# Combinations of Bernoulli Trials

I recently asked another question, which I have linked here: Combining Binomial Random Variables. I wanted to add onto that question, so I am asking in a different thread.

Brief recap of previous question: I was trying to calculate the probability that a player wins a game involving coin flips. The game was played as follows: Player 1 flips followed by Player 2. The first person to flip a "head" wins.

I am trying to extend this to a case where you must flip $2$ heads to win the game. Again, I wrote a Monte Carlo simulation which provides a different answer to my theoretical approach.

Right now I have the probability function: $P(X=k) {{2+x-1} \choose {x}}p_1^2 (1-p^2)^x [(1-p_2)^k+k(1-p_2)p_2]$ where $p_1$ is your probability of flipping heads and $p_2$ is your opponents probability of flipping heads, $x$ is the number of failures, and $k$ is the round.

The part that is throwing me off is the interaction with player 2. It seems to be consistent with a negative binomial distribution for player 1 in that they need two hits with probability $p_1 p_1$ and $x$ failures with probability $(1-p_1)^x$. These can occur in ${{2+x-1} \choose {x}}$ different ways.

However, player 2 can either hit 0 or 1 times and each of these have different probabilities of occurring. I believe player 2 will have 0 hits with probability $(1-p_2)^k$ and this can only occur 1 way or they can have 1 hit with probability $(1-p_2)^{k-1}p_2$ occurring $k$ ways, because the hit could come during any of the rounds.

The way I have written the probability function seems to take all of this into account, but I am computing differing probabilities between my computer simulation of this game and the theory.

Assuming $p_1=3/8$ and $p_2$ = 2/7, I compute $P(p_1 \text{Winning})$ = 0.75 with the simulation and $P(p_1 \text{Winning})$ = 0.57 with the above equation.

Can anyone point me where I am heading astray? Thanks.