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Math Help - dice game

  1. #1
    Super Member Random Variable's Avatar
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    dice game

    Players A and B roll a pair of dice in turn, with player A rolling first. A's objective is to obtain a sum of 6 (on one roll), while B's ojective is to obtain a sum of 7. The winner is the first player to reach his or her objective. What's the probability that player A wins?

    Does my approach make any sense? I'm probably making this way too hard.


    I let A_{W} be the event that A wins, and A_{i} be the event that player A rolls a sum of i (i=2,3,...,12) on his first roll.


    so  P(A_{W}) = P(A_{W}|A_{2})P(A_{2})+ P(A_{W}|A_{3})P(A_{3}) + P(A_{W}|A_{4})P(A_{4})  + P(A_{W}|A_{5})P(A_{5}) + P(A_{W}|A_{6})P(A_{6}) + P(A_{W}|A_{7})P(A_{7}) + P(A_{W}|A_{8})P(A_{8})  + P(A_{W}|A_{9})P(A_{9}) + P(A_{W}|A_{10})P(A_{10}) + P(A_{W}|A_{11})P(A_{11}) + P(A_{W}|A_{12})P(A_{12})

    when i \not= 6, P(A_{W}|A_{i}) = \frac {30}{36}P(A_{W}) , which is the probability that player B does not win on his first roll times the probability that A wins (assuming independence of course)

    then  P(A_{W}) = \frac {30}{36}P(A_{W})\frac{1}{36} + \frac {30}{36}P(A_{W})\frac{2}{36} + \frac {30}{36}P(A_{W})\frac{3}{36} + \frac {30}{36}P(A_{W})\frac{4}{36} + \frac{5}{36} + \frac {30}{36}P(A_{W})\frac{6}{36} + \frac {30}{36}P(A_{W})\frac{5}{36}  + \frac {30}{36}P(A_{W})\frac{4}{36} + \frac {30}{36}P(A_{W})\frac{3}{36} + \frac {30}{36}P(A_{W})\frac{2}{36} + \frac {30}{36}P(A_{W})\frac{1}{36}

    and then solve for P(A_{W})

    I get 30/61, or about 49%
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  2. #2
    MHF Contributor matheagle's Avatar
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    I would add the probability of winning on his/her first attempt plus his/her second....

    To win on the first attempt would be  {5\over 36} .

    To win on the second attempt would be  \biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr) \biggl( {5\over 36}\biggr) .

    The third would be  \biggl( {31\over 36}\biggr)^2\biggl( {5\over 6}\biggr)^2 \biggl( {5\over 36}\biggr) .

    Sum this via a geometric series should produce the answer.

     {5\over 36}\sum_{k=0}^{\infty}\biggl( {31\over 36}\biggr)^k\biggl( {5\over 6}\biggr)^k={5\over 36}\sum_{k=0}^{\infty}\Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^k.

    I just did the math and I got 31/60 too.

    {5/36\over 1- (31/36)(5/6)}=30/61
    Last edited by matheagle; May 30th 2009 at 08:42 PM.
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  3. #3
    Super Member Random Variable's Avatar
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    Yeah, I made the problem way, way too complicated. I was conditioning on the outcome of the first roll.
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  4. #4
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    Hello, Random Variable!

    You got the right answer . . . Great work!
    There are, as you suspected, other approaches.


    Players A and B roll a pair of dice in turn, with player A rolling first.
    A's objective is to obtain a sum of 6, while B's ojective is to obtain a sum of 7.
    The winner is the first player to reach his or her objective.
    What's the probability that player A wins?
    We have: . \begin{array}{ccc}P(\text{sum 6}) \:=\: \frac{5}{36} & & P(\text{sum 7}) \:=\:\frac{1}{6} \\ \\[-4mm]<br />
P(\text{not 6}) \:=\: \frac{31}{36} & & P(\text{not 7}) \:=\:\frac{5}{6} \end{array}

    Consider the ways that A can win . . .


    A wins on his first roll.
    P(\text{A, 1st roll}) \:=\:\tfrac{5}{36}

    A wins on his second roll.
    . . A does not get 6 on his first roll: . \tfrac{31}{36}
    . . B does not get 7 on his first roll: . \tfrac{5}{6}
    . . A gets 6 on his second roll: . \tfrac{5}{36}
    P(\text{A, 2nd roll}) \:=\:(\tfrac{31}{36})(\tfrac{5}{6})\tfrac{5}{36} \;=\;(\tfrac{155}{216})\tfrac{5}{36}

    A wins on his third roll.
    . . A does not get 6 on his first roll: . \tfrac{31}{36}
    . . B does not get 7 on his first roll: . \tfrac{5}{6}
    . . A does not get 6 on his second roll: . \tfrac{31}{36}
    . . B does not et 7 on his second roll: . \tfrac{5}{6}
    . . A get 6 on his third roll: . \tfrac{5}{36}
    P(\text{A, 3rd roll}) \:=\:(\tfrac{31}{36})(\tfrac{5}{6})(\tfrac{31}{36}  )(\tfrac{5}{6})\tfrac{5}{36} \:=\:(\tfrac{155}{216})^2\tfrac{5}{36}


    And we see the pattern . . .


    P(\text{A wins}) \;=\;\tfrac{5}{36} + (\tfrac{155}{216})\tfrac{5}{36} + (\tfrac{155}{216})^2\tfrac{5}{36} + (\tfrac{155}{216})^3\tfrac{5}{36} + \hdots

    . . . . . . . = \;\tfrac{5}{36}\underbrace{\left[1 + \tfrac{155}{216} + (\tfrac{155}{216})^2 + (\tfrac{155}{216})^3 + \hdots \right]}_{\text{geometric series}}

    The geometric series has: . a = 1,\;r = \tfrac{155}{216}
    . . Its sum is: . \frac{1}{1-\frac{155}{216}} \:=\:\frac{1}{\frac{61}{216}} \:=\:\tfrac{216}{61}


    Therefore: . P(\text{A wins}) \;=\;\frac{5}{36}\cdot\frac{216}{61} \;=\;\frac{30}{61}


    Edit: I was almost ready to post this long long disseration
    . . . and I was called away for supper.
    When I returned, I posted it without checking for earlier replies . . . *blush*

    Nice work, matheagle!
    .
    Last edited by Soroban; May 30th 2009 at 08:48 PM.
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  5. #5
    Super Member Random Variable's Avatar
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    The second part of the problem asks for the expected number of rolls until there is a winner. To answer that you would probably have to use conditioning. Otherwise you would have an infinite series without an obvious sum (I think).
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  6. #6
    MHF Contributor matheagle's Avatar
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    Sorobanned, I gave you over an hour to solve this.
    Seeing your past work I was waiting for you to do this.
    I had dinner, watched a movie and still was surprised I didn't see your geometric sum.
    So, I figured I had to answer this.

    P(X=k)= \biggl( {5\over 36}\biggr) \Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^{k-1}  for k=1,2,...

    So EX={5\over 36}\sum_{k=1}^{\infty}k\Biggl(\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)\Biggr)^{k-1}

    The trick here is to notice the derivative of x^k and after differentiating

    letting x=\biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr)
    Last edited by matheagle; May 30th 2009 at 09:44 PM.
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  7. #7
    Super Member Random Variable's Avatar
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    But  P(X=2) = (\frac {31}{36}) (\frac{6}{36}) , no?
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  8. #8
    Super Member Random Variable's Avatar
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    This is what I came up with in the past 30 minutes:

    E(X) = E[X|A wins on his first roll]P(Awins on his first roll) + E[X|B wins on his first roll]P(B wins on his first roll) + E[X|neither A or B win on their first roll]P(neither A or B win on their first roll)

     E[X] = 1* \frac{5}{36} + 2* (\frac {31}{36}) (\frac {1}{6}) + (2+E[X])(\frac {31}{36}) (\frac {5}{6})

    solving for E(X) I get 402/61 or about 6.6
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Random Variable View Post
    But  P(X=2) = (\frac {31}{36}) (\frac{6}{36}) , no?

    Nope, I worked it out above, you're missing that B must fail once too.

    For A to win on the second attempt that would be  \biggl( {31\over 36}\biggr)\biggl( {5\over 6}\biggr) \biggl( {5\over 36}\biggr) .

    That's A fails on first try AND B FAILS ON THE FIRST TRY and then A wins on second try.
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  10. #10
    Super Member Random Variable's Avatar
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    The second question asks for the expected number of rolls until there is a winner. So P(X=2) is the probability B wins on his first roll.
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  11. #11
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Random Variable View Post
    The second question asks for the expected number of rolls until there is a winner. So P(X=2) is the probability B wins on his first roll.
    My rv X was the trial on which A won.
    So I didn't answer your question.

    LET W be the trial in which there is a winner.
    For W to be 1 we need either {A wins immediately} OR {A losses and B wins}

    So P(W=1)=(5/36)+(31/36)(1/6).

    and the word 'rolls' is confusing, there are 2 rolls per play.
    Actually 4, 2 by each player.
    So I figure you mean how many times they play this game.
    Last edited by matheagle; May 31st 2009 at 07:07 AM.
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  12. #12
    Super Member Random Variable's Avatar
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    I thought you were responding to my second question. Sorry.
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  13. #13
    Super Member Random Variable's Avatar
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    If player A wins on his first roll, then there is a winner after 1 roll.

    If player B wins on his first roll, then there is a winner after 2 rolls.

    If player A wins on his second roll, then there is a winner after 3 rolls.

    And so on.
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  14. #14
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Random Variable View Post
    If player A wins on his first roll, then there is a winner after 1 roll.

    If player B wins on his first roll, then there is a winner after 2 rolls.

    If player A wins on his second roll, then there is a winner after 3 rolls.

    And so on.
    That's not how I intrepreted it.
    But that's easier.
    So W=1 means A won immediately on the FIRST round
    W=2 means A didn't win but B won immediately
    W=3 means neither won on first round, but A won on his second try.

    It's 2 am, bedtime
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  15. #15
    Super Member Random Variable's Avatar
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    I'm so stupid. I just realized that all I had to do was condition on whether or not player A rolls a sum of 6 on his first roll.


    Let  A_{W} be the event that player A wins

    Let  A_{6} be the event that player A rolls a sum of 6 on his first roll.

    Let  \bar{A}_{6} be the event that player A does not roll a sum of 6 on his first roll.


     P(A_{W}) = P(A_{W}|A_{6})P(A_{6}) + P(A_{W}|\bar{A}_{6})P(\bar{A}_{6})

     P(A_{W}) = 1* \frac {5}{36} + \frac {5}{6} P(A_{W}) \frac {31}{36}

     P(A_{W}) = 30/61


    And I'm sticking with my answer to the second question.


    Let X be the total number of rolls until there is a winner.

    Let  A_{6} be the event that A rolls a sum of 6 on his first roll.

    Let  B_{7} be the event that B rolls a sum of 7 on his first roll (which means A didn't roll a 6 on his first roll).

    Let  \bar{B}_{7} be the event that B doesn't roll a 7 on his first roll.


    E[X] = E[X|A_{6}]P(A_{6}) + E[X|B_{7}]P(B_{7}) + E[X|\bar{B}_{7}]P(\bar{B}_{7})

     E[X] = 1* \frac{5}{36} + 2* (\frac {31}{36}) (\frac {1}{6}) + (2+E[X])(\frac {31}{36}) (\frac {5}{6})

    EDIT:  E[X] = \frac {402}{61}
    Last edited by Random Variable; May 31st 2009 at 09:05 AM. Reason: correct final answer
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