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Player A needs to flip Heads twice to win (does not need to be consecutive Heads), while Player B only needs to flip Heads once to win. Player A goes first. They take turns flipping coins until one player wins. How many games on average would player A win after 100 games? Each win counts as one game, regardless of how many rounds the game lasts.

Here are some examples:

First Game:

(Round 1) Player A flips Heads; Player B flips Tails.

(Round 2) Player A flips Heads (Player A wins)

Second Game:

(Round 1) Player A flips Heads; Player B flips Heads (Player B wins)

Third Game:

(Round 1) Player A flips Tails; Player B flips Heads (Player B wins)

Fourth Game:

(Round 1) Player A flips Heads; Player B flips Tails

(Round 2) Player A flips Tails; Player B flips Tails

(Round 3) Player A flips Heads (Player A wins)

The purpose of the problem is to determine your likelihood of winning in a real game-theory situation. There are many sports example and one of them is pick-up basketball. For example, the first team that scores 21 points wins. Your team has 17 points while the other team has 19. Assuming both sides have a 50% chance of scoring a basketball when any player shoots, how often would the team with 17 points win? The coin flip analogy is a simplified version of this problem.

I've tried doing simulations (below) but don't think that is the right way to solve the problem.

HT HTH HTTH

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    $\begingroup$ "...after 100 games?" tosses maybe? $\endgroup$ – pointguard0 Jan 30 at 22:53
  • $\begingroup$ Does "flip a head twice" mean "twice in a row" or "twice overall"? $\endgroup$ – Graham Kemp Jan 30 at 23:42
  • $\begingroup$ @pointguard0, good question, after 100 games $\endgroup$ – Sam Nguyen Jan 31 at 23:34
  • $\begingroup$ @GrahamKemp, twice overall $\endgroup$ – Sam Nguyen Jan 31 at 23:34
  • $\begingroup$ @JoséCarlosSantos, thanks for letting me know. I've updated the original post so hopefully it qualifies this time. $\endgroup$ – Sam Nguyen Jan 31 at 23:35
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First, let us assume A gets heads on the first throw. Now the game is first to get heads wins with B tossing first. Let $p$ be the chance B wins this subgame. If B gets heads he wins, otherwise he is in A's shoes and wins with chance $1-p$, so we have $$p=\frac 12\text{(heads)} + \frac 12\text{(tails)}(1-p)\text{(wins after tails)}\\p=\frac 23$$ That means if A flips heads he wins $\frac 13$ of the time.

If A flips tails, B wins outright with a head. Otherwise we are back to the start of the game. Let A win the overall game with probability $q$. We have $$q=\frac 12\text{(heads)}\cdot \frac 13\text{(wins after heads)} + \frac 12\text {(tails)} \cdot \frac 12\text{(B gets tails)} q\text{(wins new game)}\\\frac 34q=\frac 16\\q=\frac 29$$ Of $100$ games, A should win a little over $22$.

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  • $\begingroup$ Thank you for your help Ross! $\endgroup$ – Sam Nguyen Feb 2 at 0:48

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