Originally Posted by

**awkward** a) Suppose the players toss N fair coins. Let X be the number of heads tossed by A and Y be the number of heads tossed by B. Then both X and Y have Binomial(N, 1/2) distributions, so

$\displaystyle P(X = Y) = \sum_{i=0}^N P(X=i) \; P(Y=i) =\sum_{i=0}^N \left( \binom{N}{i} (1/2)^N \right) ^2$

$\displaystyle = \sum_{i=0}^N \binom{N}{i}^2 (1/2)^{2N}$

$\displaystyle =(1/2)^{2N} \sum_{i=0}^N \binom{N}{i}^2$

$\displaystyle =\binom{2N}{N} (1/2)^{2N}$

where we have used a well-known identity, $\displaystyle \sum_{i=0}^N \binom{N}{i}^2 = \binom{2N}{N}$.

b) First, let's continue a) by finding $\displaystyle P(X > Y)$ and $\displaystyle P(X \geq Y)$.

By symmetry, $\displaystyle P(X < Y) = P(Y < X)$. Then

$\displaystyle 1 = P(X < Y) + P(X=Y) + P(X >Y) = P(X=Y) + 2 P(X>Y)$

so

$\displaystyle P(X>Y) = (1/2) [1 - P(X=Y)] = (1/2) \left(1 - \binom{2N}{N} (1/2)^{2N} \right) $ .....(1)

and

$\displaystyle P(X \geq Y) = P(X>Y) + P(X=Y) = (1/2) \left( 1 - \binom{2N}{N} (1/2)^{2N} \right) + \binom{2N}{N} (1/2)^{2N}$

$\displaystyle = 1/2 + \binom{2N}{N} (1/2)^{2N+1}$ ....(2)

Now consider the case where A has N+1 coin tosses. As before, let X = the number of heads in the first N tosses, and let's say Z is the total number of heads in all N+1 tosses.

We find $\displaystyle P(Z > Y)$ by conditioning on A's last toss. If A's last toss is a tail, then $\displaystyle Z > Y$ if and only if $\displaystyle X > Y$. If A's last toss is a head, then $\displaystyle Z > Y$ if and only if $\displaystyle X \geq Y$. So

$\displaystyle P(Z > Y) = (1/2) P(X > Y) + (1/2) P(X \geq Y)$.

I think that if you substitute the expressions for $\displaystyle P(X > Y)$ from (1) and $\displaystyle P(X \geq Y)$ from (2) in the above equation and do a little algebra, you will find that $\displaystyle P(Z > Y)$ reduces to 1/2.