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
