
Re: probability question about the dice game
Posted:
Feb 14, 2013 12:31 PM


Ray Vickson writes:
> On Thursday, February 14, 2013 5:29:06 AM UTC8, 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 singletoss 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 momentgenerating > 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 = XY 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.

