Tricky Bayesian question using posterior predictive distributions

[lauren325]

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 Prob(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)

 

[mmwave]

To compute Prob(Z>x), we can use the cumulative distribution function (CDF) of the geometric distribution. The CDF gives the probability that Z is less than or equal to a certain value, so we can use it to calculate the probability that Z is greater than a certain value (in this case, x).

The CDF of the geometric distribution is given by:

F(z;p) = 1 - (1-p)^z

So, to calculate Prob(Z>x), we can use:

Prob(Z>x) = 1 - F(x;p)

Substituting this into the formula for Prob(Person A wins|x), we get:

Prob(Person A wins|x) = 1 - F(x;p) + F(x;p)*0.5

= 0.5 + 0.5*F(x;p)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(p^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(1 - (1-p)^x)

= 0.5 + 0.5*(