In the game Pogémon, Pogé Balls contain Pichagus with a $50 \%$ chance and otherwise contain nothing. The new Pogémon universe started with one Pichagu and has existed for five hours. Every hour, every Pichagu at the time opened a new Pogé Ball.

a) What is the expected number of Pichagus created in the last hour?

b) Every two currently existing Pichagus will now battle each other. What is the expected number of battles?
This problem was inspired by https://xkcd.com/1516/.

a) Every hour, the expected number of Pichagus increases $50 \%$. Thus, the expected number of Pichagus after 4 hours is $\left(\frac{3}{2}\right)^{4}$. Each of these Pichagus has a $50 \%$ chance of creating a new Pichagu in the last hour, so the answer is $\frac{1}{2}\left(\frac{3}{2}\right)^{4}=\frac{81}{32}=2.53125$.
b) We show via induction that the answer for $n$ hours is $E(n):=\left(\frac{3}{2}\right)^{2 n-1}-\left(\frac{3}{2}\right)^{n-1}$. This is clearly true for $n=0$; no battles would happen at that point.

Now we show that this is true for $n+1$ given that it's true for $n$. After one hour, there are two cases: either there is one Pichagu or there are two.

In the first case, the expected value is just $E(n)$ as there are $n$ hours left.

Otherwise, say that the Pichagus are divided into two "factions": the Pichagus that descend from the original Pichagu and the ones that descend from the second one. Each faction has an expected value of $E(n)$ internal battles (battles between two members of the faction). Additionally, as in part (a), each faction has an expected value of $\left(\frac{3}{2}\right)^{n}$ members, so there will be an expected value of $\left(\left(\frac{3}{2}\right)^{n}\right)^{2}$ battles between factions. Thus in this case there will be an expected value of $2 E(n)+\left(\frac{3}{2}\right)^{2 n}$ battles.

Combining the cases we have

$$
\begin{aligned}
E(n+1) & =\frac{1}{2} E(n)+\frac{1}{2}\left(2 E(n)+\left(\frac{3}{2}\right)^{2 n}\right) \\
& =\frac{1}{2}\left(\frac{3}{2}\right)^{2 n}+\frac{3}{2} E(n) \\
& =\frac{3}{4}\left(\frac{3}{2}\right)^{2 n-1}+\frac{3}{2}\left(\left(\frac{3}{2}\right)^{2 n-1}-\left(\frac{3}{2}\right)^{n-1}\right) \\
& =\left(\frac{3}{2}\right)^{2 n+1}-\left(\frac{3}{2}\right)^{n},
\end{aligned}
$$

as desired. Plugging in $n=5$ gives $\left(\frac{3}{2}\right)^{9}-\left(\frac{3}{2}\right)^{4}=\frac{3^{9}-2^{5} \cdot 3^{4}}{2^{9}}=\frac{17091}{512}=$ 33.380859375 .