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(AHSME 1994) When $n$ standard six-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $S$. What is the smallest possible value of $S$?

(I've been trying to use generating function, but without success. I took this one from The Art and Craft of Problem Solving - Paul Zeitz, second ed, pag. 8, chapter 1 exercise 1.3.6.)

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up vote 4 down vote accepted

When you roll $n$ dice, the total must be an integer in the interval $[n,6n]$. For $n\le k\le 3n$, the probability of getting a total of $k$ is the same as the probability of getting a total of $7n-k$: each roll that gives you $k$ corresponds to one that gives you $7n-k$ by turning each die over. For $n>1$ these pairs are the only totals with equal probability.

In order for the total $1994$ to be possible, you must have $333\le n\le 1994$. The complementary sum $S$ will be $7n-1994$, and you want to choose $n$ to make this as small as possible. Clearly this is achieved when $n$ is as small as possible, i.e., $333$, in which case $S=7\cdot333-1994=337$.

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I understood, thanks. – Hugo C. Botós Sep 16 '12 at 20:54
Shouldn't first inequality be $n\le k\le 6n$? One-character edit is not allowed. – olek Sep 18 '12 at 14:37
@Oleksandr: Either works: when I wrote it I was thinking specifically of matching the lower number of a pair with the higher, so I used the range that covered the lower numbers only. – Brian M. Scott Sep 18 '12 at 17:56

You need $n \geq333$ to roll $1994$. Imagine a row of $6$s. You need to subtract $1$ from some of them $4$ times to get $1994$. Action of the same number of possibilities would be if we replaced $6$ with $1$ and subtracting to adding. Then the sum would be $337$.

In general, $n$ satisfies $6n > 1994$ and the sum is $n+(6n-1994) = 7n-1994$ so there is no point in considering larger $n$.

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hummm. Very nice way to solve it, i like it. thanks – Hugo C. Botós Sep 16 '12 at 21:02
Pedantry: $n$ satisfies $6n \ge 1994 \,$ rather than $6n \gt 1994$, though it makes no difference here. – Henry Sep 17 '12 at 17:37

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