# Understanding a coin throwing game

Suppose players A and B take turns to throw a fair coin. The one that obtain $$Tail$$ first wins. Suppose $$A$$ starts. What is the probability that $$A$$ wins the game? How about the probability that $$A$$ wins given that he did not obtain "tails" on her first two trials. Finally, given that $$A$$ lost the game, what is average number of tosses?

## approach

If $$A$$ starts, notice that if he obtains $$T$$, than the game is done and this occur with probability $$1/2$$ If not then for $$A$$ to win it gotta be in the third round so we want something like HHT and $$P(HHT) = 1/8$$ and simiarly, if she dont win in third round then she got a change to win in fifth round and $$P(HHHHT) = \frac{1}{2^5}$$ and so on. Thus, we have

$$P(A \; wins) = \frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + ... = \sum_{i \geq 1} \frac{1}{2^{2i-1} } = 2 \left( \sum \frac{1}{4^i} \right) = 2 (1/[1-(1/4)] - 1 ) = \boxed{2/3}$$

Now, for the second case We know for $$A$$ to win he must have the string HHHHT which means that $$A$$ to win we want to calculate :

$$P(HHHHT) + P(H^6 T) + P(H^8 T ) + ... = \frac{2}{3} - \frac{1}{2} - \frac{1}{8}$$

using result from previous part. this equals $$\boxed{1/24}$$ in clear discrepancy with answer key which gives the asnwer as $$1/3$$. Did I misunderstood the problem?

Finally, as for the expectation if $$A$$ losses. We are looking at patterns of the form $$HT$$, $$HHHT$$, $$HHHHHT$$ remembering that $$A$$ was the first to start the game. If we call $$X$$ to be number of tosses until game ends then we observe that $$P(X=2) = P(HT)$$, $$P(X=4) = P(HHHT)$$, ... we observe that $$P(X=3)$$ is not possible since $$A$$ is to lose the game. Thus, the expectation is

$$E(X) = \sum_{i=1}^{\infty} \frac{2i}{2^{2i}} = 2 \sum \frac{i}{4^i}$$

by using the calculus identity $$\sum n x^n = \dfrac{x}{(1-x)^2}$$, we obtain

$$E(X) = \frac{8}{9}$$

again in discrepancy with my answer key which gives the solution to be $$\frac{8}{3}$$. What is my mistake here?