# Solving for n in binomial distribution

• December 6th 2010, 12:34 AM
Mollier
Solving for n in binomial distribution
Hi,

I just came over a little problem that has me puzzled.

If we have two teams, where one of the teams has a 55% chance of winning a match, how many times must the two teams meet such that we can be 95% certain that the best team wins?

What pops into my mind is the binomial distribution, but then I would have to solve the following equation for n,

$\sum^{(n/2+1)}_{k=0}\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} > 0.95$,

which looks daunting. Perhaps I could use Markov chains in some way?
Any hints are appreciated!
• December 6th 2010, 05:14 AM
Soroban
Hello, Mollier!

The wording of the problem is quite fuzzy.
I'll give you my interpretation.

Quote:

If we have two teams, where one of the teams has a 55% chance of winning,
how many times must the two teams meet so that we can be 95% certain
that the better team wins at least one game?

Let $\,A$ be the stronger team.

Then $P(\text{A wins}) \:=\:0.55\quad\Rightarrow\quad P(\text{A loses}) \:=\:0.45$

The probability that $\,A$ loses $\,n$ games is: . $(0.45)^n$
. . and we want this to be less than 5%.

So we have: . $(0.45)^n \:<\:0.05$

Take logs: . $\ln(0.45^n) \:<\:\ln(0.05) \quad\Rightarrow\quad n\ln(0.45) \:<\:\ln(0.05)$

Divide by $\ln(0.45)$, a negative quantity: . $n \;>\;\dfrac{\ln(0.05)}{\ln(0.45)}$

. . and we have: . $n \:>\:3.751663619 \quad\Rightarrow\quad n \;\ge 4$

Therefore, if they play 4 or more matches,
. . we can be 95% certain that $\,A$ will win a game.

• December 6th 2010, 05:18 AM
Mollier
I'm so sorry, I translated it wrong.
The problem asks for the number of matches that need to be played such that we can be 95% certain that the best team has won most matches. Is that better?