9 ms·
How to explain the Monty Hall problem to a disbeliever
- arnvald 4y agoThe 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
- qumpis 4y agoOr to just imagine a 1000 boxes with the same problem formulation
- yesseri 4y agoAnd the key thing here is that all boxes except two gets removed, not only one.
- oleganza 4y agoThat was a new spin to the explanation that I didn't think of before.
- jstx1 4y agoThis doesn't do anything for me. (I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
- coldtea 4y agoBecause a 99/100 chance is much better than 2/3 to drive the point home...
- curiousgal 4y agoImagine there are 999 boxes with nothing in them and one box with the keys. After picking a box, the hosts opens 998 empty boxes. Would you still stick with your initial choice?
- jstx1 4y agoI would change my choice because I understand the problem. But I would also change my choice in the scenario with 3 boxes. I'm not arguing with the conclusion, what I don't understand is the people who have their mind changed by the argument. Extending it to 1000 boxes/doors still doesn't explain why the remaining unopened box is different from the box your picked originally.
- curiousgal 4y ago> 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.
- brigandish 4y agoIt's because it makes the initial choice so increasingly unlikely (increasing with the number of doors) to be correct that when the doors are taken away and you're left with only two, one of which must be right, it means that the other door is incredibly likely to be the right one.
- jstx1 4y ago> It's because it makes the initial choice so increasingly unlikely 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.
- brigandish 4y agoThere are 1000 doors. You choose 1. Anyone knows it's incredibly unlikely that the correct one is chosen first time. 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?
- deleted 4y ago[deleted]
- __s 4y agoI've often found it easier to understand things intuitively by putting an idea to the slippery slope test. If such & such were true, imagine changing some parameters to an extreme, how absurd does it become? For monotonic functions it's useful
- wanderingstan 4y agoThis was the one that worked when explaining it to my friends. 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.
- charlieflowers 4y agoTo me, THAT is the most powerful intuitive description.
- kasperni 4y agoI think their explanation is a lot easier to understand "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."
- jasode 4y ago>= 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.
- kasperni 4y ago> Your sentence the particular way you worded it is not the correct mathematical model. 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.
- basch 4y agoYou do get to switch to both other doors. One of the two remaining doors is a goat, and it is opened for you. Another way to phrase that is getting to pick both doors and the goat doesnt count against you. 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.)"
- justatdotin 4y agoyes, that's how I got it
- adhesive_wombat 4y agoThe 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.
- GloucesterBoy 4y agowow... what a grim approach!
- pfdietz 4y agoThat Escalated Quickly would be a great name for a game show.
- adhesive_wombat 4y agoNext on 4: Brian Cox presents Quantum Squid Game, in partnership with Academi.
- TarasBob 4y agoIn the scenario where they are shot that you described, it is still better to switch!
- masklinn 4y agoYou can’t switch if your universe has been pruned Morty.
- darkerside 4y agoThis explained it better to me than the article
- v64 4y agoI 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.
- celeritascelery 4y agoThat was how I had to convince myself. I wrote a python program to simulate it a million times, Thinking for sure I was going to disprove everyone. But the result clearly showed 2/3 probability of a good outcome if you switched. This let my mind switch from “this can’t be true” to “this is true, now I need to intuit why”.
- jstx1 4y ago1. 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.
- summarity 4y agoFor 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.
- nmdeadhead 4y agoI do not think that when it's explicitly stated it becomes intuitive for everyone and that adds a further wrinkle. Many people will still get it wrong, even when the problem is stated correctly. 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."
- em500 4y agoYup. Wikipedia contains a discussion on host behavior: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... 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.
- em500 4y agoYes, in many formulations the unstated assumptions that the host (a) will always open a door after your initial pick, and (b) that it's always one without a car behind it. Making the assumptions explicitly makes the solution and intuition much simpler. 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.
- rob74 4y agoAs a programmer, I wasn't convinced until I wrote a short program to simulate it. Then the penny finally dropped...
- melling 4y agoThe easiest way to remember. 1000 doors, choose 1. Host opens 998 doors. Do you switch?
- tutuncommon 4y agoI 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.
- pdpi 4y agoThe 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
- victorNicollet 4y agoFrom 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.
- pontus 4y agoThere'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.
- darekkay 4y agoCorrect me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen: - 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.
- n4r9 4y agoI think you're correct. Whether or not the spectator had a non-zero chance of opening the non-empty door is irrelevant after the fact.
- darkerside 4y agoIn 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?
- lapinot 4y agoThe random action of the host choosing to open one of the two loosing boxes is not distinguishable by the player: the resulting states are always part of the same information-set in game-theory parlance. As player actions are always only depending on information-sets, the table actually contains information-sets as rows. It only looks weird because most of the information-sets of the game assign deterministic values to all the variables.
- api 4y agoI got this immediately when I thought of it in terms of information theory. Each door opening gives you one more bit of information.
- xiaodai 4y agowhy explain? just walk away. not worth it. they dont need to understand. it's on them
- fwlr 4y agoConsider 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.
- acadapter 4y agoThere 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).
- jlongr 4y agoC = 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.
- thom 4y agoThe 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.
- wanda 4y agoI 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.
- mcv 4y agoThe 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...
- firstlink 4y agoThis alternate problem (commonly known as "Monty Fall") has always infuriated me more than the original. I do not believe it is possible that it matters whether the host "knows" or doesn't. The problem asks for the probability of switching resulting in a win, conditioned on the host doesn't reveal the prize. It does not matter mathematically why the host doesn't reveal the prize (i.e., whether P(host doesn't reveal the prize) < 1 or = 1), the conditional probability and thus the answer will be the same. 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.
- mojomark 4y agoThere'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.
- basch 4y agoI don't see how "would you like 1/3rd odds or 2/3rd odds of winning" is less easy to understand that adding more numbers and layers to the problem by adding in words like $1Billion and 100 boxes. If someone is willing to pick 1/3rds odds of winning, the discussion needs to go somewhere else.
- LelouBil 4y agoWhat 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
- kthejoker2 4y agoThe 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.
- mcdonje 4y agoI wonder if anyone has proven this experimentally or by logging results of the gameshow. Seeing it work in real life would help the naysayers.
- richij 4y agoThere 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.
- jdechko 4y agoI 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.
- beej71 4y agoThere 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!"
- firstlink 4y agoI think the 100 box explanation misses a subtlety in the problem. Conditional on those 98 boxes being empty, box 17 and box 68 are equally likely to have the prize. The key is that you have a large probability of backing the host into a corner as it were: forcing him to choose 98 of 98 openable boxes vs 98 of 99. This is where your large probably of winning comes from, but it's no clearer in the large (100-box) example than the small. Therefore I posit that the large example exploits a misunderstanding about the solution rather than elucidating it.