Following the idea in the video, I ran a simulation, and the result agrees with what I calculated.
Let me explain a little better what I did.
Let f(n) be the average number of jumps with n possibilities.
f(1)= 1 initial jump over the pond + f(0), since there is nothing to jump to after jumping over the pond. f(0) is obviously 0 and this gives f(1) =1, which is intuitively obvious.
f(2) = 1 initial jump + ½ f(1) + ½ f(0) = 1.5.
f(3) = 1 + ⅓ f(2) + ⅓ f(1) + ⅓ f(0) and so on.
The great thing about doing this on a computer is that I only have to say that f(n) = 1 + sum(f(i))/n i from 0 to n-1. The computer function calls itself automatically and computes the individual f(i). I do have to provide a stopping point by saying that f(0)=0. Otherwise the program would try to compute f(i) for increasingly negative numbers and would never stop.