probability problem about the probability that Team A will win this game

**notatall** · January 19th 2011, 07:36 PM

Hi,

A very simple prob question. Hope sb could help:

Two basketball teams: team A makes it shots 48% of the time and team B makes its shots 52% of the time. The outcome of the each shot is independent of all other shots and that the team who makes the most shots during the game wins. If each team takes 100 shots in a game, what is the probability that team A wins this game?

Please explain why. Thanks in advance.

**Chris L T521** · January 19th 2011, 08:05 PM

Originally Posted by notatall

Hi,

A very simple prob question. Hope sb could help:

Two basketball teams: team A makes it shots 48% of the time and team B makes its shots 52% of the time. The outcome of the each shot is independent of all other shots and that the team who makes the most shots during the game wins. If each team takes 100 shots in a game, what is the probability that team A wins this game?

Please explain why. Thanks in advance.

Let $X$ be the random variable that represents the number of shots made by team A. Then $X$ has a binomial distribution, in particular $X\sim\mathcal{B}(100,.48)$ .

What we're after here is $P(X\geq51)$ . But that's the same as saying $1-P(X\leq 50)$ .

Now recall that if $X\sim\mathcal{B}(n,p)$ then $\displaystyle P(X\leq x) = \sum\limits_{k=0}^x {n\choose k}p^k(1-p)^{n-k}$

Now, use this formula to get the answer.

Does this make sense? Can you finish this problem?

**chisigma** · January 19th 2011, 08:11 PM

For both teams the probability of 'k over n' shots is...

$\displaystyle P_{A} (k)= \binom{n}{k} p_{a}^{k}\ (1-p_{a})^{n-k}$

$\displaystyle P_{B} (k)= \binom{n}{k} p_{b}^{k}\ (1-p_{b})^{n-k}$

If we set $n=100$ , $p_{a}=.48$ , $p_{b}=.52$ the probability that A wins is...

$\displaystyle P_{A wins}= \sum_{i=1}^{100} P_{A} (i)\ \sum_{k=0}^{i-1} P_{B} (k)$

Kind regards

$\chi$ $\sigma$

**notatall** · January 20th 2011, 04:40 AM

Yes. It makes perfect sense to me. Thanks a lot.

**notatall** · January 20th 2011, 04:43 AM

Thanks

I will compare your solution with Chris's method and let you know if they agree.

**notatall** · January 20th 2011, 03:27 PM

Hi Chris,

Thanks for your answer. But I think the way you compute the probability that team A wins has problem. Since the team which makes the most shots wins, you have to consider the number of shots that team B makes when team A makes certain number of shots. Once team B makes less shots than team A, team A wins the game. So I guess chisigma's answer makes more sense. But thanks anyway

**matheagle** · January 20th 2011, 03:29 PM

I was trying to clean up the sum, but it's a mess.

$\sum_{i=1}^{100}{100\choose i} (.48)^i(.52)^{100-i} \sum_{k=0}^{i-1}{100\choose k}(.52)^k(.48)^{100-k}$

**notatall** · January 20th 2011, 03:30 PM

Hi,

I have done this problem in the way you suggested and the probability that team A wins is 0.2620. I am wondering if there is an easy way to get the answer since

\displaystyle P_{A wins}= \sum_{i=1}^{100} P_{A} (i)\ \sum_{k=0}^{i-1} P_{B} (k)

is hard to compute. I used Matlab to get the answer.

Thanks, it really helps

**notatall** · January 20th 2011, 03:33 PM

Thanks

I think Chrisigma is correct and I have to use Matlab to actually get the final answer.

**chisigma** · January 21st 2011, 11:39 AM

The computation of...

$\displaystyle P_{Awins} = \sum_{i=1}^{n} \binom{n}{i}\ p_{a}^{i}\ (1-p_{a})^{n-i}\ \sum_{k=0}^{i-1} \binom{n}{k}\ p_{b}^{k}\ (1-p_{b})^{n-k}$ (1)

... requires only a minimal knowledge of computer programming. Setting $p_{a}= .48$ and $p_{b}= .52$ I obtained...

$n=10 \rightarrow P_{Awins} = .342510814481$

$n=100 \rightarrow P_{Awins} = .261961725783$

$n=1000 \rightarrow P_{Awins} =.0349809721915$

If necessary greater value of n can be examined...

Kind regards

$\chi$ $\sigma$

**matheagle** · January 21st 2011, 11:47 AM

fine, I have less than mininal knowledge of computing, lol

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