# Choosing probabilities as to make a coin game fair

Consider a game with two players A and B. Player A throws a coin with probability $p_A$ of landing heads, in which case he wins the game. If not, player B throws another coin with probability $p_B$ of landing heads, in which case B wins. If not, it is A's turn again.

Calculate the probability of both players winning the game, as a function of $p_A$ and $p_B$. If $p_A=\frac{1}{3}$, find $p_B$ so that the game is fair.

This is what I've done so far. I think I can determine the probabilities of A and B winning the game, as a function of both $p_A$ and $p_B$ and of the number of rounds $n$:

$p(W_A)={p_A}^n\left(1-p_B\right)^{n-1}$

$p(W_B)={p_B}^{n-1}\left(1-p_A\right)^{n-1}$

The way I see it, the equations are different because A gets to go first. Intuitively, B is in an unfair position unless $p_B > p_A$.

I suppose that a fair game would mean that $p(W_A)=p(W_B)$, but for what $n$? This has led me to think that there is a way to describe those probabilities without depending on the number of rounds, but I don't know how.

This is a homework question. I'm not looking for the answer, but for guidance in getting there. Let me give you some context. I am taking a graduate-level course on Stochastic Control (my research area is robotics). The course starts with a review of probability, and this particular question is part of the exercise list. The bibliography for this part of the course is Sheldon Ross' Introduction to Probability Models. Thanks for the help!

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The probability of player A winning the game is equal to the sum of the probability of him winning each individual round. So, you get an infinite sum which can be evaluated using various techniques.

However, I would like to introduce a nicer alternative for you to try.

Using your notation of $p(W_a)$ as the probability of player A winning the game and $p(W_b)$ as the probability of player B winning the game, we can create a few formulas:

First of all, either player A or player B must win the game, so $p(W_a)+p(W_b)=1$.

Secondly, consider a typical "round" in the game. There are three possible results: A wins, B wins, or nobody wins. If nobody wins, the probability of A winning is unchanged ($p(W_a)$ and $p(W_b)$ are constants, right?). Thus, $p(W_a)=p(\text{A wins})+p(\text{nobody wins})p(W_a)$ and $p(W_b)=p(\text{B wins})+p(\text{nobody wins})p(W_b)$.

To solve, we simply need to calculate the probability of A winning in any round, the probability of B winning in any round, and the probability of nobody winning in any round. Because A goes first, $p(\text{A wins})=p_A$. What about the other possibilities?

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This is really clever. Is this kind of recursive thinking common in a lot of probability problems? – Pedro d'Aquino Mar 21 '11 at 0:02
@Pedro: This approach is typically called first step analysis since you step forward once and notice that this can be repeated. This is done to replace the infinite sum. If you actually notice and write out the infinite sum, the infinite sum $P(W_a)$ can be written as $P(A) + P(\text{Nobody}) P(W_a)$. (The way we sum up geometric series). – user17762 Mar 21 '11 at 0:16

Hint: if both A and B get tails on their first throws, you're back in the same situation as at the start of the game.

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Your initial tackle is a good one, though you might want to change it up a bit. Seeing the problem as it is, the number of rounds is not given so, up until now, you have just calculated the probability of each part to win in round $n$. Hence the probability of winning should be the sum over all rounds.

Hope this helps

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