```Date: Feb 14, 2013 12:31 PM
Author: Jussi Piitulainen
Subject: Re: probability question about the dice game

Ray Vickson writes:> On Thursday, February 14, 2013 5:29:06 AM UTC-8, starw...@gmail.com wrote:> > two players Ann and Bob roll the dice. each rolls twice, Ann wins> > if her higher score of the two rolls is higher than Bobs, other> > wise Bob wins. please give the analyse about what is the> > probability that Ann will win the game> > P{A wins} = 723893/1679616 =approx= .4309872018.> > This is obtained as follows (using the computer algebra system> Maple). First, get the probability mass function (pmf) of the max of> two independent tosses, which you can do by first getting its> cumulative distribution = product of the two single-toss cumulative> distributions. Then get the mass function by differencing the> cumulative. The pmf is p[i] = [1, 8, 27, 64, 125, 216, 235, 224,> 189, 136, 71]/36^2 on i = 2,...,12.Your final denominator is 6^8 for a problem involving four tosses.I'd've expected 6^4. Should the denominator be only 6^2 for p[i]?And surely p[i] should be defined on 1, ..., 6, not on 2, ..., 12.The latter looks like the probability of a sum instead of a max.> Let X = score of A and Y = score of B. The moment-generating> function (mgf) of X is MX(z) = sum{p[i]*z^i,i=2..12}, while the mgf> of (-Y) is MY(z) = MX(1/z). The mgf of the difference D = X-Y is> MD(z) = MX(z)*MY(z). Expanding this out we have P{D = k} => coefficient of z^k, for k = -10,...,10, and the probability that A> wins is the sum of the coefficients for k >= 1.
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