 1/3, namely 1/2 - same as if you do not switch", which does seem correct. If he blasts the door, your odds of winning with any strategy (that does not involve cheating) is 1/3.

To be clear, if the host picks randomly (whatever happens to the door he picks) your odds are the same whether you switch or not.

So it does sound like we disagree some. If the host picks randomly and does not reveal, it makes no difference. If the host picks randomly, revels, and you are still in the game. It should make a difference. (After all, you now know more than you did before your first pick. Namely, that a 1/2 chance after a 1/3 exclusion did not remove the winning door.)

Will try and run a simulation tonight or this weekend.

I haven't had a chance to do the simulation yet, but the symmetry of the odds here finally dawned on me.

Specifically, if you have a 50% chance of winning on swap, you have a 50% chance of losing by not swapping. So, yeah, by the time you get to the swap, no matter what, you are at a 50% chance of winning. Swap or not.

Nope. Here are two ways to see how this works.

First, let's make it clear exactly what variation of the game we are playing.

1. Prize is assigned to a random door with each door being equally likely. Neither you nor Monty know which door.

2. You pick a door. Because the prize was assigned randomly and you don't know where it is, it is irrelevant how you pick your door. Without loss of generality (WLOG) we can assume you always pick door #1.

3. The host picks a door and opens it. Because the prize was assigned randomly and Monty does not know where it is, it is irrelevant how Monty picks a door. WLOG we can assume he always opens door #2.

4. If Monty revealed the prize when he opened his door, the game ends and you lose.

5. If Monty did not reveal the prize, you are given the opportunity to switch to the remaining door (door #3).

6. Your door is opened. You win if the prize is behind it. Otherwise you lose.

There are three equally likely cases to consider.

1. The prize is behind door #1. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is bad.

2. The prize is behind door #2. This occurs 1/3 of the time. Monty opens #2. The prize is there and the game ends. Note that in this case, YOU ARE NOT GIVEN THE OPPORTUNITY TO SWITCH.

3. The prize is behind door #3. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is good.

Note that in the cases where you are given the opportunity to switch (#1 and #3), switching wins in one of them and switching loses in the other. Each of these cases is equally likely (occurring in 1/3 of the games of played), and so in this version of the game switching makes no difference.

Here's another way to look at it. Since neither you nor Monty know where the prize is when you pick doors, we could change the game so that the prize is not placed until AFTER Monty opens a door, and this would not change any probabilities.

So, in this modified but equivalent game, we play like this:

1. You pick a door.

2. Monty picks a door and opens it. There is nothing behind it, because the prize has not yet been placed.

3. You are asked if you want to switch to the other unopened door.

4. The prize is placed randomly.

5. If the prize is placed behind the opened door, the game ends and you lose.

6. Otherwise, your door is opened and you win if the prize is behind it.

It should be clear that you have a 1/3 chance of winning the car in this game no matter how you pick your door or whether or not you switch. At the time the prize is placed, there is a door that is now your door, and you win if and only if the prize gets randomly placed behind that door.

> Note that this depends on whether Monty's choice of door was informed by the location of the prize

That's part of the setup. If it's not, it's not being told correctly. It's usually in the form of "...and Monty opens a door that is always a goat..."

Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.

Fantastic explanation.

I think this is what the article was trying to get across with the example of the doors but I felt that just confused the matter by talking about "shifting" the probability.

Alternatively, there was a 1/3 chance you picked the winning door, which is the only scenario where you lose if you swap.

What makes the answer most obvious is to play the game with 100 doors.

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