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A fair coin is tossed $n$ times by Adam and $n$ times by Andrew. What is the probability that they get the same number of heads?

Now since there are a total of $2n$ flips, $n$ from each person, we would need to choose $k$ flips that they both have heads correct? So since this is a binomial$\sim (2n,\frac12)$ I came to find that we have $${2n \choose k }\left(\frac12\right)^{2n}$$ However my textbook says it differently. I'm wondering why the book says it is: $${2n \choose n }\left(\frac12\right)^{2n}$$ Is it the same thing? I'm just confused as to why it would say choose $n$. Should I be assuming assume that half of the total flips are are going to be the same?

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2 Answers 2

up vote 10 down vote accepted

The following solution has a minimal amount of computation, but may require some concentration.

Call our two players Alicia and Beti. They get the same number of heads iff for some $k$, they each get $k$ heads.

The probability that Beti gets $k$ heads is the same as the probability Beti gets $k$ tails, so it is the same as the probability that Beti gets $n-k$ heads.

So the probability they get the same number of heads is the same as the probability that between them they get a total of $n$ heads.

The probability that when we toss a coin $2n$ times we get $n$ heads is $$\binom{2n}{n}\frac{1}{2^{2n}}.$$

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Wow this is very nice. +1 –  EuYu Nov 30 '12 at 3:34

What is the quantity you're calling "$k$"?

If $k\in\{0,1,2,\ldots,n\}$ then the probability that they both get exactly $k$ heads is $\dbinom n k ^2 \left(\dfrac12\right)^{2n}$.

So the probability that there is some value of $k$ in that set for which they both get excatly $k$ heads is $$ \sum_{k=0}^n \binom{n}{k}^2 \left(\frac12\right)^{2n}. $$

If you expand this as $$ \left(\binom n 0 ^2 + \binom n 1 ^2 + \binom n 2 ^2 + \cdots + \binom n n ^ 2\right)\left(\frac12\right)^{2n} $$ then you don't see any number called "$k$" in that expression, upon which the answer depends. There shouldn't be such a thing.

Next there's the question of why $$ \binom n 0 ^2 + \binom n 1 ^2 + \binom n 2 ^2 + \cdots + \binom n n ^ 2 = \binom{2n}{n}. $$ I think that question's been posted here a few times. Here's a way of seeing that: You'e looking for the number of ways to choose $n$ things out of $2n$, so divide the set of $2n$ into two sets of size $n$. The number of ways to pick $k$ things from the first set and $n-k$ from the second is $\dbinom n k\dbinom n{n-k}$ $=\dbinom n k \dbinom n k$. Now add that up over all the possible values of $k$.

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In the 2nd line the probability should be $\binom{n}{k}^2 \cdots$, not $\binom{n}{k}\cdots$. –  cjackal Nov 30 '12 at 3:24
    
@MichaelHardy Ohh now I get what you're saying. That makes much more sense. Thanks! –  TheHopefulActuary Nov 30 '12 at 3:30

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