# Biased coins: Probability such that the first player throws heads

Let $$A$$, $$B$$ be the two players. Each one has a coin has a probability of getting heads of $$p_i$$. Player $$A$$ always starts first. What is the probability such that $$A$$ wins?

Ex. The both coins land 'heads' on average 1 out of 2 times. The solution says $$\frac{1}{2}$$ as the result.

My approach was to draw a probability tree and compute the probability such that $$A$$ throws heads. We know, that this is geometrically distributed, but don't know how to get the result.

• Another way to get the answer: If you let $\alpha$ be the probability that $A$ gets head first, you should be able to see that $$\alpha = p_A + (1-p_A)(1-p_B)\alpha$$ (note that the game "restarts" if the first two tosses are tails). Then solve for $\alpha$. Apr 21, 2019 at 22:24

If you drawn a tree, then you can see that either $$A$$ wins straight away, or $$A$$ flips tail and $$B$$ also flip tail and then $$A$$ gets heads, ..., or $$A$$ and $$B$$ each get $$j$$ tails in a row and then $$A$$ gets heads.

So, the probability of $$A$$ winning is given by summing all these probabilities: $$\sum_{j=0}^{+\infty} \big((1 - p_A)(1 - p_B)\big)^{j} p_A$$ Let $$c = (1 - p_A)(1 - p_B) \in [0; 1]$$. The above sum can be written as: $$p_A \sum_{j=0}^{+\infty} c^{j}$$

Can you compute the result?

Edit: For future reference, I'm adding the result for the probability of $$A$$ winning: $$\frac{p_A}{1-c} = \frac{p_A}{p_A + p_B - p_A p_B}$$

And, just for fun, I wondered which should be the probability when $$p_A = p_B$$ that makes the probability of $$A$$ winning equal to $$1/2$$.

$$\frac{p_A}{2 p_A - p_A ^2} = \frac{1}{2}$$ $$2p_A = 2 p_A - p_A ^2$$ $$p_A ^2 = 0$$ $$p_A = 0$$

Hence, $$\lim_{p_A \to 0} \frac{p_A}{2 p_A - p_A ^2} = \frac{1}{2}$$ So, if the game goes on forever (in the limit $$p_A \to 0$$) then both players have $$1/2$$ probability of winning.

Furthermore, if $$p_A = p_B$$ the first player to toss the coin has the highest probability of winning. I.e., for $$p_A \in [0; 1]$$, $$\frac{p_A}{2 p_A - p_A ^2} \geq \frac{1}{2}$$

• You're welcome. Apr 20, 2019 at 12:05