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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?

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  • $\begingroup$ Do you expect dollars to be a function of $n$? $\endgroup$ – Lacklub Apr 6 '16 at 19:14
  • $\begingroup$ Yes, it's a function of n. And you don't need sums to express it! $\endgroup$ – ffao Apr 6 '16 at 19:20
  • $\begingroup$ (1) I find the title misleading.  If you mean for this to be a fair coin, you should say so.  (2) This seems a little heavy for Puzzling.SE; questions like this might be more appropriate on Mathematics.SE. $\endgroup$ – Peregrine Rook Apr 6 '16 at 22:07
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    $\begingroup$ I'm pretty sure math puzzles are on-topic in this site. And I can't possibly think why someone would think this is heavy after reading the answer below. $\endgroup$ – ffao Apr 6 '16 at 22:23
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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: $\frac{2^{3n}}{3n\choose n}$.

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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):

$$\frac{1}{P_{A=B}} = \frac{2^{3n}}{\binom{3n}{n}} = \frac{2^{3n} (2n)! n!}{(3n)!}$$

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    $\begingroup$ Note that by Vandermonde's identity, $\sum_{i=0}^n\binom{n}{i}\binom{2n}{i} = \sum_{i=0}^n\binom{n}{n-i}\binom{2n}{i} = \binom{3n}{n}$. $\endgroup$ – Anon Apr 6 '16 at 19:25
  • $\begingroup$ You dropped the $\frac{1}{2^{3n}}$ $\endgroup$ – Lacklub Apr 6 '16 at 19:30
  • $\begingroup$ @Lacklub I think its right now. Also added some parentheses. $\endgroup$ – Tony Ruth Apr 6 '16 at 19:31
  • $\begingroup$ I didn't know this was a well-known identity, I ended up proving it again when making this puzzle (i.e. what f'' did) $\endgroup$ – ffao Apr 6 '16 at 19:33

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