How to explain the Monty Hall problem to a disbeliever
michalpaszkiewicz.co.uk
michalpaszkiewicz.co.uk
Imagine a universe with many simultaneous Monty Hall clones playing at once in many studios, where Monty doesn't know and opens another door at random. If that door has the car behind it, Monty and contestant are both shot in the head and the studio burned down and erased from all records. This bloody culling of branches of the probabilities is the same as effected by giving Monty the knowledge and telling him to act on it.
The associated line of reasoning resolved the paradox for me. If I stick with my original choice, it is as if I ignored the new information. If I switch the choice, I react to the new information.
The extreme of this is to pick among countably infinite doors, having the presenter open countably infinite doors and leaving just yours and another closed. Who could reasonable suggest that the chance is still 50/50, assuming you don't flip a coin and base your choice on that?
This assumes that him opening a door is new information. It isn't unless you make an assumption about the game masters intentions.
If the host doesn't want you to win then you will lose 100% of the time if you switch, if the host wants you to win then you could win 100% of the time by switching etc. But people think that these assumptions are "obvious" so of course you must make them, but that makes the riddle bad.
If the riddle doesn't say that the host always opens an empty door after you pick a door, then the typical solution isn't correct. People does the typical "X happened once" and assumes it means "X always happens", that is a typical naive assumption but no, just because it happened once doesn't mean it always happens, you can't make that assumption here unless stated in the riddle.
66% of the time. 33% of the time you picked right first time and lost it by switching. The host never conveys "dont switch."
The host always gives you an opportunity to switch boxes.
And perhaps also, “remember: Monty will always open a door, and the contestant knows it”.
Makes me wonder if there were similar shows where the host can choose not to open a door.
If you choose the door with the car, the host will open another door to tempt you, so then you don't switch, unlike in traditional Monty Hall. And if he didn't open a door, well, that means you choose a goat, so you should switch for a 50% chance. Except the host knows you're thinking this. And you know he knows. And he knows you know he knows. And...
Perhaps, iterated infinitely, this eventually resolves into a limiting expected value for switching and not, but I have no idea how to compute such a scenario.
And yet it is the subject of endless discussions. The optimum strategy is pretty easy to work out, yet not widely believed.
* if on the 1st try you choose the correct box (33% chance), then the one you can switch to will be wrong
* if on the 1st try you choose the wrong box (66% chance), then the one you can switch to will be correct one
therefore your goal is to pick the wrong box on the 1st try and then switch, and you have 66% chance to do it
"When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3."
Your sentence the particular way you worded it is not the correct mathematical model.
The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply.
Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"
It wasn't really my sentence I just quoted the article. Nonetheless I disagree with you. The probability that the other boxes have the key is 2/3 and that is all that really matters. Opening a door doesn't change anything.
It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened!
(Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.)
Edit: an other version was to change the hosts proposal: what if he let you choose one box, and then said he would let you switch to having whatever was in the other 999 boxes? Of course you would switch! The crux is understanding that this offer is actually the same as in the first proposal, since the host is not opening the boxes at random.
(I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
But you still need to conivnce people that the one remaining unopened door is more likely than the door you originally selected. They were both unlikely to begin with, ramping up the number of doors doesn't explaing why one of them should be preferred.
Now 998 doors are removed. There is 1 door from the others and the door you choose. Given that your choice is almost certainly wrong, and that your opponent couldn't remove the correct door from amongst the 998, that means the other door is the correct one.
Is that convincing enough?
Extending it to 1000 boxes/doors still doesn't explain why the remaining unopened box is different from the box your picked originally.
Because you now know that every other box is empty. So by process of elimination you know that your box and the remaining one are different.
By opening the box he knows to be empty, it manufactures an equivalence to if he'd offered to open the box you chose, vs open the other two together.
The article hides this away by lumping those two under "Host opens B or C" without further justification, but it's important to notice that this only works because those twelve outcomes have different probabilities.
Edit: In table form,
Car Guest Monty Swap? Probability
A A B No 1/3 * 1/3 * 1/2 = 1/18
A A C No 1/3 * 1/3 * 1/2 = 1/18
A B C Yes 1/3 * 1/3 = 1/9
A C B Yes 1/3 * 1/3 = 1/9
B A C Yes 1/3 * 1/3 = 1/9
B B A No 1/3 * 1/3 * 1/2 = 1/18
B B C No 1/3 * 1/3 * 1/2 = 1/18
B C A Yes 1/3 * 1/3 = 1/9
C A B Yes 1/3 * 1/3 = 1/9
C B A Yes 1/3 * 1/3 = 1/9
C C A No 1/3 * 1/3 * 1/2 = 1/18
C C B No 1/3 * 1/3 * 1/2 = 1/18
Total Yes = 6 * 1/9 = 6/9 = 2/3
Total No = 6 * 1/18 = 6/18 = 1/3In the Monty Hall problem, you think you are choosing between one door and one other door. But in fact, you are choosing between one door and two doors.
The choice is between "this door" (1/3 probability for winning) and "all other doors" (2/3 probability for winning).
If your initial choice was a car, the host can open any of the remaining doors, but if your initial choice was a goat this forces the host to reveal extra information to you (namely which of the remaining doors contains the car). Since your initial probability of picking a goat was 2/3, there is 2/3 probability that the host will reveal the prize door for you.
This is why the puzzle is only loosely based on a TV show. No real TV or other iterated games will work like this, since the optimal strategy is pretty simple. In a real TV show, the host would mix up his strategy (never revealing the car door, but only occasionally opening a door after the candidates choice). In that case it's not possible to work out an optimal strategy without additional assumptions or clues wrt the host behavior. E.g. he might be biased to open a remaining door with higher probability when the initial choice was correct, to increase suspension for the viewers, in which the dominant strategy is actually to not switch. But in a real TV show or iterated game, the host behavior is likely not deterministic.
However, if the rule is not explicitly stated, how can the player know that the rule exists? Perhaps "Monty" is evil and will not always open a door, "evil Monty" will only open a door when he knows you've chosen correctly.
IOW, without that rule explicitly stated, the answer "Switch" is simply incorrect. Without that rule, the answer is "I don't have enough information to know."
In fact, the Wikipedia Monty Hall article discusses pretty much any aspect of the problem that anyone has ever brought up in any Monty Hall forum thread or blog post.
Like the article, take a deck of cards. Ask them to pick a card for you without looking. Set it aside.
Tell them if they have the queen of diamonds, you buy lunch, otherwise they buy lunch. Ask if they want to swap decks.
After they inevitably say yes, go through the 51 other cards and turn over 20 cards that aren't the queen of diamonds. Ask if they're sure they'd like to switch.
Remove 20 more cards and repeat. Then 9 more (leaving you with 2 cards.) Ask again, turn over one last card, ask one more time. (This last iteration is the actual Monty Hall problem.)
The key thing is they should understand now that Monty Hall knows where the queen of diamonds/ car is and turns over other cards/goats precisely because he knows they don't change the odds of the original choice, but many people incorrectly believe that it does.
Consider instead a Monte Hall scenario with 100 boxes (vice just 3), maybe we call this the "deal or no deal" variant of the problem...
The user picks 1 box and has a 1/100 chance of selecting the box with the prize. Now, the host opens 98 boxes that they know do not have keys in them, leaving two unopened boxes (the one the user picked and the one the host left unopened).
Now, pick your box from the remaining selection and claim your prize. You bet your sweet bippy I know which box I'm picking.
Information is gained by observing the winnowing of the field of options.
If someone is willing to pick 1/3rds odds of winning, the discussion needs to go somewhere else.
2. Your original choice gives you a goat 2/3 of the time and a car 1/3 of the time.
3. So by switching you get a car 2/3 of the time and a goat 1/3 of the time.
You pick a box, and another empty box is revealed. Then, the two remaining boxes are shuffled so that you no longer can tell which is which (but the game host still knows). You then choose one of four options:
- take box A,
- take box B,
- ask the host to give you the box you picked initially, or
- ask the host to give you the box you didn't pick initially.
"Which of the three boxes do you choose?"
"Box A."
"Is the prize probably in Box A?"
"No, it's probably in one of the two other boxes."
"So it's probably not in Box A."
"Correct."
"Look: Box C doesn't contain the prize. Do you want to stay with Box A which probably doesn't have the prize, or switch to one of Box B or C, which probably does have the prize?"
"Well, since Box A probably doesn't have the prize, I should switch to Box B or C. But I know Box C doesn't contain the prize, so I'll switch to Box B, which probably does."
Another approach is to make a lot of boxes.
"Here are 100 boxes. Choose one that has the prize."
"I choose box 17, which probably doesn't have the prize."
"Now I'm going to open 98 more boxes and show you that none of them contain the prize. As you see, only box 68 remains closed (as well as your box 17). Do you wish to switch to box 68?"
"Hell yes!"
The Host knows which box
has the prize.
You choose box A, leaving B and C.
The Host opens box B and,
revealing it empty, gives you
information about box C
that you don't have about your box
since the Host cannot choose your box
and must choose an empty box
from an information
perspective, box C has
better odds.1) All of these explanations end up taking this to the extreme. (Imagine playing 10,000 games. Or imagine 100 doors). The game is purposefully set up as 3 doors and one ”game”. The decision the player makes is final and they don’t get to see “averages over time”.
2) Confirmation bias (there’s probably a more correct term, but I’m going with this). A player picks the door and then switches, knowing their chance of winning is 66% by switching. But it’s also 33% losing. Switch and lose, and psychologically you feel you made the wrong choice. People who don’t understand the math will tell you that you made the wrong choice. I think that can cause a lot of people to second guess themselves, even if they know the problem.
3) Fortunately, the stakes are fairly low. Unlike some of the proposals, losing only means losing out on a car, not death.
The MHP has a mathematical solution, but it’s also very much a human-nature problem.
If, on the other hand, the presenter didn't know which box is the right one, and just opens one of the other two boxes at random (with a 1/3 chance of opening the box with the prize), then, if the opened box turns out to be empty, the chance of the remaining box being the right one drops to 50%.
The difference between these two scenarios becomes obvious if we expand it to 100 boxes: You pick a box, 1% chance of being right. Of the other 99, at least 98 are empty. The presenter knows which, opens the 98 empty boxes, and now there's a 99% chance of the remaining box being the right one.
Other scenario: The presenter opens 98 boxes, not knowing which are empty, so he has a 98% chance of opening the box with the prize. On the unlikely chance that all are empty, there prize must have been in either the box you picked, or the remaining box, but we still have no information about which it is, so there's a 50% that you're holding the right box.
Of course if you don't know whether the presenter knows, and you don't know if he was lucky that the 98 boxes he opened were all empty, or that he knew, then the situation becomes quite a bit more complicated, but the chance of switching being the best option is going to be larger than 50%.
How much? Let's say there's an a priori 50% chance that he knows or doesn't know. If he doesn't know, then opening 98 empty boxes is pretty unlikely, so it's pretty likely that he knows. It's probably possible to calculate those odds, but I'm not going to try that now.
And if the presenter's behaviour changes depending on whether you picked the right box or not, for example he uses his knowledge to actively tempt you away from the right box, then all bets are off. Or maybe him opening another box is proof that you've got the right one. Or maybe that's what he wants you to think...
Arguing Monty Fall is 50/50 is just the fallback position of unrepentant Monty Hall halfers, CMV.
ETA: I also believe the 50/50 result hinges upon an implausible interpretation of the problem. It usually goes like this: if the host accidentally opens the prize, then the game is void. Under this interpretation, 50/50 is correct, as can be easily verified through simulation. However this isn't the Monty Fall problem as stated! The problem states that the host does not reveal the prize, not that the game is void if he does. The difference is, in the voided game version, the games where you originally pick the prize (and thus are destined to lose by switching) are never voided, whereas the games where you stand a chance of winning (because you didn't originally pick the prize) are sometimes voided. This unequal voiding probability skews the game against you, reducing your 2/3 win rate to 1/2. But again, this is the wrong game, it is not the one which was actually described. If the host "does not" reveal the prize, then it really doesn't matter why.
It is because when the player has picked a goat door, the host is restricted to reveal specifically which has the only other goat, but when the player has picked the car door, the host is free to reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case.
For example, if you select door 1 and he reveals door 2, it was 100% sure that he would have taken #2 if the correct were #3, as he wouldn't have had another option. Instead, we couldn't ensure that he would have opened door 2 in case the correct were yours, as he could have preferred to reveal door 3 in that case (each of them would have had 50% chance of being removed). So, from the times that you start selecting door 1, it tends to happen with twice the frequency that he opens door 2 once the correct is #3 than once the correct is #1.
Instead, if the host does not know the locations and his revelation is random, he cannot make that distinction of revealing one door more than the other depending on the prize location, precisely because he does not know where it is. If you pick door 1, you cannot say that he will open specifically door 2 with more frequency when door 3 is the correct than when door 1 is the correct. If he decides to open door 2, that choice is independent of where the car is.
I think it matters a lot, and I suspect this is the issue that Monty Hall nonbelievers (who think the chance is 50% even if the host knows) are struggling with, but from the opposite direction of you.
Let's try to work out the probabilities:
You randomly pick box (let's call it A), 1/3 chance of being correct. The host randomly randomly opens one of the other two (let's call it B). 1/3 chance he opens the box with the prize, invalidating the game. 2/3 chance he opens an empty box. If he picks an empty box (2/3 chance), each of the other boxes (A and the remaining box, let's call it C) still has 1/2 chance of holding the prize, so 1/2 * 2/3 = 1/3 chance each a priori, but 1/2 after eliminating the 1/3 chance of invalidating the game.
So now we've got 1/3 chance of A having the prize, 1/3 of the game being invalid, 1/3 of the third box holding the prize. After eliminating the invalid game, there's still an equal chance of A and C holding the prize.
When the host knows which box holds the prize and uses that information to always pick an empty box (which he always can, because there will always be at least one unpicked empty box), there's no chance of invalidating the game, so the remaining box has a 2/3 chance of holding the prize.
> The problem states that the host does not reveal the prize, not that the game is void if he does.
Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty. If he doesn't, there's always a chance of him accidentally opening the box with the prize. If he opens an empty box by luck, then you dodged that 1/3 chance of invalidating the game. You're now in the changed probability space given that the host didn't accidentally invalidate the game. Each box had an a 1/3 priori chance of being right, but given the 2/3 chance that the host didn't invalidate the game, they now each have a (1/3)/(2/3)=1/2 chance of being the correct box.
It is the knowledge of the host that eliminates the chance of an invalid game and puts the 2/3 chance on the remaining box.
> If the host "does not" reveal the prize, then it really doesn't matter why.
It does. You dodged a bullet, and that has an influence on the remaining probabilities. In the original Monty Hall problem, the host provides extra information, in Monty Fall, he doesn't, but you dodge a bullet. That difference matters.
You're ignoring the difference by posing a non mathematical objection to the statement of the problem. This is like having a physics problem about frictionless spherical cows and objecting that cows aren't really frictionless and spherical, and giving an answer for a different problem. Sure you might come up with a valid answer for your different problem, but it isn't a valid answer for your original problem. Likewise 1/2 is a valid answer to the invalidated trials version of the problem here, but not a valid answer to the problem as stated, which remains 2/3.
These are the valid answers to the two problems. It's not clear to me what you think I'm ignoring, but if you can make that clear, perhaps I can explain my meaning a bit better.
In the trial invalidation version of the problem, but not in the problem as stated. The problem as stated provides, the host does not reveal a car (or your original door).
Let me put it another way. If the question had asked, "Conditional on the host not revealing a car (or your original door), what is the probability of winning if you switch?", then that too would be 1/2. My problem is that this is not what is asked. We don't get to answer an easier or different version of the problem. And the reason it matters is because people start spouting woo about how the host's knowledge or intentions are what mattered, when that isn't true at all. What matters, as you have demonstrated and I have tried to clarify (we aren't really disagreeing), is whether we are exploring the conditional probability in the uniform distribution over a sample space where the host might open another door (1/2), or the probability in the uniform distribution over a sample space where he cannot (2/3). If one denies the difference between these versions of the problem, one ends up in woo-space where the host's knowledge or intentions matter. They don't.
Knowledge and intentions don't matter? They most certainly do.
Imagine a third version of the problem (I'm afraid I don't have a good name for it) where the host actively tries to lure you away from the correct answer, and only opens an empty box if you've picked the correct box. Now, if you pick a box and the host opens an empty box, you should absolutely not switch, because the remaining box has 0% of being the right one.
Knowledge and intentions matter. The real problem happens when you don't know what strategy the host is using.
> Vazsonyi ran the program 100,000 times. Erdos watched the results of the simulation. The simulation results indicated that by switching, the odds of winning are indeed two out of three. Finally, he was grudgingly convinced that switching was better. He did not like it but seeing was believing. He could not argue with the results.
Apparently he later called Ronald Graham and explained to him that he finally heard a proof of the problem that made perfect sense to him, which he explained to Graham. Graham said he didn't understand the proof at all, but was happy Erdos understood where he had been mistaken.
Since there are 2 empty boxes, this means that if you pick some box out of 3, you will have 2/3 chances of picking an empty box.
However, the revealed box will always be empty, so in both of these outcomes the revealed box will be the other empty one.
So in 2/3 of cases, the pair (picked,revealed) will contain both of the empty boxes, and so switching will get you the prize
C G G
G C G
G G C
You choose a door, say Door 1. Monty opens another door with a goat behind it.
C x G
G C x
G x C
Now look at the grid. Staying in column 1 gives you the probability space C G G, a 1/3 chance of getting a car. Switching gives you C C G, a 2/3 chance of getting a car.
1000 doors, choose 1.
Host opens 998 doors.
Do you switch?
> How NOT to explain the Monty Hall problem to a disbeliever
FTFY.
Imagine that after you pick your box, Monte Hall invites an audience member up on stage and instructs them to choose one of the remaining two doors to open. This audience member doesn't know anything at all and just randomly picks one of the two doors. When their door is opened we see that it's empty. You're now given the option of switching just like in the standard game. Should you?
Cosmetically everything is identical with the standard game, but if you analyze the game carefully this time you're left with a 50/50 shot so there's no benefit of switching.
I think most of the arguments in this article would appear to work for this modified version of the game which means that they're not actually getting to the heart of the problem.
For completeness, the reason this now reduces to 50/50 is that there's also now a chance that the spectator opens the door with the car behind it, something that couldn't happen in the original Monte Hall problem. Put another way, there's actually a little bit of information that's conveyed to you when you see that the spectator happens to not open the door with the car and this extra information exactly cancels the usual benefit you get from eliminating the other empty door. In the example of "scaling up" in the article, if you did this with 20 doors and the spectator randomly picks 18 of the 19 unopened ones to open and then happen to not stumble upon the car, you might actually think that you could have been lucky all along. Ultimately you're left with a 50/50 chance.
- a car is revealed, I lose immediately (there is no option to switch anymore)
- no car is revealed, which means I again have 2/3 chances when switching, not a 50% chance as you've stated
In your example, the spectator opens an empty door, so there is no difference to the host opening an empty door in regards to the probability. Again, if the spectator opens a car, I just lose.
Door1 Door2 Door3
Car Empty Empty
Empty Car Empty
Empty Empty Car
Let's suppose your strategy is to stick with your original choice. In the first case above then you get the car. In the second case the audience member stumbles upon the car and you lose. In the third case you lose because you stick with an empty door. All three cases are equally likely and since the second one ends, and you know that your game didn't end, you know that you're either in case 1 or case 2. Your chance is thus 50/50.
The issue is that in the classic MH problem, cases 2 and 3 are collapsed into one outcome (MH opens an empty door), but that's not true here.
More mathematically, you should ask yourself p(Door1 | Game Did Not End).
Using Bayes we see
p(Door1 | GDNE) = p(GDNE | Door1) * p(Door1) / p(GDNE).
p(GDNE | Door1) = 1 p(Door1) = 1/3 p(GDNE) = p(GDNE|Door1)p(Door1) + p(GDNE|Not Door1)p(Not Door1) = 11/3 + 1/22/3 = 2/3.
This, p(Door1 | GDNE) = 1 * (1/3) / (2/3) = 1/2.
It's a completely different problem than the Monty Hall problem, which can never end in the car being discarded.
Here's another way of seeing it which I think may be effectively the same. Suppose that there are 100 doors in a circle. You pick one and then all the doors are opened except for the one you picked and the next door along. If the car hasn't been revealed yet, it seems intuitively right that you have a 50-50 chance of winning whether you stick or switch at that point.