# Probability question?

## Two teams are racing for a playoff spot. There are 12 games left, and both teams have a 75% winning percentage. One team currently has the better record by two games (for example, team 1 is 2-0 and team 2 is 0-2). What is the probability that the team behind will catch up to the team in front to take the playoff spot?

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Feb 10, 2018

#### Answer:

I get roughly 11.75%

#### Explanation:

We have two teams, A and B. A is leading the series by 2 games with 12 games remaining. We're trying to find the probability that B will overtake B for the lead and win the series.

Both teams have a winning percentage of 75%. This means that the probability that both teams will win any given game is $\frac{3}{4}$, which also means that the probability of losing any given game is $\frac{1}{4}$.

In order for B to overtake and win, it needs to win 3 more games than A wins in the remaining 12 games (winning 2 more games will result in a tie).

Looking at each individual team for a moment, say for instance A, we can look at the total range of possible outcomes of their remaining 12 games by using the following formula:

sum_(k=0)^nC_(n,k)p^k(~p)^(n-k)

with p=3/4, ~p=1/4, and n=12

And so the total result for A is (starting from losing all 12 games all the way through to winning all 12 games):

$\left(1\right) {\left(\frac{3}{4}\right)}^{0} {\left(\frac{1}{4}\right)}^{12} + \left(12\right) {\left(\frac{3}{4}\right)}^{1} {\left(\frac{1}{4}\right)}^{11} + \left(66\right) {\left(\frac{3}{4}\right)}^{2} {\left(\frac{1}{4}\right)}^{10} + \left(220\right) {\left(\frac{3}{4}\right)}^{3} {\left(\frac{1}{4}\right)}^{9} + \left(495\right) {\left(\frac{3}{4}\right)}^{4} {\left(\frac{1}{4}\right)}^{8} + \left(792\right) {\left(\frac{3}{4}\right)}^{5} {\left(\frac{1}{4}\right)}^{7} + \left(924\right) {\left(\frac{3}{4}\right)}^{6} {\left(\frac{1}{4}\right)}^{6} + \left(792\right) {\left(\frac{3}{4}\right)}^{7} {\left(\frac{1}{4}\right)}^{5} + \left(495\right) {\left(\frac{3}{4}\right)}^{8} {\left(\frac{1}{4}\right)}^{4} + \left(220\right) {\left(\frac{3}{4}\right)}^{9} {\left(\frac{1}{4}\right)}^{3} + \left(66\right) {\left(\frac{3}{4}\right)}^{10} {\left(\frac{1}{4}\right)}^{2} + \left(12\right) {\left(\frac{3}{4}\right)}^{11} {\left(\frac{1}{4}\right)}^{1} + \left(1\right) {\left(\frac{3}{4}\right)}^{12} {\left(\frac{1}{4}\right)}^{0} = 1$

The same is true for B.

Since the results for A and B are independent, for any given set of results of the two teams, we need to multiply them.

The solution set for B to overtake A and win the series are the following sets:

B wins 3, A wins 0
B wins 4, A wins 0, 1
B wins 5, A wins 0, 1, 2
B wins 6, A wins 0, 1, 2, 3
B wins 7, A wins 0, 1, 2, 3, 4
B wins 8, A wins 0, 1, 2, 3, 4, 5
B wins 9, A wins 0, 1, 2, 3, 4, 5, 6
B wins 10, A wins 0, 1, 2, 3, 4, 5, 6, 7
B wins 11, A wins 0, 1, 2, 3, 4, 5, 6, 7, 8
B wins 12, A wins 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

for a total of 55 sets of multiplications and summing them all up (keep in mind that this is the "short way" - with 144 potential results (12 for A, 12 for B), if we were to look at this problem from the perspective of A, we'd have 89 multiplications and summing them up).

I'll do the work on a separate Numbers sheet using a table format (one axis will have A probabilities and the other axis B probabilities, with the table holding the results of the multiplications.

I get roughly 11.75%

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