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Math Help - Complex Probability Problem (Bayes' Theorom?) 6 doors, 2 guesses, 1 hint

  1. #1
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    Complex Probability Problem (Bayes' Theorom?) 6 doors, 2 guesses, 1 hint

    Here's the situation. You have 6 six doors.

    A B C D E F

    Behind one of them is a prize.

    You get to choose a door.

    If it has the prize, you are told so and win immediately.

    If not, you are told whether or not the prize is in an adjacent door. (note: if you pick door A, and they say its adjacent, this means it HAS to be door B)

    You then get a second guess.

    What are the chances of winning the game if you pick door A or F? What about picking B, C, D, or E?
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  2. #2
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    Quote Originally Posted by shenanigans87 View Post
    Here's the situation. You have 6 six doors.

    A B C D E F

    Behind one of them is a prize.

    You get to choose a door.

    If it has the prize, you are told so and win immediately.

    If not, you are told whether or not the prize is in an adjacent door. (note: if you pick door A, and they say its adjacent, this means it HAS to be door B)

    You then get a second guess.

    What are the chances of winning the game if you pick door A or F? What about picking B, C, D, or E?
    If you pick door A your chance of winning is 1/3 since you win if it was in fact behind doors A or B, same reasoning applies to door F.

    CB
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  3. #3
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    Close... but that is incorrect. No matter what door you pick, you have a 50% chance of winning, assuming you use common sense regarding the adjacent hints. For example... pretend your strategy is as follows: you always choose door A. If wrong and adjacent you know its B. If not adjacent you always guess C. Thus, if the prize was behind doors A B or C you will be guaranteed a win. And there's a 3/6 or 50% chance its behind doors A B or C.
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  4. #4
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    Quote Originally Posted by shenanigans87 View Post
    Close... but that is incorrect. No matter what door you pick, you have a 50% chance of winning, assuming you use common sense regarding the adjacent hints. For example... pretend your strategy is as follows: you always choose door A. If wrong and adjacent you know its B. If not adjacent you always guess C. Thus, if the prize was behind doors A B or C you will be guaranteed a win. And there's a 3/6 or 50% chance its behind doors A B or C.
    If you choose door A there is no strategy involved, you win if the prize is behind door A or door B, which is 1/3.

    The same if you pick door F

    The above is what the first part of the question asks for.

    CB
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  5. #5
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    First of all, this theory has already been proven both by many stats majors and thinkers alike. Once you understand it, it's really not that complicated of a problem. Secondly, I made a computer simulation of this scenario, and tested it for both method A and method B 500,000 times. At any given moment from 1,000 trials to 500,000 trials, the ratio was within 0.01% of 50/50

    method A: Always choose door A, if wrong, and the prize is adjacent, choose door B. If it's not adjacent, choose door C.

    method B: Always choose door B, if wrong, and the prize is adjacent, choose door C. If it's not adjacent, choose door D.

    Let me break it down for you.

    Method A: You have a 1/6 chance of winning right off the bat. However, if you don't, there is a 1/5 chance that the prize is next to you, and a 4/5 chance that the prize is far from you, in which case you have a 1/4 chance of guessing correctly. =)

    So we take

    (1/6) + (5/6)(1/5)(1/1) + (5/6)(4/5)(1/4)
    =
    (1/6) + (1/6) + (1/6) = (3/6) = (1/2)

    Voila. Now let's analyze method B.

    Method B: You have a 1/6 chance of winning from the start. If you are wrong, and the prize is adjacent, which will happen 2/5 of the time, you have a 1/2 chance of guessing right. If the prize is far, which will happen 3/5 of the time, you have a 1/3 chance of guessing right.

    (1/6) + (5/6)(2/5)(1/2) + (5/6)(3/5)(1/3)
    =
    (1/6) + (1/6) + (1/6) = (3/6) = (1/2)

    There you have it. I was initially very frustrated with this problem as well, but once I figured it out and was then able to prove the theory by running a relatively infinite amount of trials, it made sense.

    Just in case you were curious.. here is the code for the simulation.

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  6. #6
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    Quote Originally Posted by shenanigans87 View Post
    First of all, this theory has already been proven both by many stats majors and thinkers alike. Once you understand it, it's really not that complicated of a problem.
    Oppsss, I missed out a term

    CB
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