Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(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.)

share|improve this question
add comment

2 Answers

up vote 3 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 $S=7\cdot333-1994=337$.

share|improve this answer
    
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. –  Oleksandr Kozlov 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
add comment

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 $n$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.