After finding a coin in her farm, Alice proposes the following game to Bob:
Alice will flip the coin $n$ times. Bob will then flip the coin $2n$ times. If Bob manages to get the same amount of heads as Alice, he wins. Otherwise Alice wins.
As they're best friends, Alice wants the game to be completely fair. Assuming Bob pays a dollar to play, how much should she pay Bob for a win so they're expected to break even?
Suppose instead that
Bob wins if he gets as many heads as Alice gets tails. This has the same probability as the original situation, by reversing all of Alice's flips.
But in this new situation,
Bob wins if exactly $n$ heads are flipped in total. The number of ways for this to happen is $3n\choose n$, so Bob wins with probability $\frac{3n\choose n}{2^{3n}}$.
For the game to be fair, the amount he wins should be
the inverse of this: $\frac{2^{3n}}{3n\choose n}$.
The probability of Alice getting $i$ heads after $n$ flips is:
$$ P_n^i = \frac{1}{2^n} \binom{n}{i}$$
The probability of Bob getting $i$ heads after $2n$ flips is:
$$ P_{2n}^i = \frac{1}{2^{2n}} \binom{2n}{i}$$
Then the probability of Alice and Bob getting the same number of heads is:
$$P_{A=B} = \sum_{i=0}^n P_n^i P_{2n}^i = \frac{1}{2^{3n}} \sum_{i=0}^n\binom{n}{i}\binom{2n}{i} $$
Finally, the amount Alice should pay Bob if he wins is (Thanks McFry):
$$\frac{1}{P_{A=B}} = \frac{2^{3n}}{\binom{3n}{n}} = \frac{2^{3n} (2n)! n!}{(3n)!}$$
