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> The probabilities change, even when a door you didn't pick [and doesn't hold the prize] is opened. That's the common misunderstanding of the problem. Most p
by adekok 11y ago
> The probabilities change, even when a door you didn't pick [and doesn't hold the prize] is opened.
That's the common misunderstanding of the problem. Most people think that the probabilities go fro 1/3, 1/3, 1/3 to 1/2, 1/2, after choosing a door and having Monty Hall open one of the others. The probabilities don't change.
The probabilities are 1/3, 1/3, 1/3 at the start. After you choose a door, they're still 1/3, 1/3, 1/3. Or a put a different way: 1/3 (your choice) 2/3 (what you didn't choose). When Monty Hall opens one of the other doors, the probabilities don't change. It's still 1/3 (your choice) 2/3 (what you didn't choose).
However, because he's removed one of the doors from the problem, the 2/3 probability is now applied to the remaining door.
That can be viewed as "probabilities changing" for the remaining door, but it's better described as the probabilities being shared between the doors you didn't choose.
- Retric 11y agoThis assumes you always open a door, that door never holds the prize, and there is exactly 1 prize. You could run (ed:a similar game with different rules) such that the odds go 1/3,1/3,1/3 to 1/2, 1/2 if you flipped a coin to chose the second door and sometimes show the prize. Alternatively, if you chose when to open the second door, you could make swapping a very good (100%) or very poor choice (0%). Worse, you could swap what's behind the doors after they chose. Which IMO is why people find this so confusing. In the real world the odds presented may or may not line up with the actual odds. aka unknown unknowns.
- ghshephard 11y agoThis is incorrect. Flipping a coin to "to chose the second door and sometimes show the prize." has no impact on the probabilities. The parent is correct - there is a 2/3 chance the prize exists under one of the other two doors, and if one of them is shown not to have the prize (through random flipping, deliberate selection, whatever) - then the final door now has a 2/3 chance to have the prize.
- function_seven 11y agoIf one of the doors is shown to not have the prize, the odds are now 1/2 when the choice of reveal door is made randomly. That's because 1/3 of the time, you're not even offered the chance to switch, since it's pointless. I'll run the trials here. Assume the first two doors have the goat, and the third door has the car. Pick Reveal Switch? Outcome ---- ------ ------- ------- 1 2 Y 3 Car 2 1 Y 3 Car 3 1 N 3 Car 3 2 N 3 Car 1 2 N 1 Goat 2 1 N 2 Goat 3 1 Y 2 Goat 3 2 Y 1 Goat 1 3 - No chance to switch 1 3 - No chance to switch 2 3 - No chance to switch 2 3 - No chance to switch Meaning, given that you've made it to the point that an offer to switch doors is presented, there is no advantage to do so (or not do so).
- carleverett 11y agoBut in the actual game, you are always given the chance to switch. Here are the possible outcomes of the game, with the third column being what the contestant should do to win: You Pick Car Switch? ------- ----- ------ 1 1 N 2 1 Y 3 1 Y 1 2 Y 2 2 N 3 2 Y 1 3 Y 2 3 Y 3 3 N Without knowing anything else, switching means that you win 2/3 of the time.
- function_seven 11y agoYour comment (EDIT: Sorry, not yours, but parent) was talking about a random flip of coin determining which door Monty shows. That's not the actual game. If Monty is randomly showing a door without regard to goat/car, then you have to also consider the scenarios where he reveals a car and you lose early.
- ghshephard 11y agoIf, while randomly flipping a coin, he reveals the car then you have a 0% chance of winning. If he doesn't reveal a car then you have a 2/3 chance of winning if you switch doors.
- richmarr 11y ago>> The probabilities change > That's the common misunderstanding of the problem... You've got a choice here... you can either interpret "The probabilities change" as meaning the probabilities of a given door being correct at the start (which would be a misunderstanding) or you could interpret it from the pragmatic angle; the probability of success if you switch doors after one is taken away. Given the context I'm inclined to believe that wrsh07 intended the latter.
- tyleraland 11y agoI only started to understand how to look at this problem once I realized the host was going to eliminate a bad door _no matter what was chosen_. Of course if you have already selected a bad door he won't eliminate it. Another useful exercise is to stretch the number of doors to some large N and suppose the choice was posed repeatedly with a decreasing number of doors. If you can have anywhere from 2 to N guesses and as N decreases the number of bad doors decreases, it seems more obvious that you should always switch.
- ustolemyname 11y agoNot switching until the end should work out better than switching multiple times. 4 doors, in 8ths to simplify fractions. 2/8,2/8,2/8,2/8 => initial conditions, choose leftmost 2/8,3/8,3/8 => random bad door removed, choose a random 3/8 3/8,5/8 => remove another door. Next choice is only 5/8, whereas if we had waited it would have been 6/8.
- zodPod 11y agoThis is a great clarification of how this problem works. I STILL can't grasp why you'd switch. Since you have no idea which door it is, couldn't the 2/3 probability be applied to either his door or your door? For all you know, the one that he removed was just one random one of the goat doors. Your chance of picking the car was 1/3 before, if you could have the car already, why would it be better to switch now that he removed which ever would definitely be a goat? It seems like you still have a 50/50 chance of getting the car and your chance of the one that remains (unpicked/unremoved) is just as likely to be the other goat as it is to be the car.. That said, I made a simulation of it where it does the following: 1. Assign random locations behind doors, 2. Player randomly chooses door, 3. Host excludes door (random if both are goats, goat if one is a car), 4. Player either switches, or doesn't switch based on what's chosen for that simulation (the full set gets one or the other) After running 50,000,000 then dividing success/tries, my number is nearly exactly the same between the two. Is my simulation working incorrectly?
- wrsh07 11y agoI find it clearer if you have 100 doors. Pick one. The host eliminates 98 doors [with no prize behind them]. Do you switch? Or do you stick with your original guess? As the parent mentioned, switching gives you a 99/100 chance of being correct, staying gives you a mere 1/100 chance.
- ricardobeat 11y agoI don't see why the other door has 99x more probability of being correct. It's still 50/50, original door or remaining door, the other 98 don't change the odds.
- pklausler 11y agoYour initial pick has a 1/100 chance of being right. If you switch, and you were right, you lose. On the other hand, your initial pick has a 99/100 chance of being wrong, and if you were wrong, and you switch, you win. So switch.
- wrsh07 11y agoYou are, of course, correct. Sorry about the careless wording [I think the analogy is still interesting, though!]
- pklausler 11y agoThis is good reasoning. Unlike other probability puzzles, the new information isn't useful in this one. P(my first choice was right) = P(my first choice was right | given that I've been shown a goat in another door) It's my second-favorite probability puzzle; my favorite is the one in which the new information has no relevance at all but still changes the probabilities, the "not both daughters" problem.
- pklausler 11y agoI should have just given the other problem, sorry for the laziness. Your friend tells you that she has two children. They're not both boys, she adds. What are the odds that they're both girls? (Obviously 1/3). When she tells you that one child is named Mary, the odds that they're both girls change (to 1/2)! Even though you knew that at least one of the children was a girl, and the name might as well be arbitrary, and the new information seems useless... the odds have changed. It's nontrivial to work this one out; exercise for the reader.
- logfromblammo 11y agoAfter first datum (there are two children): Presuming A and B to be independent events (first child is a girl, second child is a girl) P(A) = 1 - P(!A) = 1/2 P(B) = 1 - P(!B) = 1/2 P(A and B) = P(B|A) * P(A) = 1/2 * 1/2 = 1/4 P(A and !B) = P(!B|A) * P(A) = 1/2 * 1/2 = 1/4 P(!A and B) = P(B|!A) * P(!A) = 1/2 * 1/2 = 1/4 P(!A and !B) = P(!B|!A) * P(!A) = 1/2 * 1/2 = 1/4 Good so far. But the probability later on depends on how you find out that at least one of your friend's children is a girl. If your friend tells you a random fact about her whole family, the probability is 1/3. If she tells you a fact about a randomly chosen child, the probability is 1/2. You can presume from "they're not both boys" that the statement is equivalent to "there is at least one girl". When she tells you the name, you now know "a specific child of mine is a girl". "Not both boys" gives you the precondition "!(!A & !B)" which equates to "A or B". P(A or B) = P(A and B) + P(A and !B) + P(!A and B) = 3/4 P(A and B|A or B) = P(A or B|A and B) * P(A and B) / P(A or B) = 1 * 1/4 / 3/4 = 1/3 "Mary is a girl" gives you either the precondition "A" or the precondition "B" P(A and B|A) = P(A|A and B) * P(A and B) / P(A) = 1 * 1/4 / 1/2 = 1/2 P(A and B|B) = P(B|A and B) * P(A and B) / P(B) = 1 * 1/4 / 1/2 = 1/2 These are both the same number, so that works out fine.