# Negative binomial problem?

• Jul 18th 2011, 01:35 AM
downthesun01
Negative binomial problem?
Suppose that two teams are playing a series of games, each of which is independently won by team A with probability p and by team B with probability 1-p. The winner of the series is the first team to win i games.

If i=4, find the probability that a total of 7 games are played. Also show that this probability is maximized when p=0.5

I thought that this is a pretty straight forward negative binomial problem, but I don't seem to be getting the correct answer. Here's my work:

\$\displaystyle pr(X=7)=\$\$\displaystyle {7-1}\choose{4-1}\$\$\displaystyle (0.5^{4})(0.5^{3})=0.15625\$

The problem is that the probability of 7 games being played doesn't seem to be maximized when p=0.5.

For example, if I set p=0.6 then I get:

\$\displaystyle pr(X=7)=\$\$\displaystyle {7-1}\choose{4-1}\$\$\displaystyle (0.6^{4})(0.4^{3})=0.16589\$

I assume that I'm reading the question wrong, using the wrong distribution, committing a math error, or some combination of the three. Any help is appreciated.
• Jul 18th 2011, 02:28 AM
islam
Re: Negative binomial problem?
hello there
actually every function of the form
m*p^4*(1-p)^3
where m is an integer has amaximum at
x=4/7
now in ur case m is abolinomial coefficient so it must be an integer
• Jul 18th 2011, 02:54 AM
downthesun01
Re: Negative binomial problem?
Thank you for replying, but I don't understand what you mean.

Is this not a negative binomial problem? Is my math wrong?
• Jul 19th 2011, 08:34 PM
downthesun01
Re: Negative binomial problem?
I figured out the problem. I wasn't taking into account that either team can win the series. Therefore it's:

\$\displaystyle {6}\choose{3}\$\$\displaystyle p^{4}(1-p)^{3}+\$\$\displaystyle {6}\choose{3}\$\$\displaystyle p^{3}(1-p)^{4}\$

Where
p is the probability of team A winning a game

and

(1-p) is the probability of team B winning a game