For part (a)

Define Y = number of turns played after

**4th** turn.

$\displaystyle \mathcal{P}(Y=1) = (1-p)$

$\displaystyle \mathcal{P}(Y=2) = p(1-p)$

$\displaystyle \mathcal{P}(Y=3) = p^2(1-p)$

...

$\displaystyle \mathcal{P}(Y=y) = p^{y-1} (1-p)$

Find the expected value:

$\displaystyle E(Y) = \sum_{i=1}^{\infty} i \mathcal{P}(Y=i)$

$\displaystyle E(Y) = \sum_{i=1}^{\infty} i p^{i-1} (1-p)$

$\displaystyle E(Y) = (1-p) \sum_{i=1}^{\infty} i p^{i-1} $

the sum is a hypergeometric progression, which hopefully you have been taught. i got:

$\displaystyle \sum_{i=1}^{\infty} i p^{i-1} = \frac{1}{(1-p)^2} $

if you dont believe that, see the spoiler:

and hence

$\displaystyle E(Y) = \frac{(1-p)}{(1-p)^2} = \frac{1}{1-p}$

So the expected number of games:

$\displaystyle E(4+Y) = 4+E(Y) = 4 + \frac{1}{1-P}$

for (b), note that you are certain to lose exactly 1 game after your

**4th** turn has finished. So find the expected number of looses in the first 4 turns and add one.