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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?

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    $\begingroup$ Is the coin fair? It might be easier by imaging them flipping $X+Y-1$ times $\endgroup$
    – Henry
    May 30, 2021 at 12:33
  • $\begingroup$ @gunes I agree that this may be self-study but isn't the OP supposed to add the tag? $\endgroup$ May 30, 2021 at 16:38
  • $\begingroup$ I'm voting to reopen since there is no definite evidence that this was self-study. $\endgroup$ May 30, 2021 at 17:53
  • $\begingroup$ @congriUQ you can add the answer you deleted to the above question as your effort. $\endgroup$
    – gunes
    May 30, 2021 at 19:50
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    $\begingroup$ @gunes I did see the other answer but didn't notice it was by the OP. I have undeleted my own answer. In my opinion the difficulty doesn't suggest self-study (although the solution is trivial once you see it). The OP might have deleted his/her answer given that it was incorrect. $\endgroup$ May 30, 2021 at 20:23

1 Answer 1

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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.

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