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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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RGVickson@shaw.ca

Posts: 1,643
Registered: 12/1/07
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:
>
>
>

> > 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.

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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