Consider this simple 2-person game I just made up:

Player A goes gets to draw a uniform U[0,1] number up to X times. At any time, he may either keep his number, or draw a brand new uniform number. However, if he draws all X times, he must keep his last draw as his score.

Player B goes second, and gets to draw up to Y times. Same rules apply to him.

The winner is the player with the highest score. Assuming both A and B play optimally, the probability that player A wins the game is given by F(X,Y). F(X,Y) can be given by the following recurrence relationship (I'll leave it as a separate exercise to derive this):

$F(0,Y) = 0$ $\forall$ Y>0

$F(X,Y) = \left(\frac{Y}{Y+1}\right)*\left(F(X-1,Y)\right)^{\frac{Y+1}{Y}} + \frac{1}{Y+1}$

Here's the question: Find the limit $\lim_{n\rightarrow \infty}$ $F(n,n)$. Obviously, this can be approximated numerically directly from the recurrence definition, but I'm wondering if this limit has an elegant closed-form solution (or perhaps can be derived as a unique root of an implicit equation).

The solution has a nice interpretation to it. Basically, as you let each player have more and more turns, how much does this neutralize B's advantage of getting to go second?