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How to explain the Monty Hall problem to a disbeliever

by anupj on 3/14/23, 10:55 AM with 93 comments

by adhesive_wombat on 3/14/23, 11:23 AM
The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious.
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.
by arnvald on 3/14/23, 11:15 AM
The most intuitive and simple explanation that worked for me is:
* 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
by fwlr on 3/14/23, 11:32 AM
Consider the Honty Mall problem: it’s like the original problem, except after you pick a box, Honty offers you both of the other boxes. It’s much easier to see swapping is better in this problem, and it’s also easier to see that the chance is 2/3 if you swap. Then you just have to show that the Honty Mall problem is equivalent to the Monty Hall problem, by stipulating that Monty will always open a box that’s empty.

by pdpi on 3/14/23, 11:27 AM

The article gets the 50/50 case subtly wrong. The really naive analysis is that you have two doors so it's 50/50. A less naive, more interesting analysis, is that there are twelve possible outcomes, of which six are favourable (if the correct option is A, pick A -> presenter picks B is different from pick A -> presenter picks C).

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 = 1/3

by acadapter on 3/14/23, 11:33 AM
There is a linguistic illusion at work here.
In 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).
by summarity on 3/14/23, 11:23 AM
For me the hangup was always the hidden rule: host won’t open a door with a car. That is unstated and remains unstated even in modern discussions of the problem (see Pinker’s “Rationality”). Once explicitly states the outcome becomes intuitive.
by kthejoker2 on 3/14/23, 1:24 PM
The easiest way is to convince them is with real money on the line.
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.
by mojomark on 3/14/23, 11:57 AM
There's actually an easier way to remove the subtleties of this statistical problem by using a hyperbolic example. For kicks and shiggles we'll up the stakes to be a prize of $1B USD.
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.
by jstx1 on 3/14/23, 11:23 AM
1. Changing your mind gives you the opposite of your original choice. << once you get this the rest is obvious
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.
by victorNicollet on 3/14/23, 11:29 AM
From my experience, the issue is that the 50% intuition is hard to overcome (and it takes an overwhelming amount of evidence for the 33%/66% reality in order to overcome that intuition). Here's an alternate version of the game which aims to dismantle the 50% intuition directly, instead of trying to argue for 33%/66%.
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.
by tutuncommon on 3/14/23, 11:25 AM
I like the information point that after one is shown to be wrong you now have new information because they could not show your selected box or the one with the prize so the one remaining is more likely.
by beej71 on 3/14/23, 7:37 PM
There are a couple approaches I like:
"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!"

by wanda on 3/14/23, 11:47 AM

I usually explain it to people like this:

  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.

by jdechko on 3/14/23, 2:07 PM
I completely agree with the math on this (math is math). However here are my observations on why this is still such an interesting problem.
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.
by mcv on 3/14/23, 11:52 AM
The vital part about this problem is that the presenter knows which box is the right one. By opening an empty box, he's giving you extra information. And that means there's 2/3 chance of the remaining box having the prize.
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...
by v64 on 3/14/23, 11:23 AM
I love the Monty Hall problem because it's so unintuitive that even Paul Erdos struggled to believe it for a while.
> 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.
by darkerside on 3/14/23, 11:28 AM
In the table, why does the host picking one of the other boxes get combined into a single probability row? If the host could pick either box, and both of those choices result in a loss, should we count that as additional possibilities?
by LelouBil on 3/14/23, 12:19 PM
What made it clear for me was this :
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
by rob74 on 3/14/23, 11:25 AM
As a programmer, I wasn't convinced until I wrote a short program to simulate it. Then the penny finally dropped...
by jlongr on 3/14/23, 11:36 AM
C = car, G = goat. Column 1 = Door 1, etc. The car could be behind any of the three doors, hence the three rows showing each possibility.
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.
by thom on 3/14/23, 11:46 AM
The intuitive explanation I've found useful is: two-thirds of the time, he's telling you exactly where the car is. One-third of the time he can't help you either way. If you have a friend whose advice is more often right than wrong, you should generally take it.
by melling on 3/14/23, 11:25 AM
The easiest way to remember.
1000 doors, choose 1.
Host opens 998 doors.
Do you switch?
by mcdonje on 3/14/23, 1:34 PM
I wonder if anyone has proven this experimentally or by logging results of the gameshow. Seeing it work in real life would help the naysayers.
by api on 3/14/23, 11:32 AM
I got this immediately when I thought of it in terms of information theory. Each door opening gives you one more bit of information.
by richij on 3/14/23, 1:53 PM
There seems to be an error in the headline. It seems like it's missing a word.
> How NOT to explain the Monty Hall problem to a disbeliever
FTFY.
by xiaodai on 3/14/23, 11:30 AM
why explain? just walk away. not worth it. they dont need to understand. it's on them
by pontus on 3/14/23, 11:27 AM
There's another version of the Monte Hall problem that highlights why this is such a counterintuitive problem.
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.