# Coin tossing problem between two players. Find probability that one player tosses heads first.

Player $1$ tosses a biased coin with the probability of getting $H$ (a head) being $p$, for some $0<p<1$, and Player $2$ tosses a biased coin with the probability of getting $H$ (a head) being $q$, for some $0<q<1$. They toss their coins at the same time. The first player to get $H$ wins. If they both get $H$, the game ends in a draw. What is the probability that Player $1$ wins?

Here is what I have:

Let $P(A_k)=P(\{\text{Player$1$wins on$k$th toss} \})=P(\{\text{Player$1$gets$H$on$k$th toss and Player$2$gets$T$on$k$th toss})=P(\{\text{Player$1$gets$H$on$k$th toss}\})P(\{\text{Player$2$gets$T$on$k$th toss}\}) =(1-p)^{k-1}(1-q)^{k-1}p.$

Then $P(\{\text{A wins}\})=P(\cup_{k\geq 1} A_k)=\sum_{k\geq 1} P(A_k)= \sum_{k\geq 1} (1-p)^{k-1}(1-q)^{k-1}p=\frac{p}{1-p} \left(\frac{1}{1-(1-p)(1-q)}-1\right).$

Can someone tell me if I am on the right track? Thank you!

• I believe the question should replace "If they both get H, the game ends in a draw" with "If both get H or neither gets H, the round ends in a draw." – true blue anil Oct 2 '16 at 7:06
• @trueblueanil I think the intent is that if they both get a T, the game goes to another round. Otherwise, it would be a simple one-shot coin toss, not a "first player to get H" game. – Patricia Shanahan Oct 2 '16 at 7:18
• @PatriciaShanahan: The way I have interpreted the question, it is not a simple one-shot coin toss, the game extends if both get H or both get T, and continues until one or the other wins. – true blue anil Oct 2 '16 at 7:35
• @PatriciaShanahan Well, I have removed the answer based on my interpretation of the question to reduce confusion, but my answer is different from that of others. – true blue anil Oct 2 '16 at 8:17
• @trueblueanil Your interpretation conflicts with "If they both get H, the game ends in a draw." Note "game ends". – Patricia Shanahan Oct 2 '16 at 8:20

To supplement user362325's answer, I will provide another method of solving this problem:

Note that, if neither player tosses heads in each of their first flips, the problem essentially reverts back to the original. Let $x_1$ be the probability that Player 1 wins.

We know that

$$x_1 = p+(1-p)(1-q)x_1$$

where the first $p$ represents the first flip being heads and the other part is the probability that neither player flips heads, $(1-p)(1-q)$, times the original $x_1$.

Can you take it from there?

• The first flip being H doesn't guarantee a win. Simultaneously, the opponent's must be T. Remember, it clearly says "They toss their coins at the same time." – true blue anil Oct 2 '16 at 7:50

Thanks for the suggestion from Carl Schildkraut.

From his expression,$$\ x_1=p+(1−p)(1−q)x_1$$ making $x_1$ as th subject, $$\ [1-(1-p)(1-q)]x_1=p$$ $$\ x_1=\frac{p}{1-(1-p)(1-q)}$$

Is this numerically identical to my original answer?

• This answer is correct. By my calculations, it is $1-$ your original number. – Carl Schildkraut Oct 2 '16 at 3:00
• The first flip being H doesn't guarantee a win. Simultaneously, the opponent's must be T. Remember, it clearly says "They toss their coins at the same time." – true blue anil Oct 3 '16 at 6:03

$\ P(A_w)$ = $\ P(A_H)$+$\ P(A_T,B_T,A_H)$+$\ P(A_T,B_T,A_T,B_T,A_H)$+...

$\ P(A_w)$ = $\ p$+$\ (1-p)(1-q)p$+$\ (1-p)^2(1-q)^2p$+$\ (1-p)^3(1-q)^3p$
+...

$\ P(A_w)$ = $$\sum_{k=0}^\infty p[(1-p)(1-q)]^k$$

Note that it forms a geometric sequence.

Thus $\ P(A_w)$ = $$\frac{p}{1-(1-p)(1-q)}$$

• Okay, so my approach is correct, but I should change the sum to start at $k=0$. – Michael Oct 2 '16 at 1:46
• The first flip being H doesn't guarantee a win. Simultaneously, the opponent's must be T. Remember, it clearly says "They toss their coins at the same time." – true blue anil Oct 3 '16 at 6:03

$\underline{Revised\;answer}$

I will label player $1$ as $A$ for clarity.
I also take as per the exact wording of the question, that both getting heads on a round is a draw , whereas both getting tails on a round extends the game, and that we want to find $P(A\;eventually\;wins)= \Bbb P\; (say)$

$A$ can win in the $1^{st}$ round by getting H while $B$ gets $T$, and in subsequent round(s) if and only if it extends to subsequent round(s)

• $\underline{Using\; geometric\; series}$

$A$ can win in first round with $Pr = p(1-q),$ or move into next round with $Pr = (1-p)(1-q),$ and again win with $Pr = p(1-q),$ and so on.

Thus $\Bbb P = p(1-q) + (1-p)(1-q)p(1-q) + ((1-p)(1-q))^2p(1-q)\; + ...$

This a G.P. with $a = p(1-q),\; r = (1-p)(1-q)\;\;\; S(\infty) = \dfrac{a}{1-r}$

$so\; \Bbb P= \dfrac{p(1-q)}{1 - (1-p)(1-q)}$

• $\underline{Using\; recursion}$

$A$ wins in first round with $Pr = p(1-q)\;$ or gets back to square one with $Pr = (1-p)(1-q)$

So $\Bbb P = p(1-q) + (1-p)(1-q)\Bbb P$

which again yields, $\Bbb P = \dfrac{p(1-q)}{1 - (1-p)(1-q)}$