Let's imagine a sport in which two archers shoot at the same time on two different targets, the speed of shooting is at the same rate for each archer. The targets are selected before each competition individually by each archer. These targets vary in size, color and number of points, needed to be scored for the win by hitting it; that's such a strange game. In order to win, the archer must hit the target required times before before a similar is done by the opponent. If such conditions hold for each archer at the same time then we have a draw.
The 1st archer chose the 1st target; it is necessary to hit the 1st target 10 times to win, the probability of hitting the 1st target by 1st archer is 75%.
The 2nd archer chose the 2nd target; it is necessary to hit the 2nd target 12 times to win, the probability of hitting the 2nd target by the 2nd archer is 86%.
What are the coefficients to win the game by the 1st archer, by the 2nd or to get a draw?
Thanks in advance for the help!
Best Answer
Call player 1 $A$, and has a probability of winning $a = 0.75$
Call player 2 $B$, and has a probability of winning $b = 0.86$
Now, the recurrences for winning in exactly $n$ steps:
\begin{align*} P_A\left(n\right) &= \sum_{i=1}^{10} \left(1-a\right)a^{i-1}P\left(n-i\right) \\ P_A\left(10\right) &= a^{10} \\ P_A\left(k\right) &= a^{10}\left(1-a\right)\hspace{20pt}\text{for $k=11\ldots 20$} \end{align*}
and
\begin{align*} P_B\left(n\right) &= \sum_{i=1}^{12} \left(1-b\right)b^{i-1}P\left(n-i\right) \\ P_B\left(12\right) &= b^{12} \\ P_B\left(k\right) &= b^{12}\left(1-b\right)\hspace{20pt}\text{for $k=13\ldots 24$} \end{align*}
Hence, the probability of a draw is:
\begin{align*} \mathbb{P}\left(\text{Draw}\right) &= \sum_{n=10}^{\infty} P_A\left(n\right)\cdot P_B\left(n\right) \\ &\approx 0.0108907854748 \end{align*}
Probability that $A$ wins is: \begin{align*} \mathbb{P}\left(A<B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n>k\right) \\ &\approx 0.328106229131 \end{align*}
Probability that $B$ wins is: \begin{align*} \mathbb{P}\left(A>B\right) &= 1- 0.328106229131 - 0.0108907854748 \\ &= 0.6610029853942 \end{align*}
The probabilities $P_A(n)$ and $P_B(n)$ have the following probability generating functions, which can be obtained from a markov chain, or from the recurrences. They were used for calculating the above probabilities:
\begin{align*} G_A\left(x\right) &= \frac{a^{10} x^{10}}{{\left(a^{10} - a^{9}\right)} x^{10} + {\left(a^{9} - a^{8}\right)} x^{9} + {\left(a^{8} - a^{7}\right)} x^{8} + {\left(a^{7} - a^{6}\right)} x^{7} + {\left(a^{6} - a^{5}\right)} x^{6} + {\left(a^{5} - a^{4}\right)} x^{5} + {\left(a^{4} - a^{3}\right)} x^{4} + {\left(a^{3} - a^{2}\right)} x^{3} + {\left(a^{2} - a\right)} x^{2} + {\left(a - 1\right)} x + 1} \\ G_B\left(x\right) &= \frac{b^{12} x^{12}}{{\left(b^{12} - b^{11}\right)} x^{12} + {\left(b^{11} - b^{10}\right)} x^{11} + {\left(b^{10} - b^{9}\right)} x^{10} + {\left(b^{9} - b^{8}\right)} x^{9} + {\left(b^{8} - b^{7}\right)} x^{8} + {\left(b^{7} - b^{6}\right)} x^{7} + {\left(b^{6} - b^{5}\right)} x^{6} + {\left(b^{5} - b^{4}\right)} x^{5} + {\left(b^{4} - b^{3}\right)} x^{4} + {\left(b^{3} - b^{2}\right)} x^{3} + {\left(b^{2} - b\right)} x^{2} + {\left(b - 1\right)} x + 1} \end{align*}
If you want an exact solution and have enough patience, proceed as described in the similar problem here.
Update
Alright, here are the exact values!
Building the simultaneous equations as described in that linked answer, we will end up with 120 equations of 120 variables.
This is the required equation: $$ A_{m,n} = a\, b\, A_{m+1,n+1}+a\, (1-b)\, A_{m+1,0}+(1-a)\, b\, A_{0,n+1}+(1-a)\, (1-b)\, A_{0,0} $$ and use suitable conditions for the $A$ and $B$ to get the simultaneous equations.
On solving, we end up with these fractions:
Probability that $A$ wins:
Probability that $B$ wins:
Probability of a draw:
Update 2
I read the question again. It's not mentioned that they must be consecutive hits!
Anyway, the equation won't change much:
$$ A_{m,n} = a\, b\, A_{m+1,n+1}+a\, (1-b)\, A_{m+1,n}+(1-a)\, b\, A_{m,n+1}+(1-a)\, (1-b)\, A_{m,n} $$
and set the boundary conditions.
Therefore, without the constraint of consecutive hits:
Probability that $A$ wins:
$\approx 0.538220344220283$
Probability that $B$ wins:
$\approx 0.311708148717965$
Probability of a draw:
$\approx 0.150071507061752$
Update 3
We may also compute it from the summations, by fixing the last shot as a hit. \begin{align*} \mathbb{P}\left(A<B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n>k\right) \\ &= \sum_{k=10}^{\infty} \binom{k-1}{9}a^{10}(1-a)^{k-10} \cdot \left(\sum_{i=k+1}^{\infty} \binom{i-1}{11} b^{12}(1-b)^{i-12}\right) \\ &= \frac{534150450266477653874216085760600428121292900457}{992438238358118763882506280744260200076845920193} \end{align*}
\begin{align*} \mathbb{P}\left(A=B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n=k\right) \\ &= \sum_{k=10}^{\infty} \binom{k-1}{9}a^{10}(1-a)^{k-10} \cdot \binom{k-1}{11} b^{12}(1-b)^{k-12} \\ &= \frac{148936702096113277233830214165033296478353815143}{992438238358118763882506280744260200076845920193} \end{align*}