JaKiri Posted July 17, 2004 Posted July 17, 2004 That explanation makes more sense to me. But what about the the fact that after one door has been eliminated, the chances of choosing the right door is 50%? It isn't a fact... I'm not sure how (in mathematics) to prove something that isn't true. Forgive me.
aommaster Posted July 17, 2004 Posted July 17, 2004 Well, after they narrow it down to two doors, you have 50% chance of picking the right door. So, what use is changing the door?
Cap'n Refsmmat Posted July 17, 2004 Posted July 17, 2004 That's what I want to know. If you have the wrong door, you switch, then you get the right one. You have the right one, you switch, you lose.
aommaster Posted July 17, 2004 Posted July 17, 2004 After I get this straight, I may be able to understand why changing the door gets you a higher chance of winning!
JaKiri Posted July 17, 2004 Posted July 17, 2004 Well, after they narrow it down to two doors, you have 50% chance of picking the right door. So, what use is changing the door? You have a 50% chance of picking the right door if you have no other information. The main problem here is that you're interpreting it (which is the standard misinterpretation) as 'choosing a door'. You're not, you're choosing which probability subgroup you want to be a part of. That's what I want to know. If you have the wrong door, you switch, then you get the right one. You have the right one, you switch, you lose. Yep. You appear to have missed that there's twice the likelihood of choosing the wrong door in the first place, though.
Cap'n Refsmmat Posted July 17, 2004 Posted July 17, 2004 The main problem here is that you're interpreting it (which is the standard misinterpretation) as 'choosing a door'. You're not' date=' you're choosing which probability subgroup you want to be a part of.[/quote'] What? I'm not a mathematician Yep. You appear to have missed that there's twice the likelihood of choosing the wrong door in the first place, though. So? They eliminate the other one, so then it's down to 50/50.
aommaster Posted July 17, 2004 Posted July 17, 2004 ok. So, what you are trying to say is that, in order to get the right interpretation, we need to look at it in a probability tree-sort of way. We have to relate to the fact that one of the doors has been shown to be incorrect when we are left with two doors. Is that correct?
JaKiri Posted July 17, 2004 Posted July 17, 2004 So? They eliminate the other one, so then it's down to 50/50. Oh for the love of god. Lets go through this again: Please tell me the probability of NOT picking the winning door in the first choice.
JaKiri Posted July 17, 2004 Posted July 17, 2004 ok. So, what you are trying to say is that, in order to get the right interpretation, we need to look at it in a probability tree-sort of way. We have to relate to the fact that one of the doors has been shown to be incorrect when we are left with two doors. Is that correct? Yep, that's the basic idea. Draw out the probability tree of each possible choice of door, assuming a given one is the winning door (rather than doing the tree for all possibilities of winning door; it just overcomplicates things, as the doors are all identical for all intents and purposes, and thus interchangeable).
aommaster Posted July 17, 2004 Posted July 17, 2004 I think that the set of replies that go with this one: Oh for the love of god. Lets go through this again: Please tell me the probability of NOT picking the winning door in the first choice. would give me a better insight on how to work that out. I have to go, so, deal with it with Cap'n and I'll be back to read the threads. I think I am starting to understand it now!
JaKiri Posted July 17, 2004 Posted July 17, 2004 2/3 Even though you're not him, yep. Now, If you have the wrong door, you switch, then you get the right one. combine these two facts! Probability of getting wrong door = 2/3 Probability of winning if you change with the wrong door = 1 Probability of winning if you change with the right door = 0, probability of picking right door = 1/3 Chance of winning if you change = chance of choosing wrong door * chance of winning with wrong door + chance of choosing right door * chance of winning with right door = 2/3 * 1 + 1/3 * 0 = 2/3 This has been demonstrated SO MANY WAYS. It's TRUE.
Cap'n Refsmmat Posted July 17, 2004 Posted July 17, 2004 Let's assume the winning door is 2. You choose 1. They reveal 3. You switch to 2, you win. Same but you don't switch. You lose. You choose 2. They reveal 3. You switch to 1, you lose. Same but you don't switch. You win. You choose 3. They reveal 1. You switch to 2, you win. Same but you don't switch. You lose. Am I missing anything? That's even. oops, missed this page.
Cap'n Refsmmat Posted July 17, 2004 Posted July 17, 2004 = 2/3 * 1 + 1/3 * 0 = 2/3 This has been demonstrated SO MANY WAYS. It's TRUE. Something * 0 = something other than 0?
Sayonara Posted July 17, 2004 Posted July 17, 2004 Something * 0 = something other than 0? (2/3 * 1) + (1/3 * 0)
Cap'n Refsmmat Posted July 17, 2004 Posted July 17, 2004 Oh. He forgot the parentheses. I kinda get it now, but only kinda.
Dave Posted July 17, 2004 Posted July 17, 2004 He didn't forget anything. Basic order of addition/subtraction/etc that I learnt when I was back in primary school.
Hailstorm Posted July 18, 2004 Posted July 18, 2004 Visual, four trials of each to make it easy: ||_| = Door one = Car = Trial one selection |_| = Door two = Goat = Trial two selection |_| Door three = Goat = Trial three selection |_| = Goat released In trial one: The door with the car behind it - First set of four lines In trial two: The door with a goat behind it - Second set four lines In trial three: The door with a goat behind it - Third set four lines Please note: your door selection is bolded ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins Hence: Switch wins 8/12 (2/3) and Stay wins 4/12 (1/3) The logical flaw seems to come in with the first selection. I felt for some reason that there should be four of each goat door opened (instead of two of each), but that would have left me choosing the first car door twice as often. My original question of whether or not he can remove a goat from the door you chose would only further bolster the viability of switching (staying on a door with a goat in a game-show setting would be simply stupid, whereas when scripting a program, it seems more reasonable). Viewing it as a 50/50 chance would require: ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Stay wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins ||_| |_| |_| | Switch wins But when looking at it, you'd have to choose the door with the car in it (in this case door #1) far more often than random chance allows (100% more likely). In order for a true 50/50 chance, you would have to be able to see the car... or perhaps smell the goats with fairly good accuracy on the first selection only, but not the second. Perhaps the release of the first goat would skew goat-sense.
aommaster Posted July 18, 2004 Posted July 18, 2004 I love Jakiri's explanation, it really makes sense to me. Thanx alot! Now I understand it. Also, thanx to hailstorm, who really put it in detail! Thanx!
Dave Posted July 18, 2004 Posted July 18, 2004 If it makes anyone feel better, this problem (the Monty Hall problem) was debated over and over again (and still is) for about 10 years before it was generally accepted.
Cap'n Refsmmat Posted July 18, 2004 Posted July 18, 2004 I think I get it now. You're more likely to get a losing door in the first place, so if you switch, you're more likely to win. Right?
aommaster Posted July 18, 2004 Posted July 18, 2004 yup! You got it. Or at least, that's the way I understand it!
Hailstorm Posted July 19, 2004 Posted July 19, 2004 Another easy way to look at it is as follows: There are infinity +2 doors, there is one car behind one door, and one goat behind each additional door. If you choose any number, and then the infinity goats are removed, you're clearly better switching. Your chance of getting a car on your first guess is zero (since this is the math forum it is [math] 1/ (infinity+2)[/Math). Whereas the chance of you getting a car should you switch your choice is [math] (Infinity +1)/(Infinity+2).
Guest Anenkefali Posted September 18, 2004 Posted September 18, 2004 The key to this problem is that the man who opens the door for you knows exactly where the car is. He will NEVER open the door that contains the car, so the probability for that is exactly 0. If we, for a second, ignore this fact there can only be 6 possible outcomes. They are as follows. Good door = door with car Bad doors = doors with goats [A] You pick the good door. The host opens bad door 1. You pick the good door. The host opens bad door 2. [C] You pick bad door 1. The host opens the good door. [D] You pick bad door 1. The host opens bad door 2. [E] You pick bad door 2. The host opens the good door. [F] You pick bad door 2. The host opens bad door 1. Now, if we include the fact that the host will NEVER open the good door, we see that the event "The host opens the good door" will have probability 0. This gives us: P(A) = 1/3 * 1/2 P(B) = 1/3 * 1/2 P© = 1/3 * 0 P(D) = 1/3 * 1 P(E) = 1/3 * 0 P(F) = 1/3 * 1 P(A) + ... + P(F) = 1 The situations in which we would have won if we choosed not to switch doors are A and B. P(A) + P(B) = 2/6 = 1/3 The situations in which we would have won if we switched doors are C, D, E and F. P© + ... + P(F) = 4/6 = 2/3
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