Dice game with a player advantage

Discussion in 'Probability and Statistics' started by Jonathan, Dec 13, 2021.

  1. Jonathan

    Jonathan

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    Alice and Bob throw 2 dice each, but Alice gets a +1 bonus to her roll.
    What is the probability Alice got a higher total than Bob?

    I've done a little bit of this on a spreadsheet, but don't have the answer. And maybe there's a better approach.
     
    Jonathan, Dec 13, 2021
    #1
  2. Jonathan

    Country Boy

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    With a pair of standard 6 sided dice the probability of rolling a total of
    2, (1, 1), is 1/36
    3, (1, 2) or (2, 1), is 2/36= 1/18
    4, (1, 3) or (2, 2) or (3, 1), is 3/36= 1/12
    5, (1, 4) or (2, 3) or (3, 2) or (4, 1), is 4/36= 1/9
    6, (1, 5) or (2, 4) or (3, 3) or (4, 2) or (5, 1), is 5/36
    7, (1, 6) or (2, 5) or (3, 4) or (4, 3) or (5, 2) or (6, 1), is 6/36= 1/6
    8, (2, 6) or (3, 5) or (4, 4) or (5, 3) or (6, 2), is 5/36
    9, (3, 6) or (4, 5) or (5, 4) or (3, 6), is 4/36= 1/9
    10, (4, 6) or (5, 5) or (6, 4), is 3/36= 1/12
    11, (5, 6) or (6, 5), is 2/36= 1/18
    12, (6, 6), is 1/36.

    So, including the bonus, the probability of Alice getting a 3= 2+ 1 is 1/36. To win, Bob must get 4 or higher, probability 3/36+ 4/36+ 5/36+ 6/36+ 5/36+ 4/36+ 3/36+ 2/36+ 1/36= 33/36= 11/12.
    (That could have been done quicker as 1- 1/36- 2/36= 1- 3/36.)
    The probability of this happening is (1/36)(11/12)= 11/432.

    The probability of Alice getting a total of 4= 3+ 1 is
    2/36= 1/18. To beat that Bob must get a 5 or better. The probability of that is 4/36+ 5/36+ 6/36+ 5/36+ 4/36+ 3/36+ 2/36+ 1/36= 30/36= 5/6.
    The probability of this happening is (1/18)(5/6)= 5/108.

    The probability of Alice getting a total of 5= 4+ 1 is 1/12. To beat that Bob must get 6 or higher. The probability of that is 5/36+ 6/36+ 5/36+ 4/36+ 3/36+ 2/36+ 1/36= 26/36= 13/18.
    The probability of this happening is (1/12)(13/18)= 13/216.

    The probability of Alice getting a total of 6= 5+ 1 is 1/9. In order to beat this Bob would have to get 7 or better. The probability of that is 6/36+ 5/36+ 4/36+ 3/36+ 2/36+ 1/36= 21/36= 7/12.
    The probability of this happening is (1/9)(7/12)= 7/108.

    The probability of Alice getting a total of 7= 6+ 1 is 5/36. In order to beat that Bob must get 8 or better. The probability of that is 5/36+ 4/36+ 3/36+ 2/36+ 1/36= 15/36= 5/12.
    The probability of this happening is (5/36)(5/12)= 25/432.

    The probability of Alice getting 8= 7+ 1 is 5/36. To beat that Bob must get 9 or better. The probability of that is 4/36+ 3/36+ 2/36+ 1/36= 10/36= 5/18.
    The probability of this happening is (5/36)(5/18)= 25/648.

    The probability of Alice getting 9= 8+ 1 is 5/36. To beat that Bob must get 10 or more. The probability of that is 3/36+ 2/36+ 1/36= 6/36= 1/6.
    The probability of this happening is (5/36)(1/6)= 5/216.

    The probability of Alice getting a total of 10= 9+ 1 is 1/9. To beat that Bob must get 11 or 12. The probability of that is 2/36+ 1/36= 3/36= 1/12.
    The probability of this happening is (1/9)(1/12)= 1/108.

    The probability of Alice getting a total of 11= 10+ 1 is 1/12. To beat that Bob would have to roll a 12 which has probability 1/36.
    The probability of this happening is (1/12)(1/36)= 1/432.

    If Alice rolls 11 or 12, she will have a total of 12 and 13 with her bonus and Bob cannot beat those.

    The overall probability that Bob beats Alice is 11/432+ 5/108+ 13/216+ 7/108+ 25/432+ 25/648+ 5/216+ 1/108+ 1/432,

    648= 2(324)= 2(2)(162)= 2(2)(2)(81)= 2^3(3^4)
    216= 2(108)= 2(2)(54)= 2(2)(2)(27)= 2^3(3^3)
    432= 2(216)= 2(2)(108)= 2(2)(2)(54)= 2^4(27)= 2^4(3^3)
    108= 2(54)= 2(2)(27)= 2^2(3^3)

    The least common denominator is 2^4(3^4)= 16(81)= 1296.

    11/432+ 5/108+ 13/216+ 7/108
    +25/432+ 25/648+ 5/216+ 1/108+ 1/432= 33/1296+ 156/1296+ 78/1296+ 42/1296+ 75/1296+ 50/1296+ 30/1296+ 12/1296+ 3/1296= 479/1296. That is about 37%.
     
    Country Boy, Dec 26, 2021
    #2
    nycmathguy likes this.
  3. Jonathan

    nycmathguy

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    I'm very impressed with your reply. A job well-done!
    I will need your help in the future should I decide to review basic probability.
     
    nycmathguy, Dec 26, 2021
    #3
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