Fdtxjr

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

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 $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ 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 $a_5.$

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 $(1/3)^4cdot (2/3)$.