AN. More Sucker Bets

April 9, 2006 · The Mathcast, errata, math puzzles · Permalink

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Is there always a better choice?

 
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2 Comments »

  1. jeffisom said,

    July 23, 2007 at 11:46 am

    I discovered your podcast on iTunes last week and have really enjoyed going through the past episodes and catching up. I do have a question about this one.

    The 3 piles that you gave were as follows:
    1: A 6 8
    2: 3 5 7
    3: 2 4 9

    So when I pick from the first pile and you pick from the second pile here are the possible outcomes:
    A 3 - You Win
    A 5 - You Win
    A 7 - You Win
    6 3
    6 5
    6 7 - You Win
    8 3
    8 5
    8 7

    So you win only 4 out of 9 times

    When I pick from the second pile and you pick from the third pile here are the possible outcomes:
    3 2
    3 4 - You Win
    3 9 - You Win
    5 2
    5 4
    5 9 - You Win
    7 2
    7 4
    7 9 - You Win

    So you win only 4 out of 9 times again

    When I pick from the third pile and you pick from the first pile here are the possible outcomes:
    2 A
    2 6 - You Win
    2 8 - You Win
    4 A
    4 6 - You Win
    4 8 - You Win
    9 A
    9 6
    9 8

    So you win only 4 out of 9 times again

    So it seems that the piles should have been in the opposite order
    1: 2 4 9
    2: 3 5 7
    3: A 6 8

    Am I missing something?

    Thanks!

  2. strauss said,

    July 27, 2007 at 9:42 am

    You are correct! You, know, it ain’t always easy to keep things straight when we’re taping. I can only hope that the fact we’re not always completely on the ball is part of our charm.

    As another listener wrote:

    Pile x beats Pile ((x +1) mod 3) + 1, five out of nine times.

    So the second player should take from the pile before, below, or the the left of the first player.

    Didn’t you say it the other way around?

    Thanks for the show. I wouldn’t have memorized the trick if you had gotten it exactly right.

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