Infinite probability loop in a basketball minigame Two brothers A and B challenge each other in a basketball minigame . They take turns shooting getting a point for every basket and the first to reach his target score wins. Brother A is the older and more skilled of the two and decides to give is brother a series of advantages. A shooting percentage is in fact 50% whereas B shooting percentage is 40%. The first advantage that B has is the ability to take 3 shots per turn, while A can take 2 shots per turn. The second advantage is that B target score is 5, while A target score is 7. The third and final advantage is that B will be the one starting the challenge.
What is the probability that A wins the minigame?
I know how to solve this problem when the target score is one for both and they both have one shot(there are also answers on this site for that kind of problem), but when those numbers are anything that is not equal to 1 I can't solve it. The main complication for me is the fact that there are target scores and not a final round when scores are compared, and this gives an infinite loop of both players missing continuously.
Any kind of help is apreciated
 A: Relying solely on intuition, rather than formal probability theory, I approach the problem as follows:
B can not win in less than $2$ turns.
Let $f(b)$ denote the probability that B reaches his goal in exactly $b$ turns $~: ~b \in \{2,3,\cdots \}.$
A can not win in less than $4$ turns.
Let $g(a)$ denote the probability that A has not reached his goal after $a$ turns: $~: ~a \in \{4,5,\cdots \}.$
The probability that A wins is $1$ minus the probability that B wins.
The probability that B wins is
$$f(2) + f(3) + f(4) + \sum_{k=5}^\infty \left[ ~f(k) \times g(k-1) ~\right]. \tag1 $$
Messy as the resulting computations will be, the problem has been reduced to the computations of $f(k)$ and $g(k-1).$  I will derive the necessary closed form expressions, but I will not do the actual computations.  The Math is simply too messy.
There may well be an elegant alternative approach that someone more knowledgeable in Probability Theory can derive.  Further, this alternative approach may well facilitate the routine calculation of an explicit final answer.

$\underline{\text{Computation of} ~f(k)}$
For B to reach his goal in exactly $k$ turns, two things have to happen:

*

*In $(k-1)$ turns, B must be $1$ less, $2$ less, or $3$ less than his goal.


*On turn $(k)$, B must make up the deficit.
Let $h(k,r)$ denote the probability that after $k$ turns, B is $r$ shy of his goal of $(5) ~: ~r \in \{1,2,3\}.$
Let $i(r)$ denote the probability that B scores at least $(r)$ goals on a given turn.
Then,
$$f(k) = \sum_{r = 1}^3 \left[h(k-1,r) \times i(r)\right].$$
For B to have score of $(5-r)$ after $(k-1)$ turns, B must have $(5-r)$ successes and $[3 \times (k-1)] - (5 -r) = (3k + r - 8) ~$ failures.
Therefore,
$$h(k-1,r) = \binom{3[k-1]}{5-r} (0.4)^{(5-r)} \times (0.6)^{(3k + r - 8)}.$$
In order for B to score at least $r$ goals on one turn, the number of goals that he scores must be some element in $\{r, r+1, \cdots, 3\}.$
Therefore, $~\displaystyle i(r) = \sum_{m=r}^3 \binom{3}{m} (0.4)^m \times (0.6)^{3-m}.$
Putting the components of the computation of $f(k)$ together:
$$f(k) = \sum_{r = 1}^3 \left[h(k-1,r) \times i(r)\right]$$
$$ = \sum_{r=1}^3 \left\{ ~\binom{3[k-1]}{5-r} (0.4)^{(5-r)} \times (0.6)^{(3k + r - 8)} \times \left[\sum_{m=r}^3 \binom{3}{m} (0.4)^m \times (0.6)^{3-m} \right] ~\right\}. \tag2 $$

$\underline{\text{Computation of} ~g(k)}$
In order for A to have not reached his goal in $k$ turns, A must have scored exactly $s$ shots in $2k$ attempts, $~: s \in \{0, 1,2,\cdots, 6\}.$
Therefore,
$$g(k) = \sum_{s=0}^6 \binom{2k}{s} (0.5)^{(2k)}. \tag3 $$

$\underline{\text{In Summary}}$
The probability that A wins is $1$ minus the probability that B wins.
$f(b)$ denotes the probability that B reaches his goal in exactly $b$ turns $~: ~b \in \{2,3,\cdots \}.$
$g(a)$ denotes the probability that A has not reached his goal after $a$ turns $~: ~a \in \{4,5,\cdots \}.$
The probability that B wins is
$$f(2) + f(3) + f(4) + \sum_{k=5}^\infty \left[ ~f(k) \times g(k-1) ~\right].$$

$h(k,r)$ denotes the probability that after $k$ turns, B is $r$ shy of his goal of $(5) ~: ~r \in \{1,2,3\}.$
$~\displaystyle h(k-1,r) = \binom{3[k-1]}{5-r} (0.4)^{(5-r)} \times (0.6)^{(3k + r - 8)}.$
$i(r)$ denotes the probability that B scores at least $(r)$ goals on a given turn.
$~\displaystyle i(r) = \sum_{m=r}^3 \binom{3}{m} (0.4)^m \times (0.6)^{3-m}.$
$$f(k) = \sum_{r = 1}^3 \left[h(k-1,r) \times i(r)\right]$$
$$ = \sum_{r=1}^3 \left\{ ~\binom{3[k-1]}{5-r} (0.4)^{(5-r)} \times (0.6)^{(3k + r - 8)} \times \left[\sum_{m=r}^3 \binom{3}{m} (0.4)^m \times (0.6)^{3-m} \right] ~\right\}. $$

$$g(k) = \sum_{s=0}^6 \binom{2k}{s} (0.5)^{(2k)}. $$
