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Topic: probability question about the dice game
Replies: 21   Last Post: Feb 18, 2013 2:47 PM

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Jussi Piitulainen

Posts: 355
Registered: 12/12/04
Re: probability question about the dice game
Posted: Feb 14, 2013 12:31 PM
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Ray Vickson writes:

> On Thursday, February 14, 2013 5:29:06 AM UTC-8, 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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