20 ms·
Monty Hall Problem
- leeoniya 7y ago> Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999). this tidbit is pretty wild
- turc1656 7y agoWhat's even wilder is that the people working for game shows utilized this strategy long before it was a known, publicly verified thing. So someone who wasn't a world famous mathematician figured it out and was confident in their work - the same work that a renowned, industry-leading professional refused to believe.
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- ghaff 7y agoOnce you figure it out, it's pretty obvious though. And the game show people don't need a rigorous mathematical proof. You can simulate it with some playing cards and understand it in a hurry. What made it very clear to me is if you just increase the number of incorrect doors that you open by a lot. (Making the assumption that the host knows where the prize is.) In that case, it's obvious that you're improving the odds by showing where the prize isn't.
- emilga 7y agoIt's also amusing to see a mathematician be more convinced by the computer simulation than the proof.
- CobrastanJorji 7y agoReminds me of the Knuth quote: "Beware of bugs in the above code; I have only proved it correct, not tried it."
- abeppu 7y agoWell, it's also amusing to me that people want to jump to simulations that generate a large number of examples. In the original 3-door version of it, it's reasonable to exhaustively list all the possible sequences and just tally up the outcomes under each strategy. The simulation is just sampling from this not-that-long list.
- tcmb 7y agoAlso this: > Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans (Herbranson and Schroeder, 2010).
- Hamuko 7y agoDid some academic rewrite the article or is there some explanation as to why so many of the citations on that page are so un-Wikipeda-like?
- stupidcar 7y agohttps://en.wikipedia.org/wiki/Wikipedia:Citing_sources#Citation_style https://en.wikipedia.org/wiki/Wikipedia:Citing_sources#Citat... Basically, Wikipedia does not mandate a citation style, it just has to be consistent throughout an article. And once set, another editor can't change it simply for personal preference. Presumably the original author, or whichever editor first added citations, was more familiar with this style, and so it has stuck.
- jwilk 7y agoWhat's un-Wikipeda-like about them?
- drfuchs 7y agoI found the easiest way to convince non-mathematical types is by suggesting we do the million-door version: “Ok, pick a door between 1 and 1,000,000. We’ll call that door N. Now, I’ll open 999,998 bad doors. So, now only door N and door 223,205 are still closed, and one of them has the good prize. Want to keep door N, or swap?”
- avn2109 7y agoGosh this makes it much easier.
- turc1656 7y agoThis is an excellent way to make it click for people.
- graeme 7y agoThis is exactly how I conceived of it when I first heard of the problem. Pretty sure I remember demonstrating it to people using 10+ playing cards and flipping over all but one of the cards the person didn’t choose.
- CydeWeys 7y agoThe flaw with this thought experiment as a way of teaching the original problem is that it's not at all the same to reveal 998,000 bad doors as it is to reveal just 1, and when revealing just 1 in this situation the effect is much smaller than just revealing 1 of 3 doors.
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- nwallin 7y agoPlease explain to me why it is different whether to switch doors after opening all but one door vs after opening all but one door? (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.
- mycmtwilbedead 7y agoSo very cool. A random wiki with a few points gets on the front page Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
- teej 7y agoI saw an interesting perspective on the Monty Hall problem from judegomila earlier this month - https://twitter.com/judegomila/status/1183425802399969282 https://twitter.com/judegomila/status/1183425802399969282 "Your brain will compute monty hall problem in fast casual causal mode AND/OR compute slow mode formally for the exact answer. Probability predictions that accurately map reduce to efficient intuitive representations in our brain dictate direction in our evolutionary biology." "[since] we don't solve the monty hall problem effectively with our intuition, an evolving life form would be poor at resource collection from such problem sets without knowledge of axiom representation and formal probability."
- godson_drafty 7y ago"Sire. Shall we hunt in the valley, in the forest, or in the mountains?" We shall hunt in the forest. "Sire! Our scouts have just returned from the mountain. They report that the herds are not there at the present time! Shall we try the valley, or continue on to the forest?" A) Stick to the plan, and hunt in the forest B) Hunt in the valley instead
- greenshackle2 7y agoThe Monty Hall paradox hinges on the fact that the host knows in advance where the car is. If your scouts don't know where the herds are, but just picked an area at random to recon, then it is a different problem, and switching makes no difference.
- bzbz 7y agoAnd I think you stumbled upon the crux of the problem. I think we have a hard time grasping the fact that Monty not just picks one of the other two doors, but also specifically guarantees that it won't be the winning door. Without that guarantee, it makes no difference. Additionally, there are not many situations in real life that would replicate the conditions of that guarantee. So our "intuition" is simply geared towards situations where an actor can't read our thoughts (door pre-selection) before we act them out, and in such a scenario, pre-selecting/switching selection of doors doesn't matter.
- myheliguh 7y agoSo very cool. A random wiki with a few points gets on the front page Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
- gen19600_ 7y agoSo very cool. A random wiki with a few points gets on the front page Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
- jhncls 7y agoSomething that really adds to the confusion, is that "in a not mathematically perfect" world, the observer doesn't know whether the host will always open a door. When the candidate chooses the winning door, the host could always open another door, and when the candidate chooses a loosing door, only open another door with a reduced probability. Life teaches people to always be suspicious when free money is promised.
- mrob 7y agoThe "Monty from Hell" scenario (Monty only offers the choice if you picked the car, and otherwise does nothing), which the original statement of the problem allows, is IMO the most likely in real life because it would make for the most entertaining television.
- kangnkodos 7y agoYou have real world knowledge of what makes entertaining TV, and perhaps how this particular show actually operated. And you are correct. In the real world, Monty Hall always revealed the worst of the two prizes that the contestant did not select. In the real world, you can use common sense and assume that Monty Hall will use that algorithm. But in a mathematical probability problem, you can't do that. For example, if there's a slightly smaller object and a slightly larger object concealed in two different sized boxes, common sense says that a person would put the larger object in the larger box. But if it's a abstract mathematical probability problem, and you have no information on the algorithm used, then you have to assume the concealer flipped a coin when they decided which box to use. If I'm the one setting up the big-small game, and I use the random algorithm, the smaller object is going to be in the bigger box about half the time. If my mom is the one setting up the game, the smaller object is going to be in the smaller box every time. In abstract mathematical probability problems, if you don't know what algorithm was used, you aren't allowed to bring human behavior in, unless it's stated in the problem.
- mrob 7y agoThe problem depends on human behavior, so as originally stated it is unsolvable. You need to clarify that Monty always opens a door, and always reveals a goat.
- mhb 7y agoA Bridge from Monty Hall to the Hot Hand:The Principle of Restricted Choice https://pubs.aeaweb.org/doi/pdfplus/10.1257/jep.33.3.144 https://pubs.aeaweb.org/doi/pdfplus/10.1257/jep.33.3.144
- ivanb 7y agoThis is only a paradox due to very poor definition. "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." Where does it state that the host always opens a door with a goat? From the statement it looks like the goat was accidental. No wonder Paul Erdős got confused by this too. The translation of the paradox to his language could have been even more confusing.
- 0x5345414e 7y agoExactly. It made perfect sense to me once I realized that the host knows where the prize is and thus is giving you information by opening a door.
- kangnkodos 7y agoYes. If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.
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- Kiro 7y ago> If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. Isn't it the other way round?
- twiceaday 7y agoI don't think you are correct. The host does not give you any useful information by opening a goat door. Opening a goat door only serves to confuse you into not realizing that the switch-choice is actually between one of the doors and a grouping of the remaining two doors. Opening a goat door in the grouping of the other two only changes the groups appearance, making it look like the single door that you picked. It does not change the functional state of that group in terms of its probability of containing a car.
- Syzygies 7y agoAs a math professor recalling reactions back in 1990, there were various forms of "Marilyn vos Savant" bias in play. A good cover for gender bias was the absurdity of the world's highest IQ writing newspaper columns. There are plenty of other examples of narrow spikes in particular forms of intelligence having a crippling affect, while the most creative people I know work to change the world with less spectacular but more broad-spectrum cognitive advantages. Still, it's a bit of a leap to conclude she was actually wrong about this problem.
- kangnkodos 7y agoThe key was... did Marilyn vos Savant state in her version of the problem that the host always uses an algorithm to open a door with a goat? (The alternative was that the host randomly picked a door, and this one time it happened to be a goat.) I read some of her articles on this topic, and my opinion was that in some of her articles, she did not state what the host's algorithm was. So, yes, even a person with the world's highest IQ can make a mistake.
- mundo 7y agoHer column was written for an audience that was familiar with the show. Monty Hall had, at that point, been opening doors and revealing goats-but-not-cars to contestants five days a week for more than twenty years.
- junar 7y agoBut as Monty Hall himself notes, the game show rules did not actually require him to open a door. > Hall clarified that as a game show host he did not have to follow the rules of the puzzle in the vos Savant column and did not always have to allow a person the opportunity to switch (e.g., he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door). If the host uses an adversarial strategy, such as only offers a switch when the remaining door is a goat, switching is not necessarily wise.
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- Doubl 7y agoI think most people get it that prize is twice as likely to be behind one of the two other doors rather than behind the door you pick. So one time out of three you'll get it right while two times out of three the presenter indirectly indicates the winning door. Edit. Be nice if my downvoter would explain what's wrong with the above.
- GistNoesis 7y agoHere is the scam version to make money out of math people who know the Monty-Hall problem. You look at the 3 cards then put them on the table and propose a fair triple or nothing, if they can find the red queen. You make them point toward the card they chose. If they point a wrong card you go toward the card and reveal -> you won. If they have pointed the right card, you return one of the card they didn't pick and ask them if they are sure or want to change. Because they think they are facing a Monty Hall problem they will switch. -> you won again. :)
- twiceaday 7y agoThis would only work on people who were made aware of the problem but spent no time understanding it and just memorized 'switch.' Even if you didn't cheat them, this game has zero expectation value... I guess if you are working with people who wouldn't know this you are fine.
- GistNoesis 7y agoThis is not cheating, you don't have to hide the fact that you know where the good card is. It is entertainment teaching a valuable lesson. It is, as I said before, a fair game so it has zero expectation value for both players, better odds than a casino for the player and there is no reason for a gambler to refuse this bet. In expectation you won't lose money, but people who think they know better will happily give you their money by making a bad move while being convinced they are getting an edge. Most math-type people don't know the hypothesis of the Monty-Hall Problem and can't state it properly, but most have seen it played on TV-shows and know they must switch. In the heat of the moment, especially if you have played a few follow the red queen games before, the brain is still in fast mode, and will jump to using heuristics. Mind trick.
- vladislav 7y agoIt's simple: choosing again switches from a losing outcome to a winning outcome and vice versa, thus the probability of winning by choosing again is 2/3.
- Myrmornis 7y agoThat's a clean explanation. Maybe needs one introductory sentence: "Note that your first choice has probability 2/3 of being a losing choice".
- mself 7y agoHere’s another version that helps clarify it... There are three doors, one has a prize and the other two are empty. You pick a door (say, Door 1). Monty then says that his assistant will move whatever is behind Door 2 (if anything) to Door 3. He then opens Door 2 to show you it is empty. Would you switch to Door 3? Of course you would. But Monty’s assistant is lazy, so instead of moving the contents from behind one of the closed doors to the other, he just asks Monty to open the one that he knows is already empty. The result is the same.
- rq1 7y agoChange the 3 doors to 10^6 doors and it becomes obvious.
- Myrmornis 7y agoNo, it becomes less obvious, since the host's help has less influence the more doors there are. EDIT: My reply is based on a misunderstanding. I was assuming that the host continues to open one door, whereas parent intended that the host would open all-but-one doors.
- Vaslo 7y agoNot sure why you picked the Wikipedia entry - about 1000 more clearer explanations out there.
- melonkidney 7y agoYou probably picked a goat. When the host shows you where the other goat is, you switch to the car. Job done.
- _def 7y agoThis is the best explanation I've ever read. Thank you!
- Myrmornis 7y agoInstead of asking yourself "should I switch?", ask yourself "At the time of making my first guess, would I have bet that it got the car (1/3) or a goat (2/3)?" Obviously the answer is that your first guess is more likely to have been a goat. So if you had to bet, you'd have bet on the car being in one of the ones you didn't choose. After the host reveals a goat, it's no longer "one of the ones you didn't choose", but rather "the one you didn't choose" (since the host has taken the other out of consideration). So the choice is easy now (as is the probability calculation).
- dang 7y agoPrevious threads for the curious: https://hn.algolia.com/?dateRange=all&page=0&prefix=false&query=Monty%20Hall%20Problem%20comment https://hn.algolia.com/?dateRange=all&page=0&prefix=false&qu...
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- scoot 7y agoHere's a great explanation: https://mathigon.org/course/probability/monty-hall https://mathigon.org/course/probability/monty-hall Had it in a browser reading list (that I'd forgotten was there), and came across it by chance yesterday.
- nstart 7y agoSince my brain could not accept this long ago, I wrote a tiny script to see what the result would be if this was to be emulated. Works out perfectly! (Might need to update script for python 3). It's weird. Still feels so unintuitive. Which tells a lot about intuition I guess :) https://gist.github.com/kiriappeee/4316871 https://gist.github.com/kiriappeee/4316871
- hanoz 7y agoIt's bedtime here so I'm just going to chuck this in and run away: when you've got Monty Hall sussed, try Tuesday's Child, i.e. "I have two children, one of them is a boy born on a Tuesday, what is the probability I have two boys?"
- earenndil 7y agoIs it not 50%?
- hanoz 7y agoIt is not.
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- nwallin 7y agoHaving once male child eliminates the possibly of having two female children. You take a four entry truth table and scratch one out. The truth table is male->female, female->male, male->male. Tuesday is a red herring. Note that there's some ambiguity between exactly one of vs at least one of which changes the probability but not as much as 1/2->1/3. (exactly one male on a Tuesday excludes the possibility of two males born on a Tuesday) Tuesday is not a red herring. It's truth tables all the way down.
- AnotherGoodName 7y agoAhh i see. It's because the ordering doesn't matter. If the question made the ordering matter; "I have two children, the youngest is a boy, what is the probability that i have two boys" Then it's definitely 50%. The truth table would be Youngest: Boy Oldest: Girl Youngest: Boy Oldest: Boy Youngest: Girl Oldest: Girl Youngest: Girl Oldest: Boy With the bottom two being ruled out by the question. I think that's the part that catches people out here.
- alexhutcheson 7y agoThis alternative formulation from a Mathematics StackExchange answer[1] makes the decision to switch much more obvious: "There are three boxers. Two of the boxers are evenly matched (i.e. 50-50, no draws!); the other boxer will beat either them, always. You blindly guess that Boxer A is the best and let the other two fight. Boxer B beats Boxer C. Do you want to stick with Boxer A in a match-up with Boxer B, or do you want to switch?" [1] https://math.stackexchange.com/questions/96826/the-monty-hall-problem/3360686#3360686 https://math.stackexchange.com/questions/96826/the-monty-hal...
- jedimastert 7y agoI'm gonna have to do a lot of thinking to make sure this matches the problem at hand, but if it does this is an incredible translation.
- bobbylarrybobby 7y agoMonty Hall: if you pick one of the bad doors, which has probability 2/3, then you should switch after shown a bad door. Otherwise (probability 1/3) you should stick with your door. Boxers: if you pick one of the bad boxers, which has probability 2/3, then you should switch after shown a bad boxer which happens when the first round fight ends. Otherwise (probability 1/3) you should stick with your boxer.
- Vadoff 7y agoIt's not quite the same being shown a bad door vs a bad boxer that you were told would always result in a draw otherwise. That means the other boxer is always the better one (or the car), it doesn't take into account that the winning boxer could be what you have.
- zaksoup 7y agoHere's the explanation I like to use every year when this makes it to HN: You pick door 1. Out of three doors you have a 1/3 chance of having a car behind your door. The chance that the car is behind one of either of the other two doors is 2/3. Let's pretend that instead of opening a door and revealing a goat, Monty instead says to you "You can switch to both of the remaining doors. If the car is behind either one of them you get to keep it." Your likelyhood of getting the car by switching is now 2/3. Instead, Monty does the statistical equivalent: He allows you to know for sure which of the other doors definitely has a goat.
- RedComet 7y agoIt is famous for a reason. One should neither be shamed nor feel ashamed for not having intuition for this.
- overdrivetg 7y agoA fun way to play with people who won't believe you is to say OK then, let's play 10 games @ $10/game: Get 3 cards and pick one to be the prize. Your mathematically-challenged partner gets to put them down each time. You pick a card, they remove one of the other cards and then you always switch. I made like $50 from someone this way one time =) I'm not totally heartless - I took him to lunch on me afterwards to lessen the blow ;-) If you really want to drill the point home, offer to keep playing more & more batches of 10 until they've had enough...
- jaimex2 7y agoThe key to explaining the Monty Hall problem to someone is to reverse the logic. When you picked a door at the start you may have excluded the winning door from the host. This has a 1 in 3 chance of happening. So that means the host has a 2 in 3 chance of being left with the winning door. The host who has knowledge can't pick the winning door, so those 2 in 3 odds are passed on to you if you switch.
- raffy 7y agoInitial pick: 1/3 chance of winner Question: did you pick the correct door? - If correct, both remaining doors are losers, eg. {loser, loser}. - Otherwise, the remaining doors are {loser, winner}. And the loser must be revealed according to the game rules. If you don't change your answer, you're stuck with the initial chance: 1/3 If you change your answer: (1/3) * 0 + (2/3) * 1 = 2/3 - (1/3) you initially picked correct, then change to a guaranteed loser. - (2/3) you initially picked incorrect, then change to a guaranteed winner.
- knzhou 7y agoI'm gonna pile on the hundreds of comments already existing and say that the real key to the Monty Hall problem isn't about pure probability, it's about human interaction. The answer can be either 1/2 or 1/3 depending on Monty's motivations, his knowledge, and the way he decides what information to give you. This is true for Bayesian updating on any information, and is a good lesson for life in general.
- dunmalg 7y ago>depending on Monty's motivations, his knowledge, and the way he decides what information to give you. The missing piece of the puzzle that seems to frequently be left out is we KNOW Monty's motivations, knowledge, and how he decides. At the time the problem was formulated, when Let's Make a Deal was still on television, all this was common knowledge: Monty is the host of a game show, he knows where the two goats and the car are, and he decides which door to open by picking one of the unchosen doors he KNOWS has a goat behind it. The problem suffers now, 40-odd years later, by the fact that people often don't know who Monty Hall IS, and the problem is usually worded as if it's still 1963-1976, so this foundational knowledge is missing. There is no QUESTION what Monty Hall knew, or what his motivation on the show was... except people who don't know the game show don't know that. Granted, the actual game show was a bit more varied, but the presumption in the thought experiment is that Monty ISN'T trying to goad you into un-choosing the correct door, which he frequently did on the show because the tragedy of it was more entertaining than just giving away a car.
- knzhou 7y agoSure, if you fix the motivations it's easy, but that's missing the point of the problem -- the only reason it generates so much disagreement between smart people is that people make different implicit assumptions about the motivations. And that's the key to many other probability paradoxes as well.
- smohare 7y agoThis subtlety, to me, only comes into play when you cast it in a Bayesian context, because then you are forced to consider the probability that Monty reveals a goat. But the approach fundamentally breaks down because you can’t actually compute this, due to lack of any data. All it suggests to me is that this line of reasoning is insufficient, not that Monty’s motivations and knowledge actually matter.
- dukoolio 7y agoI’ve had some luck explaining it by taking advantage of the sensitivity to fairness that people seem to have. You are the contestant. You choose a door. Monty reveals a goat from another door. Now, instead of giving you a switching choice, Monty picks [insert person you find annoying] from the crowd. Monty gives them the option of choosing your door (duplicating any prize) or choosing the other door. The few times I’ve tried, people have seen it as unfair. But they don’t necessarily see the 2/3. They just see someone getting 1/2 vs their 1/3.
- DamnInteresting 7y agoI wrote about the Monte Hall problem back in 2005, and I made a simple simulator to demonstrate the success rate of door-switchers versus non-switchers: https://www.damninteresting.com/appendices/monte-hall-simulator/ https://www.damninteresting.com/appendices/monte-hall-simula...
- flashman 7y agoI have been reading your website since the heyday of Digg. Thanks for the link - I thought you must have stopped writing around 2010.
- lucas_membrane 7y agoHowever, the details of the problem matter more than you might think. On the original TV program, there were sometimes two contestants who picked doors for the grand prize. In such cases Monty would sometimes reveal a small prize behind the door selected by one contestant, and then offer the other contestant the option to switch. In that situation, the frequentist logic works out exactly the opposite, and the logic that says 'switch is better' when there is only one contestant says 'switch is worse.'
- beefield 7y agoMonty Hall Problem is to me the single most bewildering and depressing thing when I wonder how come mainstream/basic/MBA economics still keeps rational agency as its central tenent. I mean, if people can't sort this out, how on earth someone that is not either extremely stupid or extremely evil (or both) can seriously think that people can figure out their optimal health insurance even if the terms were not incomprehensible legalese designed to trick you?
- DonHopkins 7y agoNot to be confused with the Monty Haul Problem from D&D. https://dungeonsdragons.fandom.com/wiki/Monty_Haul https://dungeonsdragons.fandom.com/wiki/Monty_Haul
- simplesleeper 7y agoI wrote a post that explains this as well, including a playable game with cards (uses js) https://www.michalpaszkiewicz.co.uk/blog/montyhallexplained/index.html https://www.michalpaszkiewicz.co.uk/blog/montyhallexplained/...
- goto11 7y agoIt is really important to make it clear that: 1) Monty always opens another door (regardless of what is behind the door initially chosen by the contestant). He will never open the door the contestant chose. 2) Always opens a door with a goat behind. 3) These rules are known to the contestant These premises are not intuitive, and some versions of the problem does not make them explicit, which is why it is more of a "gotcha" riddle. In particular, the version quoted in Wikipedia does not make any of that clear. It just mentions the particular case of the host opening another the door with the goat. But without knowing the above "rules", this just means there is 50% chance per door now.
- WilliamEdward 7y agoThis ambiguity turned a really easy problem into one that people debated for decades for some reason
- cafard 7y agoThere was a Tcl for Windows distribution that came with the simulation. A co-worker ran it out to some number of tries, and was satisfied. By the way, you can get old Monty Hall shows on satellite TV to this day (or at least week)--a friend noticed it playing in a restaurant a couple of days ago.
- injb 7y agoEven though I think I understand the maths behind this, I'm one of those people who has never been able to fully accept it. When you pick a door, the odds are that the prize is behind one of the other 2 doors (2/3 chance). Therefore when all but one of the other doors are eliminated, the remaining one is twice as likely to have the prize as the one you picked. Is that it? I think my problem comes down to trying to reason about a past event using probability. I just can't get my head around that. Yes you can predict what is likely to happen if you repeat the experiment multiple times, but you're not being asked about that. You're being asked about this particular instance of the experiment. How can you know that you're better off switching this time?
- nkrisc 7y agoHere's how I finally understood it (I think): Switching is only bad if you pick correctly the first time. Since you only pick correctly 1/3 of the time, switching must be no worse or better 2/3 of the time. Since Monty will never open the door you picked nor the door with the prize, the remaining door after he opens one is the prize if you initially picked wrong, or a goat if you initially picked right. Since you initially pick right only 1/3 of the time, you switch.
- peteforde 7y agoThe one I have a hard time explaining to my friends is the Martian Potato Paradox. It's hard because I actually cannot wrap my head around the truth, even though I can explain why it's true. https://medium.com/i-math/matt-damon-s-martian-potatoes-1bcde7c6f77d https://medium.com/i-math/matt-damon-s-martian-potatoes-1bcd...
- zaro 7y agoI think this solution can be very counter intuitive or natural depending how you picture the problem in your head. If you are thinking about you playing the game then of course it is very counter intuitive, because you are playing the game and you have only one chance. Doesn't matter that by switching your chances rise to 2/3 because you play the game only once, and in the end you either win or lose. On the other hand if you think about it from a statistic/probabilistic point of view then it's all natural that the chance is 2/3 but there is the notable exception that you don't play in this case and you neither win or lose.
- timtas 7y agoI wrote a simulation that proves it pretty convincingly. http://rubyfiddle.com/riddles/7066f/4 http://rubyfiddle.com/riddles/7066f/4
- johnthescott 7y agochange the number of doors to 100 and the reasoning is obvious.