Monty Hall Problem
en.wikipedia.org
en.wikipedia.org
(note: when you do this in real life, you stop talking now. let them explain it.)
When you have N>2 doors, the host will with 100% certainty open N-2 doors with goats behind them. It doesn't matter whether N is 3 or 1,000,000. After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2.
The key phrase is "host's hidden knowledge". Once the host starts opening the wrong doors, the fundamental nature of the problem is changed from being an independent probability problem to a dependent probability one.
No, your numbers are wrong.
In your initial choice, the probability of picking the winning door is 1/N.
When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N).
So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2.
If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.
That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.
I didn't believe it until I simulated it myself.
Monte has to show you what's behind a different door than the one you initially picked. The game doesn't make sense otherwise.
Corrected simulation gives 50/50 when Monty picks at random among 3 doors.
switchWins=33044 switchLoses=33581
#include <stdio.h> #include <stdlib.h> #include <time.h> #include <string.h>
#define NUM_RUNS 100000 #define NUM_DOORS 3 int main(int argc, const char* argv[]) { srand(time(NULL)); char doors[NUM_DOORS]; int switchLoses = 0; int switchWins = 0; for(int i = 0; i < NUM_RUNS; i++) { memset(&doors, 0, sizeof(doors)); doors[rand() % NUM_DOORS] = 1; int myPick = rand() % NUM_DOORS; int montysPick = rand() % (NUM_DOORS - 1); if(montysPick >= myPick) montysPick++; if(doors[montysPick]) continue; if(doors[myPick]) switchLoses++; else switchWins++; } printf("switchWins=%d switchLoses=%d\n", switchWins, switchLoses); return 0; }
EDITED TO ADD:
Just for further info:
3 doors: switchWins=33111 switchLoses=33470
4 doors: switchWins=50182 switchLoses=24698
8 doors: switchWins=74876 switchLoses=12687
100000 doors: switchWins=99998 switchLoses=2
const pickCar = (switchChoice) => {
const doors = [0, 0, 0];
doors[(Math.random() * doors.length - 1) | 0] = 1
let pick = (Math.random() * doors.length - 1) | 0
let goat
for(let i = 0; i < doors.length; i++) { if (doors[i] == 0 && i != pick) goat = i }
let result = pick
if (switchChoice) for(let i = 0; i < doors.length; i++) { if (i != pick && i != goat) result = i }
return result
}
let switchWins = 0;
let switchLosses = 0;
for(let i = 0; i < 100000; i++) {
pickCar(true) == 1 ? switchWins++ : switchLosses++
}
console.log('wins:', switchWins, ', losses: ',switchLosses)"Picks a random door out of all closed doors, but that door is never the winner" is a contradiction. It doesn't give you a different chance of winning because it's not a coherent scenario in the first place.
This is a game where you start off with a certain layout, and then proceed forward. If you want conditions, they have to be conditions that you can apply before the game starts. You can't retroactively remove a significant chunk of samples.
If you added an actual outcome to him picking the winner, you could make a valid filter where the answer actually is 50:50. Perhaps they restart the game. Perhaps they never air the episode. And if you calculate only for finished/aired games then it's very clear then that you're solving a different math problem. You can't apply that answer to the original problem.
If he was opening doors randomly then you're left with a one-in-two chance. But if he was giving you information by opening doors he knew were bad, you're left with two possibilities: the door you started with just happened to be the right door (a tiny chance), or (much more likely) the door you started with was not the right door - in which case Monty was forced to open every other door so as to avoid showing you the car, and he has eliminated all the non-car options.
i initially thought this too, but in both cases you're gonna see 2/3 doors. the thing that makes the difference is when he has the car (2/3 of the time), he has to tell you where it is. so the choice becomes between your unknown door and his 2/3 selectively chosen car door.
No. Or, well, it depends on what happens if Monty opens a door with the car. If I lose if he does that, then switching doors doesn't change the probability.
It's easy to prove. Let's say that I always pick door #1, and Monty always picks door #2. Then we get the following probabilities:
car-goat-goat: I stay, I win. I switch, I lose.
goat-car-goat: I stay, I lose. I switch, I lose.
goat-goat-car: I stay, I lose. I switch, I win.
Staying: 1/3 chance of winning. Switching: 1/3 chance of winning.
Imagine you're taping episodes of this game show for broadcast. The contestant picks a door, and then Monty picks one of the two remaining doors at random. Monty has 1/3 chance of accidentally picking the car, and if he does, we're scrapping the entire episode.
So to enforce your constraints, we have to scrap 1/3 of all episodes, we're only ever broadcasting the 2/3 of all episodes taped where Monty, by chance, didn't pick the door with the car behind it.
Your numbers are correct, but that interpretation of the problem doesn't make a lot of sense if you think it through.
As for the game show, nothing says that he needed to give this choice every time, it was just some random thing he did to make it more interesting not an important part of the show.
> The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2.
If the episode just ends when Monty picks the car, then I lose if he picks the car. I don't just vanish into thin air such that the universe erases my run. I was on the show, and I lost. And that means my odds of winning are still just 1/3, not 1/2 as you claimed. Yes, the game has changed such that it doesn't matter if I switch doors or not with this addition, but the underlying odds don't change.
The original said: "the host, who knows what's behind the doors, opens another door, which has a goat."
Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities.
And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car.
It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.
The whole premise of the problem is that you are already on the show looking at an open door with a goat behind it.
One little letter and the whole thing is a completely different problem.
Upthread I see that you are correctly writing about the different ways you can model Monty's behaviour, and how each different way changes the nature of the problem, the outcomes, and the odds of switching vs. staying.
So I don't understand what you are arguing against right now?
I'm saying that "Monty reveals a goat" is what happens in the correct interpretation of the problem, i.e. Monty knows where the car is, and always chooses a door with a goat in every single instance of game play. In this case switching gives you 2/3 odds of winning the car.
And I'm saying that "Monty revealed a goat" is what happens in the incorrect interpretation of the problem, i.e. Monty does not know where the car is, and in this single instance of game play that you happen to be participating in, he happened to pick a door with a goat. And if you have gotten to this point, switching is 50/50. Just like you say.
Do you understand my way of describing these two different models of Monty?
I'm still not sure what the difference between these models has to do with my point about goat-car-goat. That can't happen in either model when we say that I'm picking door 1 and Monty is picking door 2.
But I'm saying that since that is a wrong interpretation of the original problem, any other wrong interpretation of the problem is equally valid. For example, if Monty picks the car, it's game over and the contestant never gets to choose. Or, if Monty picks the car, the show goes "oops", closes the door, shuffles the prizes, and lets Monty pick again until he picks a goat.
There's not enough information in "Monty picks a door at random" to infer a single interpretation of the problem.
"Monty picks a door at random AND he picked a goat". That's enough information to get down to your universe where goat-car-goat could never happen.
Nitpicky semantics, sure.
But to roll back to the original, original question: I don't think people get the Monty Hall problem wrong because they're interpreting it as "Monty picks a door at random and he picked a goat". They interpret the problem correctly, i.e. "Monty knows where the car is and always picks a goat", and they still get it wrong.
I'm not saying that goat-car-goat could never happen. I'm saying that when you're evaluating what to do after Monty Hall has revealed a goat, the one piece of information you know is that goat-car-goat didn't happen.
I guess you're analyzing from the point of view of someone who's trying to figure out their odds of winning with different strategies before playing.
I agree with you that most people don't get the problem wrong because of interpretation. I'd go even further and say that many people who think they understand it don't get it right because of the interpretation.
Most arguments I see for the correct answer don't use the assumption that Monty knows which door has the goat, and therefore work equally well under a model of Monty picking the door at random. They're just picking different ways to model the problem and assuming a uniform distribution. You get the answer right if you picked a model where the distribution is uniform under the assumptions in the problem.
The real issue seems to be much simpler. Probability is not intuitive. Most people don't have the tools to figure out problems like this reliably, and those that do tend not to apply their tools when they think they see a simple obvious answer.
No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3.
So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat.
This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability:
A. car-goat-goat
B. goat-car-goat
C. goat-goat-car
A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.
He knows exactly where the prize is. He just had a good poker face and head game. Which is why he hosted the show for 15+ years.
> Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The "opens another door, say No. 3, which has a goat" could be misinterpreted as opening the goat door being an example rather than a requirement.
The bizarre scenario some people try to propose is that Monty doesn't know what's behind the doors, but if he happens to reveal the winning door, then we...ignore that entire Universe, or...rewind the simulation and try again, until he "by chance" reveals a non-winning door.
Enough to say that if Monty opens the winning door then he apologizes, closes it, and his helpers place the car to a new random place and they try again from the start.
Or say, ignorant Monty may open the winning door accidentally, but here I have a stack of DVDs of all lucky episodes where Monty got a goat. When binge watching these videos, would you expect to see switchers win at a higher rate?
Answer to both: it becomes 50-50 if Monty was ignorant, even if we post hoc filter out those cases where he got it right.
Bonus philosophical question: you don't know whether Monty knew the car's location. He may say he does, or he may say he doesn't but you don't trust him. Or maybe he says "phew, how lucky to be a goat, I actually had a short amnesia an bluffed it". How can such a murky thing whether a human knows something or not impact the very, financially real world, even when all outward appearances are equal: I saw with my eyes that Monty opened a goat door with his hands. Yes it might have been an accident/luck but what is the precise definition of luck anyway? How can it have an influence whether his brain contained this information, given that he ended up opening this goat door anyway. How can counterfactual Universes influence ours? I have my own answers but it's an interesting thing to ponder.
It should sound confusing. People who justify their claim that switching doors doesn’t change your odds in this way are proposing a very bizarre and confusing revision of the problem.
The original statement of the problem is a very straightforward description of a simple and realistic game show, that happens to have a solution that is counterintuitive to many people.
It's not 50/50 because he is ignorant, it's 50/50 because you've discarded 1/3 of the outcomes, and all the outcomes you've discarded were ones where the contestant initially picked a goat.
Another "more serious" related "paradox" is in scientific hypothesis testing. You can have different p values even if you observed the exact same experimental outcome but your stepping criterion was different. The only difference is the speculation of what you would have done if the experimental outcome had been different. [0]
These things are quite unexpected unless you've already worked through such examples.
If Monty knows what's behind the doors, we arrive at the scenario immediately where the door that's left has 2/3 chance of hiding the car, because the door you picked has 1/3 chance of hiding the car.
If Monty doesn't know what's behind the doors, and goes into a cycle of opening doors and shuffling the prizes if it was the car, or rewinding the universe, or whatever, at the end of that sequence, you will still hold on to the same door you initially picked that has 1/3 chance of hiding the car. And the door that you get the offer to switch to therefore has a 2/3 chance of hiding the car.
"For example, he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door."
[1] https://en.wikipedia.org/wiki/Monty_Hall#Monty_Hall_problem
- Let them pick one card of three
- Offer to switch their card for the other two (closed!) cards. Most people take this deal, in my experience everyone takes this deal.
- Then, show the "goat"/useless card of the two remaining cards
- Offer them (again) to switch their original choice of card for the remaining card.
If they accept to switch, let them explain why (this is when people get it). If they don't switch, tell them that one of the two cards they could have switched for must have been a goat, so I could always show them one goat between these two.
After this, most people (all my test subjects, anyway) understand that in the long run, they're best off switching to the "two card" deal. They now understand that one must be a goat but that the probability of the "car" being among the two-card offer is twice as likely.
N<10 so YMMV
When a casual observer assesses the situation they see an equal set of unknowns behind each outcome, since that matches the set of conditions.
While the probability would have remained the same, it's the elimination of all (but one) remaining false choices that tips the scale. The weight that the selected door was the correct one remains at 1/3rd, but the probability that the only remaining door was the correct one is the inverse, 2/3rds.
Meanwhile in the 1,000,000 doors set, if instead one door was picked, one other door was opened, and the choice was between keeping the current door and picking one of the other 999,998 doors, the odds increase so infinitesimally that both ego and a need to know would bias me to staying the course.
I was debugging my monty-hall simulator to try and work out why my intuition was so wrong when I came across my comment "Iterate through doors, open first door without prize" when this point finally struck me.
Or perhaps better stated in the Wikipedia article:
> "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
a) Staying wins (Monty opens to a goat) b) Swap wins (Monty opens to a goat) c) First pick lost, no chance to swap (Monty opens to the car)
each of equal probability.
[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Variants
You can sidestep this by playing a slightly different but equivalent game.
The player chooses a door. The host then gives the player the choice of sticking with that door, or instead taking all of the other doors.
I think the intuition behind is that, when you initially choose one door out of 1000000, you are almost sure not to have picked the right one, so you ignore the fact that you even chose one. Then, the host opens 999998 doors, but leaves one closed. That door is now really special!
Or a different way to look at it: The host gives you partial information about what's between the 999999 doors you didn't choose. Doesn't it seem really unlikely that your intial choice matters given that there are so many doors?
This explanation offered above makes sense to my "gut" however: https://news.ycombinator.com/item?id=21349295
The problem to me is more that when you go back down from 1M to 3 doors one can just as well say again "but now we only have three doors". Meaning that a proof of the original problem is the only satisfactory answer; the 1M comparison is maybe for people that don't like proofs and want "evidence".
We had it beaten into us not to use truth tables to solve problems in college, because it doesn't scale. But the Monty Hall problem has a finite set of states you can easily write on a piece of paper. And then you'll see that it jumps from 2/6 successes to 3/6.
I tried to replicate that table under duress once (see initial comment) and I couldn't get it sorted. I should try it again and write it down this time.
Edit: Why do work when you can find someone else who already has:
https://www.statisticshowto.datasciencecentral.com/probabili...
and particular to my statement, this table:
https://www.statisticshowto.datasciencecentral.com/wp-conten...
When switching the number of lose scenarios is lower than a naive expectation of outcomes.
Wait, really? When did you go to college?
In mid 2000s, we used truth tables every day studying boolean algebra, logic gates, circuits, etc.
Statistics? (and this is a statistics problem) no.
It jumps to 4/6! Even better.
I couldn't get my mind to accept the explanation. So I coded it and ran the simulation 100,000 times.
I got about 50,000 instances where switching won the prize. Biases confirmed.
... until I found the bug in my code where Monty was allowed to open the door containing the car as the one showed to the player.
Fixed the bug, and switching won the prize in about 66,666 instances.
For example, we'll label the three doors P, L, R, for the prize door, the leftmost goat door and the rightmost goat door. This gives possibilities
1. You choose P, Monty chooses L - switching loses
2. You choose P, Monty chooses R - switching loses
3. You choose L, Monty chooses R - switching wins
4. You choose R, Monty chooses L - switching wins
This seems to indicate that switching gives you a 50/50 likelihood of winning. In fact, this is the correct truth table to use if Monty doesn't know which door has the prize. In that case, these are all equally likely.
Instead, if he does know where the prize is and won't open that door, this truth table is misleading because the cases where switching loses are less likely. If you choose L, he'll definitely choose R, but if you choose P, he has two options to split the probability.
Similarly, if you model Monty as not knowing what's behind the doors, then the truth table you're thinking of doesn't work. Either way, the truth table on its own isn't convincing.
The thing I find more clear is using Bayes' rule, as it shows exactly how the model of Monty's behavior affects the probability that switching is beneficial.
Let H be the event that your door has a prize and G be the event that Monty shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) = 1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it just comes down to what P(G) is.
If he always shows a goat, then P(G) 1 and you should switch. If he doesn't know what's behind the doors, it's 2/3 and it doesn't matter whether you switch. If he knows but prefers to spoil the prize, it's 1/3 and you definitely have the prize door.
┌─────────────┼─────────────┐
G G C
┌┴┐ ┌┴┐ ┌┴┐
Stay│ │Switch Stay│ │Switch Stay│ │Switch
G C G C C GThat was the part that always tripped me up, and I see it trip up plenty of others so I'm in good company, but it's definitely part of my mathematical heartbreak. I loved math in high school and a series of unfortunate events and terrible curricula ruined calculus for me. All I have to show for it is that I have more sympathy for people who 'hate' math (even though I think it's fear sublimated into frustration), and a deep and abiding hatred for Stephen fucking Wolfram.
Those italics on Mathematica became my proxy nemesis and I figured out how to break the autocorrect by editing the word just so.
Which is funny because that's a group that's very capable of simulating it for themselves.
The trick is in the wording. With 3 you are literally saying "let me remove all the cards except for a random one, the one you chose, and the one I chose, of which, one is the correct card to choose. Do you want to change your choice?" So it plays out as expected with 3, and the obviousness of the solution when done with 52 cards.
Your first choice is either right or it's wrong. If you switch, right becomes wrong and wrong becomes right.
Your first choice is 1/3 right 2/3 wrong.
I think there's a psychological component too. So many probability questions are about random events so you (very naturally) take a position that the revealed door is randomly chosen. It is NOT - you always see one of the goats.
You may be on to something.
Perhaps part of it is not wanting to switch from a winning position to a losing one. Thats harder to take so inertia sets in.
This is reminiscent of the endowment effect: https://en.wikipedia.org/wiki/Endowment_effect
I struggled with this because I didn't see the first choice as relevant to the final outcome. It is easy to see the game as restarted somehow when the 50/50 choice is presented.
He illustrated it with the problem of what happens to the level of water on a pool when you throw overboard a stone from a floating boat. Does the water level rise or drop? Well, imagine that your small stone is made of neutron star material...
Forgive me if I'm missing something obvious, I have a newborn and am underslept.
I assume neutron star material is incredibly dense and so the rock would be unbelievably heavy. To the point that the mass would actually likely sink the boat, and then we can't throw the rock out.
OK but let's assume that the boat is somehow massive enough to support the weight of the rock. Now we've changed the scale of the other variables of the experiment in proportion to the rock/neutron star, and we're back to ground zero, are we not?
It's only when the person opening the door knows where the prize is, that I should always switch?
"You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." Where does it state that the host always opens a door with a goat? From the statement it looks like the goat was accidental. No wonder Paul Erdős got confused by this too. The translation of the paradox to his language could have been even more confusing.
If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.
Isn't it the other way round?
The only scenario in which the host doesn't give you any useful information by opening a door is when the host chooses the door at random. Assuming the host didn't choose the door with the prize, your odds are 50:50.
Of course all of this hinges on you being aware of the host's decision making paradigm, and that you want a car more than you want a goat.
The first time I read about this problem, it was formulated in a way which made it clear that the host opens a door with a goat behind it. If you learn about it like this, you literally cannot believe how anyone (even Paul Erdős) could believe that switching is not better. After all, you choose a door with a goat behind it with probability 2/3 on your first try, and if you do this, than switching will always win you the car. It is so trivial that the only "paradox" here is that it is classified as a paradox.
E: the explanation on Wikipedia is unnecessarily elaborate and hard to follow:
"The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 that was not available at the beginning of the game, when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. "
What?
I only took a single class of statistics at university, but my biggest take-away from that class is that you absolutely cannot trust your gut, because the human bias is crazy strong.
Humans naturally over-connect. We see causation where only correlation is present. Gamblers believe that something random that has happened frequently (like roulette landing on a specific number) is more likely to happen again. Educated people learn about dependent and independent probability, which corrects this impulse. But sometimes it over-corrects.
The Monty Hall problem is hard because people see can't explain the dependent probability link clearly and conclude that the probabilities are independent and their brain is acting up - first guess 1 in 3, second guess 1 in 2. After all, the game host is always capable of opening a door with a goat, and the player doesn't know anything. Put differently, what did the player learn between the first and the second guess that should make the player now believe one door is more likely than the other of the remaining two?
It also takes advantage of our natural thinking. The problem is artificial, because, outside of a game show, the host could also just not offer to switch. Emotional then, brains consider the information that the host made the second offer even if it's not logically part of the problem.
The paradox disappears when you think about probability as a tool for reasoning from incomplete information, not as anything to do with "physical" property of the system under investigation. It then makes perfect sense that after receiving new information from the host we should reassign our probabilities.
"There are three boxers. Two of the boxers are evenly matched (i.e. 50-50, no draws!); the other boxer will beat either them, always.
You blindly guess that Boxer A is the best and let the other two fight.
Boxer B beats Boxer C.
Do you want to stick with Boxer A in a match-up with Boxer B, or do you want to switch?"
[1] https://math.stackexchange.com/questions/96826/the-monty-hal...
Boxers: if you pick one of the bad boxers, which has probability 2/3, then you should switch after shown a bad boxer which happens when the first round fight ends. Otherwise (probability 1/3) you should stick with your boxer.
You pick door 1. Out of three doors you have a 1/3 chance of having a car behind your door. The chance that the car is behind one of either of the other two doors is 2/3. Let's pretend that instead of opening a door and revealing a goat, Monty instead says to you "You can switch to both of the remaining doors. If the car is behind either one of them you get to keep it." Your likelyhood of getting the car by switching is now 2/3.
Instead, Monty does the statistical equivalent: He allows you to know for sure which of the other doors definitely has a goat.
monty has the car 2/3 of the time. when he has the car, he must tell you where it is (by showing you where it isn't).
that last part is key. he is not making a random selection. so if you go with his remaining door, it will have the car the same 2/3 of the time.
this tidbit is pretty wild
What made it very clear to me is if you just increase the number of incorrect doors that you open by a lot. (Making the assumption that the host knows where the prize is.) In that case, it's obvious that you're improving the odds by showing where the prize isn't.
1) Monty always opens another door (regardless of what is behind the door initially chosen by the contestant). He will never open the door the contestant chose.
2) Always opens a door with a goat behind.
3) These rules are known to the contestant
These premises are not intuitive, and some versions of the problem does not make them explicit, which is why it is more of a "gotcha" riddle.
In particular, the version quoted in Wikipedia does not make any of that clear. It just mentions the particular case of the host opening another the door with the goat. But without knowing the above "rules", this just means there is 50% chance per door now.
There are three doors, one has a prize and the other two are empty. You pick a door (say, Door 1). Monty then says that his assistant will move whatever is behind Door 2 (if anything) to Door 3. He then opens Door 2 to show you it is empty. Would you switch to Door 3? Of course you would.
But Monty’s assistant is lazy, so instead of moving the contents from behind one of the closed doors to the other, he just asks Monty to open the one that he knows is already empty. The result is the same.
Edit. Be nice if my downvoter would explain what's wrong with the above.
https://twitter.com/judegomila/status/1183425802399969282
"Your brain will compute monty hall problem in fast casual causal mode AND/OR compute slow mode formally for the exact answer. Probability predictions that accurately map reduce to efficient intuitive representations in our brain dictate direction in our evolutionary biology."
"[since] we don't solve the monty hall problem effectively with our intuition, an evolving life form would be poor at resource collection from such problem sets without knowledge of axiom representation and formal probability."
We shall hunt in the forest.
"Sire! Our scouts have just returned from the mountain. They report that the herds are not there at the present time! Shall we try the valley, or continue on to the forest?"
A) Stick to the plan, and hunt in the forest B) Hunt in the valley instead
Without that guarantee, it makes no difference.
Additionally, there are not many situations in real life that would replicate the conditions of that guarantee. So our "intuition" is simply geared towards situations where an actor can't read our thoughts (door pre-selection) before we act them out, and in such a scenario, pre-selecting/switching selection of doors doesn't matter.
The scouts can depart after you’ve chosen the valley, find nothing in the mountain, give that information to you, and you should still switch, even if the scouts didn’t confirm where they are.
Monty(or the scouts), knowing if your original pick was right has no impact. The critical piece is that they are forced to reveal to you a bad choice. 2/3 times it’s the only bad choice to reveal because you’ve picked the other.
The entire gain in odds is predicated on Monty being forced to reveal a bad choice after you have chosen.
Monty knowing ahead of time has absolutely no impact on the outcome.
If you are objecting to how I described that, in terms of who knows what when, then you may be right and I probably could have been more precise.
But if you are disagreeing that scouts sent to investigate an area and return an answer (which this time happens to be negative) differs materially from the original question then you are wrong.
Nope, it does not. The fact that he revealed a dud is what matters. The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
That's simply wrong. This has been discussed countless times on this forum. The most effective way to proceed is for you to spend 5 minutes writing the simulation that you think will prove you right. When I was on the other side of this, that experience is what convinced me.
> The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
That's not a very strong objection - game shows are weird sometimes and could often apparently be easily improved.
But more importantly, it's an objection to the wrong thing. My point was that the presented "find the herds" problem differed from the game show. "That would have been a bullshit game show" is just confused.
Right, of course. I just meant he knows ahead of opening his door, it doesn't matter if he knows when you pick your door.
The scouts did not know that the herds are not in the mountain when they chose to scout the mountains.
Monty, on the other hand, does know that the car is not behind door C, when he chooses to open door C.
The difference is that Monty will never pick the door where the car is, whereas your scouts will sometime recon a region and find the herds there.
(Unless you assume that your scouts know where the herds are, and are purposefully not telling you and scouting empty areas, which would make them terrible scouts. We can expect this sort of crafty behaviour from a game show host, sure, but not from your own scouts.)
This results in different odds. With the scouts, switching makes no difference. You can run simulations or draw a full probability tree to convince yourself of it.
Note that there's some ambiguity between exactly one of vs at least one of which changes the probability but not as much as 1/2->1/3. (exactly one male on a Tuesday excludes the possibility of two males born on a Tuesday) Tuesday is not a red herring.
It's truth tables all the way down.
Then it's definitely 50%. The truth table would be
Youngest: Boy Oldest: Girl
Youngest: Boy Oldest: Boy
Youngest: Girl Oldest: Girl
Youngest: Girl Oldest: Boy
With the bottom two being ruled out by the question. I think that's the part that catches people out here.
For instance, let's say you tell me you have two children. One was born on a Tuesday, the other on a Wednesday. I ask about the one born on Tuesday, and you tell me it's a boy. With the information I have available to me at this point, I am correct to judge that the odds you have two boys are 1/2.
This story differs from the one you provided in ways that are insignificant for story telling.
The way you tell the story depends on the reader picking up on a rather tiny detail to reach the conclusion you intend. Specifically, the reader must come to the conclusion that you are someone playing games with them who has intentionally used very careful wording to only eliminate the girl/girl entry from the outcome chart.
That's not the way people communicate naturally. If someone says "I have two kids. One's a boy...", they aren't playing probability games, they're thinking of a child they'd like to tell you a story about. In that case, when they have a distinguished child in mind before making that statement, the odds are in fact 1/2 that the other child is a boy and hence you have two boys. (Modulo all the legitimate criticisms that can be raised that there are more possible outcomes than boy/girl and that even between those two the odds aren't equal. This is a simplification of reality for the sake of a puzzle.)
In the end, problems like this mostly hinge on poor communication that abuses common patterns to convey less information than the reader interprets from their experiences communicating in the real world. As such, they're really rather boring.
[1] https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Analysis_o...
I most certainly am not. I'm going to say it's 13/27.
I agree the imprecise nature of the English language muddies the water somewhat, but the question can be re-framed to remove this aspect and yet still have people give the wrong answer, and as such is not just a "really rather boring" case of "intentional" "poor communication", but offers an interesting insight into the shortcomings of our intuition for probability.
So let's set aside the original problem and put this one instead (which may or may not be equivalent to the original depending on your interpretation, but let's take this in isolation anyway):
Woman: How may children do you have?
Man: Two.
Woman: Is at least one of them a boy who was born on a Tuesday?
Man: Yes.
The answer is in fact 13/27, because of the unlikely extra information revealed by the Tuesday condition, which I think is surprising and interesting.
P(BTB?∩B?BT | BT??∩??BT) = P(BT??∩??BT | BTB?∩B?BT) · P(BTB?∩B?BT) / P(BT??∩??BT) = 1 · (1/28 + 1/28 - 1/196) / (1/14 + 1/14 - 1/196) = 13/27
and by listing out all the combinations
Boy Tuesday │ Boy Tuesday Boy Any
────────────┼────────────────────
B Mo B Tu │B Mo B Tu
B Tu B Mo │B Tu B Mo
B Tu B Tu │B Tu B Tu
B Tu B We │B Tu B We
B Tu B Th │B Tu B Th
B Tu B Fr │B Tu B Fr
B Tu B Sa │B Tu B Sa
B Tu B Su │B Tu B Su
B Tu G Mo │
B Tu G Tu │
B Tu G We │
B Tu G Th │
B Tu G Fr │
B Tu G Sa │
B Tu G Su │
B We B Tu │B We B Tu
B Th B Tu │B Th B Tu
B Fr B Tu │B Fr B Tu
B Sa B Tu │B Sa B Tu
B Su B Tu │B Su B Tu
G Mo B Tu │
G Tu B Tu │
G We B Tu │
G Th B Tu │
G Fr B Tu │
G Sa B Tu │
G Su B Tu │This is a badly made retake of the original two problems where you had one where you just mention that you have a boy, and the other where you mention that the older is a boy. The older being a boy removes 2 cases while 1 being a boy just removes 1 case so they are different. But in this case mentioning the weekday doesn't matter at all.
To make it clearer take this script "I have 2 kids, one is a boy, that boy's birthday is {{ boy's birthday }}". In that case it is 1/3.
Take this script on the other hand "I have 2 kids {{ if 1 is a boy born on tuesday print ', one which is a boy born on Tuesday' }}". That is what you and he meant it. But it is wrong since it is not the only way to read it, and arguably it is the less correct way to read it.
This HN thread is full of people who don't know maths who think they know it and are trying to teach others...
> This is a badly made retake of the original two problems where you had one where you just mention that you have a boy, and the other where you mention that the older is a boy. The older being a boy removes 2 cases while 1 being a boy just removes 1 case so they are different. But in this case mentioning the weekday doesn't matter at all.
This doesn't make sense because if you concede that being told at least one child is a boy yields a two boy probability of 1/3, and you also concede that then being told the elder child is a boy changes that probability to 1/2, there's no reason not to concede that being told instead that at least one child is a boy born on a Tuesday gives 13/27 as Sean1708 has demonstrated. Otherwise you're applying your "decides to mention" augment differently in each case.
Anyway, it's mathematically more interesting to dispense with all the intent and instead consider the matter as an enquiry, which you may well argue is not equivalent to the original pronouncement (for reasons of intent), but nevertheless is an interesting counter intuitive observation in its own right:
(a) We have a man we know has two children. The probability he has two boys is then 1/4.
(b) If we ask him if at least one of his children is a boy and he says yes, the probability he has two boys is now 1/3.
(c) If instead we ask him if at least one of his children is a boy born on a Tuesday and he says yes, then the probability he has two boys is 13/27.
(d) If instead we ask him if at least one of his children is a boy born on Christmas Day and he say yes, the probability moves even closer to 1/2. The exact value is left as an exercise to the reader.
i.e. the probability tends towards 1/2 as the additional information narrows the boy being referred to down to a specific child. So:
(e) "is at least one child a boy with this fingerprint? - yes", or:
(f) "is the elder child a boy - yes"
will both get you to p = 1/2.
You don't see the difference between saying "the older one is a boy" and "I have a boy born on a Tuesday"? The first one selects first by the property of Age and then states their gender. The second selects by the property of gender and then states which weekday they were born.
Obviously the answer is that your first guess is more likely to have been a goat. So if you had to bet, you'd have bet on the car being in one of the ones you didn't choose.
After the host reveals a goat, it's no longer "one of the ones you didn't choose", but rather "the one you didn't choose" (since the host has taken the other out of consideration). So the choice is easy now (as is the probability calculation).
But in a mathematical probability problem, you can't do that. For example, if there's a slightly smaller object and a slightly larger object concealed in two different sized boxes, common sense says that a person would put the larger object in the larger box. But if it's a abstract mathematical probability problem, and you have no information on the algorithm used, then you have to assume the concealer flipped a coin when they decided which box to use.
If I'm the one setting up the big-small game, and I use the random algorithm, the smaller object is going to be in the bigger box about half the time. If my mom is the one setting up the game, the smaller object is going to be in the smaller box every time.
In abstract mathematical probability problems, if you don't know what algorithm was used, you aren't allowed to bring human behavior in, unless it's stated in the problem.
I think that this misrepresents mathematical reasoning.
For solving a problem in a textbook (which a mathematician does intensively at the beginning of his or her career, and seldom thereafter), you must assume neither—that the larger object goes in the larger box, nor that the concealer flipped a coin; you must use only the information contained in the problem. (From this point of view, many problems in most probability textbooks are ill posed, because they force you to make some assumption while solving them; and it then becomes a mind-reading exercise of whether you can successfully make the same assumptions as the poser, rather than a mathematical exercise.)
For a mathematician who actually wishes to use theoretical probability to do something useful, then you must again make assumptions that allow you to translate the messy real world into an idealised mathematical object; but to say that you must, or must not, assume any particular thing is otiose—different assumptions will lead to different results, and the only guide to whether you made the 'right' assumptions is whether or not the idealised mathematical results you get match the real world (to within whatever bounds of error are tolerable).
So, yes, even a person with the world's highest IQ can make a mistake.
> Hall clarified that as a game show host he did not have to follow the rules of the puzzle in the vos Savant column and did not always have to allow a person the opportunity to switch (e.g., he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door).
If the host uses an adversarial strategy, such as only offers a switch when the remaining door is a goat, switching is not necessarily wise.
When you picked a door at the start you may have excluded the winning door from the host. This has a 1 in 3 chance of happening.
So that means the host has a 2 in 3 chance of being left with the winning door.
The host who has knowledge can't pick the winning door, so those 2 in 3 odds are passed on to you if you switch.
If you are thinking about you playing the game then of course it is very counter intuitive, because you are playing the game and you have only one chance. Doesn't matter that by switching your chances rise to 2/3 because you play the game only once, and in the end you either win or lose.
On the other hand if you think about it from a statistic/probabilistic point of view then it's all natural that the chance is 2/3 but there is the notable exception that you don't play in this case and you neither win or lose.
Question: did you pick the correct door?
- If correct, both remaining doors are losers, eg. {loser, loser}.
- Otherwise, the remaining doors are {loser, winner}.
And the loser must be revealed according to the game rules.
If you change your answer: (1/3) * 0 + (2/3) * 1 = 2/3
- (1/3) you initially picked correct, then change to a guaranteed loser.
- (2/3) you initially picked incorrect, then change to a guaranteed winner.You are the contestant. You choose a door. Monty reveals a goat from another door.
Now, instead of giving you a switching choice, Monty picks [insert person you find annoying] from the crowd. Monty gives them the option of choosing your door (duplicating any prize) or choosing the other door.
The few times I’ve tried, people have seen it as unfair. But they don’t necessarily see the 2/3. They just see someone getting 1/2 vs their 1/3.
You look at the 3 cards then put them on the table and propose a fair triple or nothing, if they can find the red queen.
You make them point toward the card they chose.
If they point a wrong card you go toward the card and reveal -> you won.
If they have pointed the right card, you return one of the card they didn't pick and ask them if they are sure or want to change. Because they think they are facing a Monty Hall problem they will switch. -> you won again. :)
Even if you didn't cheat them, this game has zero expectation value... I guess if you are working with people who wouldn't know this you are fine.
It is, as I said before, a fair game so it has zero expectation value for both players, better odds than a casino for the player and there is no reason for a gambler to refuse this bet.
In expectation you won't lose money, but people who think they know better will happily give you their money by making a bad move while being convinced they are getting an edge.
Most math-type people don't know the hypothesis of the Monty-Hall Problem and can't state it properly, but most have seen it played on TV-shows and know they must switch.
In the heat of the moment, especially if you have played a few follow the red queen games before, the brain is still in fast mode, and will jump to using heuristics. Mind trick.
I mean, if people can't sort this out, how on earth someone that is not either extremely stupid or extremely evil (or both) can seriously think that people can figure out their optimal health insurance even if the terms were not incomprehensible legalese designed to trick you?
Switching is only bad if you pick correctly the first time. Since you only pick correctly 1/3 of the time, switching must be no worse or better 2/3 of the time.
Since Monty will never open the door you picked nor the door with the prize, the remaining door after he opens one is the prize if you initially picked wrong, or a goat if you initially picked right. Since you initially pick right only 1/3 of the time, you switch.
Maybe needs one introductory sentence: "Note that your first choice has probability 2/3 of being a losing choice".
https://medium.com/i-math/matt-damon-s-martian-potatoes-1bcd...
Basically, Wikipedia does not mandate a citation style, it just has to be consistent throughout an article. And once set, another editor can't change it simply for personal preference. Presumably the original author, or whichever editor first added citations, was more familiar with this style, and so it has stuck.
Had it in a browser reading list (that I'd forgotten was there), and came across it by chance yesterday.
By the way, you can get old Monty Hall shows on satellite TV to this day (or at least week)--a friend noticed it playing in a restaurant a couple of days ago.
When you pick a door, the odds are that the prize is behind one of the other 2 doors (2/3 chance). Therefore when all but one of the other doors are eliminated, the remaining one is twice as likely to have the prize as the one you picked. Is that it?
I think my problem comes down to trying to reason about a past event using probability. I just can't get my head around that. Yes you can predict what is likely to happen if you repeat the experiment multiple times, but you're not being asked about that. You're being asked about this particular instance of the experiment. How can you know that you're better off switching this time?
Get 3 cards and pick one to be the prize. Your mathematically-challenged partner gets to put them down each time.
You pick a card, they remove one of the other cards and then you always switch.
I made like $50 from someone this way one time =)
I'm not totally heartless - I took him to lunch on me afterwards to lessen the blow ;-)
If you really want to drill the point home, offer to keep playing more & more batches of 10 until they've had enough...
This is true for Bayesian updating on any information, and is a good lesson for life in general.
The missing piece of the puzzle that seems to frequently be left out is we KNOW Monty's motivations, knowledge, and how he decides. At the time the problem was formulated, when Let's Make a Deal was still on television, all this was common knowledge: Monty is the host of a game show, he knows where the two goats and the car are, and he decides which door to open by picking one of the unchosen doors he KNOWS has a goat behind it. The problem suffers now, 40-odd years later, by the fact that people often don't know who Monty Hall IS, and the problem is usually worded as if it's still 1963-1976, so this foundational knowledge is missing. There is no QUESTION what Monty Hall knew, or what his motivation on the show was... except people who don't know the game show don't know that.
Granted, the actual game show was a bit more varied, but the presumption in the thought experiment is that Monty ISN'T trying to goad you into un-choosing the correct door, which he frequently did on the show because the tragedy of it was more entertaining than just giving away a car.
If he can remove an unchosen prize door with some unknown probability p and does not reveal the contents of the removed door, then the probability of getting the prize by switching is simply (2/3)(1-p).
His motivations and knowledge just amounts to some distribution of p, but that’s a rather different problem.
When the expected value of p is 1/2, then switching is equally likely to win you the prize as your original choice, but the probability of winning is still only 1/3 in either case. Only when p=1/4 will the remaining door contain the prize with probability 1/2.
EDIT: My reply is based on a misunderstanding. I was assuming that the host continues to open one door, whereas parent intended that the host would open all-but-one doors.