# Probablility of a Dice Game

Player A rolls $m$ dice, while Player B rolls $m + 1$ dice. If Player A rolls $a$ $n$'s and Player B rolls $b$ $n$'s, then Player A wins if $a > b$ . Otherwise, Player A rolls up to $k$ of the $m$ dice to roll again. If $a'$ of these are $n$'s then Player A wins if $a + a' > b$. If $a + a' > b$, then Player B wins.

I am wondering if $m = 3$, $n = 6$ and $k = 1$ and if it was a six sided die. What would be the probability of player B winning?

I don't know how to answer this question. All I know is it has something to do with binomial distribution. I also know that there would be a $1/6$ chance of rolling a six on a six sided die.

And I think the sum of the geometric series might have to be used as well

I think for the binomial distrbution:

$f(k;n,p) = \Pr(X = k) = {n \choose k}p^k(1-p)^{n-k}$

for k = 0, 1, 2, ..., n, where

${n\choose k}=\frac{n!}{k!(n-k)!}$

That $n = 3$ and $p = 1/6$, however I do not know what k would be

Here I assume that player B wins if $a+a'\le b$. Please correct me if I am wrong.
Well, why not letting player A roll $m+k$ dice at once? Without looking at them he just sets $m$ of them apart and then it is checked whether the number of $n$'s produced by these dice allready exceeds the number of $n$'s thrown by his opponent. Only if not then the remaining $k$ become relevant. Player A will evidently win iff his original throw with $m+k$ dice has produced more $n$'s than the $m+1$ dice thrown by his opponent. If the number of sides is denoted by $s$ and $p=s^{-1}$ then to be found is $P(B\ge A)$ where $A\sim\mathbf{Bin}\left(m+k,p\right)$ and $B\sim\mathbf{Bin}\left(m+1,p\right)$ and the random variables $A$ and $B$ are independent.