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.