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.
- 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
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
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.
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 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
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?
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.
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".
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.
It's only when the person opening the door knows where the prize is, that I should always switch?
(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.