Keep it bouncing

We need the probability that the frog makes an even number of jumps in order to return to its original leaf.

The probability that the frog makes exactly $k$ jumps is ${N\choose k} p^k q^{N-k}$ , where $q=1-p$ . This is the $k^{\text{th}}$ term of the binomial expansion of $(p+q)^N$ :

$(p+q)^N=\sum_{k=0}^N {N\choose k} p^k q^{N-k} = {N\choose 0} p^0 q^{N} + {N\choose 1} p^1 q^{N-1} + {N\choose 2} p^2 q^{N-2} + \cdots$

To get the terms for even $k$ , we compare this to the binomial expansion of $(-p+q)^N$ :

$(-p+q)^N=\sum_{k=0}^N {N\choose k} (-1)^k p^k q^{N-k} = {N\choose 0} p^0 q^{N} - {N\choose 1} p^1 q^{N-1} + {N\choose 2} p^2 q^{N-2} - \cdots$

We can cancel out the terms with odd $k$ simply by adding the two expansions; hence

$2p_N = (p+q)^N + (-p+q)^N$

Of course, $p+q=1$ , so this simplifies to

$p_N=\frac{1}{2} \left(1+(1-2p)^N \right)$

For the particular question, we have $p_5 = \frac{1}{2} \left(1+\left(1-\frac{2}{3} \right)^5 \right) = \boxed{0.502\ldots}$ .

Alternatively, we can set up a recursion: $p_0=1$ , and $p_{N+1} = (1-p)p_N + p(1-p_N)$ (we just consider the two possible positions of the frog after $N$ jumps). This is somewhat simpler than the first approach!

As we'd expect, as $N \to \infty$ we have $p_N \to \frac12$ .