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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.

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  • $\begingroup$ 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 $\alpha$. $\endgroup$ Apr 21, 2019 at 22:24

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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 $p_A \in [0; 1]$, $$ \frac{p_A}{2 p_A - p_A ^2} \geq \frac{1}{2} $$

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