Probability high school math competition problem

Question

Team A and Team B are playing basketball. Team A starts with the ball, and the ball alternates between the two teams. When a team has the ball, they have a 50% chance of scoring 1 point. Regardless of whether or not they score, the ball is given to the other team after they attempt to score. What is the probability that team A will score 5 points before Team B scores any? So far I got (1/2)^9 + (1/2)^11 * ((6 choose 1) - 1) + (1/2)^13 * ((7 choose 2) - 1) + ... I don’t know how to simplify this infinite series or maybe there is better way to solve it

Answer 1

There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.

There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore, $$ a_n=\frac14a_{n-1}+\frac14a_n,\tag1 $$ holds for all $n\ge 2$. However, $a_1$ behaves a little differently; you instead have $$ a_1=\frac12+\frac14 a_1\tag2 $$ $(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$

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Here is a more direct method. Initially, there are four possibilities for how the first two shots go:

1. Both teams make it.
2. A misses, B makes it.
3. A makes it, B misses.
4. Both teams miss.

If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.

So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4\cdot (2/3)$.