# Binomial distributions: Probability that player $A$ wins $x$ games

I think this is a binomial question because it involves success/failure (win/lose) but it doesn't give that information. I'm finding this question difficult because it asks for the probability of winning $x$ games. I'm more comfortable with examples using actual numbers rather than "$x$ games".

Q: There is a $7$ game tournament between player $A$ and $B$. Player $A$ has $0.6$ chance of winning each game. Find the probability player $A$ wins the tournament in $x$ games.

So if its binomial then I'll have $p=0.6$ and $q=0.4$. Player $A$ would need to win $4$ games before s/he has won the tournament so $x$ must be $x=4,5,6,7$. So what is $n$? How do I write out my pdf for this problem? Thanks very much.

Hint: I think this is related to negative binomial distribution.
Probability of $n^{th}$ trial being $k^{th}$ success = $\binom{n-1}{k-1}p^k(1-p)^{n-k}$
Try and see if you can find the solution. Else I will write the rest.

• Negative binomial - of course! I couldn't figure out why the answer in the back of the book had n equal to x-1. Negative binomial explains it because the formula is n-1 rather than n like the formula for binomial. Answer in the book is (n−1 3)(.6)4(.4)x-4 – Helena Shaw Feb 27 '14 at 9:30
• That was supposed to be (x−1 3)(.6)4(.4)x-4 – Helena Shaw Feb 27 '14 at 9:36

It uses the Pascal Distribution.

• Yup, thats another name for negative binomial distribution. – Swapniel Feb 27 '14 at 9:06

It is not binomial, as the probability of player $A$ winning in $3$ games is $0$.

I suggest you do this task by considering values of $x=4,5,6,7$ separately. For each $x$, you need the probability of winning in $4$ out of $x$ games is simple to calculate. The events "winning after $x$ games" are also disjoint, so you can simply sum their probabilities.