2 player coin game probability Suppose two players are playing this game:
each round they flip a coin. If it's Heads, P1 gets a point. If Tails, P2 gets a point. P1 needs X points to win, while P2 needs Y. What is the probability of P1 winning?
 A: Let $x$ and $y$ denote the number of heads and tails needed to win the game by Player 1 and 2 respectively.  Let $$Y\sim \operatorname{bin}(x+y-1,1/2)$$
denote the number of tails in next $x+y-1$ rounds.  Without loss of generality we can ignore the fact that some of these rounds may not need to be played.
Letting $X$ denote the number of heads in the same rounds, the event
$$
Y\ge y. \tag{1a}
$$
is the same event as
$$
x+y-1-X\ge y
$$
which simplifies to
$$
X < x \tag{1b}
$$
The complement of (1a) and (1b) is that
$$
X \ge x \tag{2a}
$$
which is the same event as
$$
Y < y. \tag{2b}
$$
Thus, the game is decided with probability 1 during the next $x+y-1$ rounds and player 1 ultimately wins the game with probability
$$
a_{x,y}=P(X\ge x)=P(Y<y)=\frac1{2^{x+y-1}}\sum_{k=0}^{y-1}{x+y-1 \choose k}.  
$$
This can also be proved by induction using the fact that $a_{x,y}$ must satisfy the recurrence relation
$$
a_{x,y}=\frac12(a_{x-1,y}+a_{x,y-1})
$$
and using $a_{0,y}=1$ and $a_{x,0}=0$ as base cases.
