The following ingredients are enough to determine the probabilities.
$1.$ If you toss $n$ times, the probability that the highest number is $\le k$ is $\frac{k^n}{6^n}$.
$2.$ If you toss $n$ times, the probability that the highest number is equal to $k$ is $\frac{k^n -(k-1)^n}{6^n}$.
In ($1$) and ($2$), $k$ ranges from $1$ to $6$.
Added: We do the cooking, in case the list of ingredients was not enough. Assume that $m\ge 1$ and $n \ge 1$.
Player A wins if (i) her highest number is a $6$, and B's number is $\le 5$ or (ii) A's highest number is $5$, and B's is $\le 4$, or (iii) A's highest number is $4$, and B's is $\le 3$, or (iv) A's highest number is $3$, and B's is $\le 2$, or (v) A's highest number is $2$, and B's is $\le 1$.
The probability of (i) is
$$\frac{6^m-5^m}{6^m}\cdot \frac{5^n}{6^n}.$$
The probability of (ii) is
$$\frac{5^m-4^m}{6^m}\cdot \frac{4^n}{6^n}.$$
The probability of (iii) is
$$\frac{4^m-3^m}{6^m}\cdot \frac{3^n}{6^n}.$$
The probability of (iv) is
$$\frac{3^m-2^m}{6^m}\cdot \frac{2^n}{6^n}.$$
The probability of (v) is
$$\frac{2^m-1^m}{6^m}\cdot \frac{1^n}{6^n}.$$
Add up.
There are various other ways to compute the probability that A wins. If we look at the probabilities that we are adding, we can see that there are opportunities to simplify the expression for the sum. If we were dealing with $d$-sided dice, then simplification might be worthwhile, but for the case $d=6$ it probably isn't.
I think there is not a general solution. Here is my solution on 4 players, rolling 4-sided dice case. Calculation is pretty complicated. I don't know if there is any better method.
First of all, we need to figure out results of each round of rolling. 4 players roll 4 dice, then we can get:
- 0 winner: $p_0=\frac{10}{64}$
- AAAA: $p_01=4*\frac{1}{4^4}=\frac{1}{64}$
- AABB: $p_02={4\choose2}\frac{\frac{4*3}{2}}{4^4}=\frac{9}{64}$
- 1 winner: $p_1=\frac{12}{64}$
- AAAB: $p={4\choose1}\frac{4*3}{4^4}=\frac{12}{64}$
- 2 winners: $p_2=\frac{36}{64}$
- AABC: $p={4\choose2}\frac{4*3*2}{4^4}=\frac{36}{64}$
- 3 winners: $p_3=0$
- 4 winners: $p_4=\frac{6}{64}$
- ABCD: $p=\frac{4*3*2*1}{4^4}=\frac{6}{64}$
$E(n)$ is the expected number of rolls to finish a game where $n$ players have not yet won at least once. So we want to know $E(4)$.
$$E(4)=1+p_0*E(4)+p_1*E(3)+p_2*E(2)$$
$$E(3)=1+p_0*E(3)+\frac{1}{4}p_1*E(3)+\frac{3}{4}p_1*E(2)+\frac{1}{2}p_2*E(2)+\frac{1}{2}p_2*E(1)$$
$$E(2)=1+p_0*E(2)+\frac{2}{4}p_1*E(2)+\frac{2}{4}p_1*E(1)+\frac{1}{6}p_2*E(2)+\frac{4}{6}p_2*E(1)$$
$$E(1)=1+p_0*E(1)+\frac{3}{4}p_1*E(1)+\frac{3}{6}p_2*E(1)$$
And I got $E(4)=6.401612$ which is not close to the simulation result. If you find out what's wrong with it please comment. Thanks a lot!
Best Answer
Well, if you have all the individual probabilities figured out, your answer would be:
$$ P(winning)=P(A>36\mid B=36)+P(A>35\mid B=35)+\cdots+P(A>4\mid B=4) $$
Since your events are independent, we have $$ P(winning)=P(A>36)P(B=36)+P(A>35)P(B=35)+\cdots+P(A>4)P(B=4) $$
Then $P(A>n)=\sum_{k=n+1}^{36} P(A=k)$ so $$ P(winning)=0\cdot P(B=36)+P(A=36)P(B=35)+\cdots+[P(A=36)+\cdots+P(A=4)]P(B=4)\\ $$
Finally, $$ P(winning)=\sum_{n=4}^{36}\left(P(B=n)\sum_{k=n+1}^{36} P(A=k)\right). $$