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posted by janrinok on Friday July 29 2016, @10:27AM   Printer-friendly
from the something-to-think-about dept.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. 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. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Do any of you have any noteworthy experiences where knowledge of math helped you in an unusual way?

https://en.wikipedia.org/wiki/Monty_Hall_problem


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  • (Score: 2) by TheRaven on Friday July 29 2016, @05:54PM

    by TheRaven (270) on Friday July 29 2016, @05:54PM (#381632) Journal

    But picking a door to begin with is just showmanship; it doesn't affect the odds at all. If you didn't pick a door at all at the beginning and Monty randomly chooses one of the two goat doors, then you choose a door, you'd still have the same odds.

    No, because Monty's choice is different. If you didn't pick a door to begin with then you'd have a 1/2 probability of getting the correct answer, not 2/3. Monty would close one of the doors and you'd be in a trivial situation of having to pick between two doors with no knowledge. By picking one of the doors to begin with, you reduce the number of possible moves for Monty. If you (2/3 probability) pick one of the doors that doesn't have a car behind it, then Monty must close the only other car that doesn't have a car behind it. The other door is then the one that has the car. If you picked correctly the first time (1/3 probability), then the Monty can close either door.

    This is a bit easier to explain with a little bit of graph theory and game theory (two of the core bits of computer science, so hopefully not too problematic for people here), but difficult without the ability to draw the relevant pictures. The key point is that if you picked incorrectly to start with then Monty will close the door that doesn't have a car behind it and you then have a 100% probability of getting the car if you switch. For the three doors, A, B, and C, with the car behind C, if you picked A or B then he will close the other, if you pick C then he will close either A or B. The set of doors that are not closed and are not the ones that you picked then becomes:

    • If you picked A, then C.
    • If you picked B, then C.
    • If you picked C, then A or B (which are equivalent, as neither contains a car).

    In contrast, the set of doors that are the one that you picked the first time are:

    • If you picked A, then A.
    • If you picked B, then B.
    • If you picked C, then C.

    Picking one from the first set gives you a 2/3 probability of choosing the correct one, picking one from the second set gives you a 2/3 probability of picking the correct one. Or, to put it more simply, if you picked correctly the first time then you win by staying but if you picked incorrectly the first time then you win by switching. You have a 1/3 chance of picking correctly the first time and a 2/3 chance of picking incorrectly, therefore it is better to bet that your first pick was incorrect.

    If this is still confusing, I present a simple table of all of the possible paths and outcomes:

    Pick | Switch | Stick
    -----+--------+-------
    A    | Win    | Lose
    B    | Win    | Lose
    C    | Lose   | Win

    From this is should be obvious that there is a 2/3 chance of winning if you switch and a 1/3 chance of winning if you stick.

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