Coin tossing with random throws

Question

Consider the following game: two people play at tossing coins; each one of them has only one coin (1 & 2) which is toss repeatedly and the first player to obtain "head" wins. However there some particular aspects for this game:

1. The coins are biased: they have probabilities $q_1$ and $q_2$ respectively to turn up head.
2. The two players have at their disposal only $m,n$ throws respectively. The throw order is random: any permutation of the $m+n$ throws has the same probability to occur. This means that there is a probability $(1-q_1)^m(1-q_2)^n$ that no one wins.

My question is: what is the probability $p_{m,n}$ that the first player wins? Unfortunately I could find a solution only for the particular case $q_1=q_2$.

My solution attempt is as follows: I condition on the number of failures (from both sides) before player 1 wins:

$p_{m,n}=\dfrac{m q_1}{m+n}\sum\limits _{m_1=0}^{m-1} \sum\limits _{m_2=0}^{n}\dfrac{\binom{m-1}{m_1}\binom{n}{m_2}}{\binom{m+n-1}{m_1+m_2}}\left(1-q_1\right) ^{m_1} \left(1-q_2\right) ^{m_2}$

I change index from $m_2$ to $M=m_1+m_2$:

$p_{m,n}=\dfrac{m q_1}{m+n}\sum\limits _{m_1=0}^{m-1} \sum\limits _{M=0}^{m+n-1}\dfrac{\binom{m-1}{m_1}\binom{n}{M-m_1}}{\binom{m+n-1}{M}}\left(1-q_1\right)^{m_1} \left(1-q_2\right)^{M-m_1}$

Now if $q_1=q_2$ the sum over $m_1$ gives just 1 and the sum over $M$ is a geometric series:

$p_{m,n}=\dfrac{m}{m+n}(1-(1-q_1)^{m+n})$

Is it possible to obtain a closed form for the general case? I also tried to set up a recurrence relation by conditioning on the outcome of the first toss:

$p_{m,n}=\dfrac{m}{m+n}(1-q_1)p_{m-1,n}+\dfrac{n}{m+n}(1-q_2)p_{m,n-1}+\dfrac{m}{m+n}q_1$

with the boudary conditions: $p_{0,n}=0$ and $p_{m,0}=1-(1-q_1)^m$, however I do not know how to solve it. Maybe using generating functions yields any useful result? Any hint and/or advice would by highly appreciated!

Answer 1

Let A be the first player with probability $p_1$ to throw a head and has n throws available and let B be the second player with probability $p_2$ to throw a head and has m throws available . Each player's throw of coming up with heads first follows geometric distribution.

and let n> m for argument sake

Then

$P(A<B)$ is the probability that A wins.

$P(A<B) = \sum_{k=1}^{n} (1-p_1)^{k-1}.p_1 \left(\sum_{i = k+1}^{m} (1-p_2)^{i-1}.p_2\right)$

$P(A=B) = \sum_{k = 1}^{\text{min(n,m)}} p_1p_2\left[(1-p_1)(1-p_2)\right]^{k-1}$

Is there something that I am missing? If you give values for n,m,$p_1$,$p_2$ and compute this, you shall verify the correctness of the solution which I will leave it to you.

I have worked out for n=10, m = 12, $p_1$ = 0.4 and $p_2$ = 0.6. Take a look at it and let me know if you have any questions.