# Thread: Solving for n in binomial distribution

1. ## 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,

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

which looks daunting. Perhaps I could use Markov chains in some way?
Any hints are appreciated!

2. Hello, Mollier!

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

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 $\displaystyle \,A$ be the stronger team.

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

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

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

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

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

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

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

3. 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?

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### formula for binominal distribution to solve for n

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