General solution to: Two players alternately flip a coin which may be biased: What is the probability of winning by getting a head?

Question

Players A and B are playing a game where they take turns flipping a biased coin, with p probability of landing on heads (and winning). Player A starts the game, and then the players pass the coin back and forth until one person flips heads and wins. What is the probability that A wins?

My solution is to do the following:

• Treat the problem as an infinite sum
• We can get the odd numbers by doubling and adding one to any integer
• We can then reduce the problem to an infinite geometric series with $a = 1$ and $r = (1-p)^2$

$$ \sum_{i \text { odd}} \Bigl(1-p\Bigr)^ip= p\sum_{i =0}^{\infty} \Bigl({1 - p}\Bigr)^{2i+1}= p(1-p)\sum_{i =0}^{\infty} \Bigl({1\over (1-p)^{2}}\Bigr)^{ i }={1\over 1 - (1-p)^{2}}\cdot{p(1-p)}. $$

Does this seem correct?

This is based on the work in this thread, which asks for a solution for the specific instance of an unbiased coin. However, I am trying to solve for the general solution when the coin may be biased.

Answer 1

Player A can win in his $k^{th}$ turn

$$\{1,2,3,4,\dots\}$$

with the following probabilities

$$\{p,(1-p)^2p,(1-p)^4p,(1-p)^6p,\dots\}$$

Thus the probability for A to win is

$$\sum_{k=0}^\infty p(1-p)^{2k}=p\sum_{k=0}^\infty[(1-p)^2]^k=\frac{p}{1-(1-p)^2}=\frac{1}{2-p}$$

Answer 2

Let $q$ be the probability that $A$ wins.

With probability $p$, $A$ wins on the first turn.

And with probability $1-p$, the turn passes to $B$, who now has the same chance $q$ of winning. So we get

$$q=p + (1-p)(1-q)$$

Solving for $q$ gives $$q=\frac{1}{2-p}$$

Answer 3

Lets condition on the first toss and define $A$ as the event that player $A$ wins. Then: $$ P(A) = P(A | H) P(H) + P(A | T) P(T) = p + (1-p) (1-P(A)) = 1 - (1-p)P(A)\\ \implies P(A) = \frac{1}{2-p}. $$

Answer 4

Since you mentioned that you mixed up probability and odds,
let me solve it using odds

Odds in favor of $A = p:(1-p)p$

Converting to probability, P(A wins) = $\dfrac{p}{p+(1-p)p} = \dfrac {1}{2-p}$

Answer 5

As the game stops only when one of them wins, before every flip that $A$ makes the game is back to where it was before $A$ made the first flip. In other words, probability of $A$ eventually winning is same, just before any flip by $A$. So if the probability of $A$ winning is $x$ then,

$x = p + (1-p)^2 x \implies x = \frac{1}{2-p}$