# Exercise C.4.6

$\star$ Professor Rosencrantz flips a fair coin $n$ times, and so does Professor Guildenstern. Show that the probability that they get the same number of heads is $\binom{2n}{n}/4^n$. (Hint: For Professor Rosencrantz, call a head a success; for Professor Guildenstern, call a tail a success.) Use your argument to verify the identity

$$\sum_{k=0}^n\binom{n}{k}^2 = \binom{2n}{n}$$

Let's do what the hint says: call tails for Professor Guildenstern a success and heads - failure. They have the same numbers of heads when they have $k + (n - k) = n$ successes out of $2n$ trials.

$$\Pr\{R=G\} = b(n;2n;1/2) = \binom{2n}{n} \frac{1}{2^n} \frac{1}{2^n} = \binom{2}{n}/4^n$$

Alternatively, we can just calculate the probability and see what happens:

\begin{align} \Pr\{R=G\} &= \sum_{k=0}^n \Pr\{R=k\}\Pr\{G=n-k\} \\ &= \sum_{k=0}^n \binom{n}{k} \cdot \frac{1}{2^n} \cdot \binom{n}{n-k} \cdot \frac{1}{2^n} \\ &= \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} \\ &= \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k}^2 \end{align}

Those are two ways we can express the same value. This let's us verify the identity.