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Re: probability question about the dice game
Posted:
Feb 14, 2013 12:47 PM
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On Thursday, February 14, 2013 9:31:48 AM UTC-8, Jussi Piitulainen wrote:
> Ray Vickson writes:
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>
>
> > On Thursday, February 14, 2013 5:29:06 AM UTC-8, starw...@gmail.com wrote:
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> > > two players Ann and Bob roll the dice. each rolls twice, Ann wins
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> > > if her higher score of the two rolls is higher than Bobs, other
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> > > wise Bob wins. please give the analyse about what is the
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> > > probability that Ann will win the game
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> >
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> > P{A wins} = 723893/1679616 =approx= .4309872018.
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> >
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> > This is obtained as follows (using the computer algebra system
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> > Maple). First, get the probability mass function (pmf) of the max of
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> > two independent tosses, which you can do by first getting its
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> > cumulative distribution = product of the two single-toss cumulative
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> > distributions. Then get the mass function by differencing the
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> > cumulative. The pmf is p[i] = [1, 8, 27, 64, 125, 216, 235, 224,
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> > 189, 136, 71]/36^2 on i = 2,...,12.
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> Your final denominator is 6^8 for a problem involving four tosses.
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> I'd've expected 6^4. Should the denominator be only 6^2 for p[i]?
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>
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> And surely p[i] should be defined on 1, ..., 6, not on 2, ..., 12.
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> The latter looks like the probability of a sum instead of a max.
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> > Let X = score of A and Y = score of B. The moment-generating
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> > function (mgf) of X is MX(z) = sum{p[i]*z^i,i=2..12}, while the mgf
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> > of (-Y) is MY(z) = MX(1/z). The mgf of the difference D = X-Y is
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> > MD(z) = MX(z)*MY(z). Expanding this out we have P{D = k} =
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> > coefficient of z^k, for k = -10,...,10, and the probability that A
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> > wins is the sum of the coefficients for k >= 1.
A single toss of TWO dice has denominator 36, so two tosses of two dice has denominator 36^2, and the difference between the two maxima has denominator 36^4. However, as stated in my previous post, I thought the question involved two tosse of 2 dice each, rather than two tosses of a single die. That mixup arose from language usage.
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