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Consider two players A and B.

Player A rolls a fair, six-sided die $m$ times and notes the highest number on the upper face out of all of the rolls.

Player B rolls the same die $n$ times and notes the highest number on the upper face out of all of the rolls. (It's clear that as $m$ and $n$ increase, the score of 6 becomes more likely).

If $m>n$, what is the probability player A will win in terms of $m$ and $n$? Note that if the numbers are equal, player B wins.

(I know the total number of outcomes is $6^{m+n}$, but I need help counting the winning games).

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I take it by "A's number" and "B's number" you mean M and N, respectively? And the part about B winning if the numbers are equal isn't something that follows from what was previously said, but is a new condition? –  Gerry Myerson Mar 21 '12 at 5:56
    
Sorry, if I wasn't clear. A's number is the largest number she receives out of all M rolls. Obviously, as M increases, she will more likely receive a score of 6. B's number follows the same idea. –  Max Lipton Mar 21 '12 at 6:02
    
@MaxLipton just as a note: the probability of getting a 6 is constant throughout. It does not increase or decrease. What you probably mean is that the probability that a 6 has been rolled increases as $m$ and $n$ increase. –  E.O. Mar 21 '12 at 7:03
    
That's what I meant, because her score is the highest number obtained out of all of her rolls. –  Max Lipton Mar 21 '12 at 7:08
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1 Answer

up vote 4 down vote accepted

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.

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For (1), I should be only interested in the probability that the highest number is greater than K, not just greater than or equal to. –  Max Lipton Mar 21 '12 at 6:19
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Which is the reason why André gave you the probability that the highest number is lower than or equal to $k$. You should be able to deduce the probability that the highest number is greater than $k$. –  Did Mar 21 '12 at 7:05
    
Out of curiosity, could you explain why the second rule works? –  Max Lipton Mar 23 '12 at 7:40
    
@Max Lipton: How can the largest toss be $k$? Precisely if all tosses are $\le k$, but not all tosses are $\le k-1$. To be concrete, let's count the number of ways max in $n$ tosses is $5$. There are $5^n$ ways to have all the tosses $5$ or below. There are $4^n$ ways for them to be $4$ or below. So $5^n-4^n$ counts number of ways everybody is $5$ or under, but not everybody is $4$ or under. This means nobody bigger than $5$, but some people above $4$, which means biggest number we got is $5$. –  André Nicolas Mar 23 '12 at 11:55
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