[Silent A]

Hmm, figured out the answer while I was away.

My argument above actually only shows that total = 50 is not a local maximum on the probability distribution. The actual probability distribution isn't really bi-modal at all.

Instead, the above argument can be extened to claim that the probability distribution of final totals doesn't have any local maxima. It goes like this ...

At a given point during the shooting, the player will have hit "i" baskets after "n" throws. The odds of him getting the next basket is therefore i/n. So after the (n+1)th shot he will have hit either i or i+1 baskets and his probability of getting the following basket will be either i/(n+1) or (i+1)/(n+1). Notice that the fraction of future successful shots will tend to diverge away from i/n. This is true so long as 0 < i < n (in the problem described, this is always true since he starts off with one miss and one hit).

Now note that if a local maxima exists in the probability distribution, then small movements away from a local maximum will tend to move back towards the local maximum (i.e. local maxima should be stable to a certain extent). However, from the above argument, no point can be stable and therefore no point can be a local maximum. We also know that because the problem is inherently symmetrical, the final probability distribution should also be symmetrical. This means that the distribution should be either perfectly flat, or shaped like a "U" (low in the middle and high on both ends).

Now if only I could come up with an argument to suggest that the ere are no local minima either, I could show that the final probability distribution is perfectly flat without making any calculations at all.

And if the probability distribution is perfectly flat then the answer is 1/99 since the final total must be at least 1 and no more than 99.

P.S. I have an analytical proof, but I'll leave it for someone else.