A friend gave me the following problem to solve. I have been stuck at it for days. With little or no progress. Well, I did get an estimated answer by running a simulation but not the formal solution.
Two players are playing a game of football. The goals are located at positions $0$ (Player A) and $100$ (Player B), and the ball is initially at position $50$.
In one move, a player draws a number from the uniform random distribution $(0, 100)$ and kicks the ball by the same amount towards the other player's goal post.
If at any point the ball crosses a goal post the game ends and the player who made the last move wins. Find the probability of player A winning.
Edit:
I tried creating an infinite series of probabilities of winning in each move like in the first move there is a 50% chance that Player A wins. This quickly turned into a integral starting at the second move, i.e $ \int_x x * P(player-B's-win | ball-is-at x) * P(ball-at-x) \;\mathrm{d}x$
Best Answer
For any $0\le x\le 1$, let
$A(x)$ be the probability A wins when it is A's turn, and the ball is $100x$ units from $A$'s goal.
$B(x)$ be the probability A wins when it is B's turn, and the ball is $100x$ units from $A$'s goal.
You want $A(\frac12)$. We then have that $$ A(x) = x + \int_x^{1} B(t)\,dt,\qquad A(1)=1\\ B(x) = \int_0^x A(t)\,dt\qquad\;\qquad B(0) = 0 $$ If we take $\frac{d}{dx}$ of both equations, you get $$ A'(x) = 1 - B(x),\\ B'(x) = A(x)\;\;\;\;\;\;\; $$ The general solution to that system of differential equations is $$ A(x) = c_1\cos x-c_2\sin x\;\;\;\;\;\;\\ B(x) = c_1 \sin x + c_2 \cos x + 1 $$ The initial conditions $A(1)=1$, $B(0)=0$ then imply $c_1=\frac{1-\sin 1}{\cos 1}$ and $c_2=-1$, which means that $$ A(\tfrac12) = \frac{1-\sin 1}{\cos 1}\cdot \cos \tfrac12 + \sin \tfrac12\approx 0.736915 $$