Binomial distribution (i think)

Question

This may seem like a simple question but I honestly don't really know how I would go about solving it. So essentially the question is as follows:

2 players play a match whereby the winner is the first person to win n games.

the probability of player 1 winning a single game is 0.835 the probability of player 2 winning a single game is 0.165

what is the smallest value for n such that the probability that player 1 is the overall winner of the match is at least 0.9

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I feel like i need to use binomial distribution to solve the question but i wasn't able to do so. Any help will be greatly appreciated.

Answer 1

Bram28's hints point in the right direction. Still, a more general formulation would be helpful for posterity.

Say that player 1 wins the match (i.e., they are the first to win $n$ games), and player 2 wins $k < n$ games. The probability that this happens (given that player 1 wins a game with probability $p$) is

$$\text{Prob}_{n, k}(p) = p\binom{n+k-1}{n-1} p^{n-1} q^{k}$$

We use $\binom{n+k-1}{n-1}$ rather than $\binom{n+k}{n}$ because if player 1 wins the match, they must win the final game and thus there are only the remaining $n-1$ game wins to be distributed amongst the $n+k-1$ possible game slots. The initial factor of $p$ is the probability of player 1 wining the final game; the $p^{n-1}q^{k}$ factor is the probability of player 1 wining $n-1$ and losing $k$ prior games.

The probability that player 1 is the first to win $n$ games is the sum of the above quantity over the possible values of $k$, that is

$$\text{Prob of player 1 winning match} = p^n\sum_{k=0}^{n-1}\binom{n+k-1}{n-1} q^{k}. $$

We could find a closed form expression for the summation by differentiating the identity for a finite geometric series, but here it's easier to just plug in values. For $n=1$, we find 0.835 which isn't sufficient. For $n=2$, we find

\begin{align} \text{Prob of player 1 winning match} &= (0.835)^2\left[\binom{2-1}{2-1} (0.165)^0 + \binom{2+1-1}{2-1}(0.165)^1\right]\\ & = 0.9273, \end{align}

which is.

Answer 2

With $p = 0.835, 2$ games are obviously not enough, because A would need to win both games, with a $Pr = {0.835}^2 <0.9$, so trials can begin from number of games played, $K = 3$

You would find computations much simpler if you focus on the losses of $A$ rather than wins.

For example, with $K=3$, $A$ can afford to lose at most one game, so the probability of an overall win is $\binom30q^0p^3 + \binom31q^1p^2$, which happily works out to

$\binom30 0.165^0\cdot 0.835^3 + \binom310.165^1\cdot0.835^2 = 0.9273$

thus the smallest value of n (games won) needed $=2$

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Response to queries

We are trying to use the smallest number of games $(K)$ so we can find the smallest number of wins ($n$) needed.

With $K=3$, we found $n=2$ is possible if we include more than $2$ wins in the series of $K$ games.

If we want to strictly say that more than $n$ wins can't be included in the probability calculations, the formula will change to $\binom{K}{n} p^nq^{K-n} \geq 0.9$

A little thought will show that with this formula, the Pr will never reach $0.9$, even for $n=2$, see here

So either we'll have to allow more than $n$ wins in the Pr computations, else it is impossible.