I forgot to say the assumption is that P(p) is uniform. Adam
On Wed, 28 Apr 2010, Adam Ralph wrote:
> Hi James,
>
> If you use
>
> P(A=5,B=3|p) = p^5 x (1-p)^3 x Choose(8,3)
>
> which is the probability of A winning 5 times and B winning 3 times
> given a known value for p. You have a combinatoric term 8 choose 3
> which will cancel out in the final equation.
>
> Substitute this into the equation bottom of p1177 you get the
> equation bottom of 1178.
>
>
> I hope that is correct (where are those flame retardant pants).
>
> Adam
>
>
>
>
> On Tue, 27 Apr 2010, James Stroud wrote:
>
> >
> > On Apr 27, 2010, at 3:08 AM, Colin Nave wrote:
> >
> > > Sean Eddy who is an author on the above link wrote what I regard as an
> > > excellent intro to Bayesian statistics
> > > ftp://selab.janelia.org/pub/publications/Eddy-ATG3/Eddy-ATG3-reprint.pdf
> >
> > To get the expression for E(Bob wins) (3rd equation on page 1178),
> > Eddy does some "algebraic rearrangement". It's not obvious what is
> > getting rearranged. Does anyone have a clue?
> >
> > James
>
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