My Math Forum Tricky Bayesian question using posterior predictive distributions, help!!!?

 February 8th, 2015, 06:32 AM #1 Newbie   Joined: Feb 2015 From: West Calder Posts: 1 Thanks: 0 Tricky Bayesian question using posterior predictive distributions, help!!!? A game is played using a biased coin, with unknown p. Person A and Person B flip this coin until they get a head. The person who tosses a head first wins. If there is a tie, where both people took an equal number of tosses to flip a head, then a fair coin is flipped once to determine the winner. Person A plays until they get a head, then Person B plays afterwards. Using the Jeffrey's Prior for the Geometric distribution and the Posterior predictive distribution for Person B's score given we know Person A's score, what is the probability that Person A will win? My working so far: X=the number of tosses until person A gets a head Z=the number of tosses until person B gets a head The Jeffrey's prior for the Geometric distribution is: p(p) = p^-1*(1-p)^-0.5 The posterior predictive distribution for Person B's score given Person A's score is: p(z|x) = 2(2x-1)/(2x+2z-1)(2x+2z-3) Finally: Prob(Person A wins|x) = Prob(Z>x) + Prob(Z=x)*0.5 Now P(Z=x) is computed by substituting z for x in the formula p(z|x). But how do I compute Prob(Z>x)? By the way, the final answer is: Prob(Person A wins|x) = 2(2x-1)^2/(2x+2z-1)(2x+2z-3) Can anyone please tell me how to do this, and if I'm doing it correctly?

### jeffrey prior geometric distribution and the posterior predictive

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