TITLE: Limiting probabilities for two-player game drawing random uniform numbers
QUESTION [10 upvotes]: 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?

REPLY [2 votes]: Not an answer, but here are some numerical results to show that Julian Rosen's asymptotic lower bound is not too far from optimal.  
$$\matrix{F(10^1,10^1) &= 0.4284225780\cr
F(10^2,10^2) &= 0.4212694133\cr
F(10^3,10^3) &= 0.4205898646\cr
F(10^4,10^4) &= 0.4205226511\cr
F(10^5,10^5) &= 0.4205159409\cr
F(10^6,10^6) &= 0.4205152700\cr}$$