# Probability of Getting a Sum in Throwing Multiple Dice

In search of Mathematics prodigies, I asked different Math personalities to do tough problems. Let’s start with this hard probability problem of throwing multiple dice. The solution presented here is from John Cyril Claur from University of St. La Salle, Bacolod City. He is a quizzer and won multiple times in Mathematics contest. He was one of the national finalists of 21st Philippine Statistics Quiz.

Problem:

Dan throws six six-sided fair dice. What is the probability of having a sum of 35?

Solution I:

A die can fall in 6 ways; therefore, 6 dice can fall in $6(6)(6)(6)(6)(6)=6^6$ or $46656$ ways (total outcomes)

Since the maximum sum that we can get from throwing 6 dice is 36(which is all are 6), so, it would be easy and not so tedious for us to get all the possible outcomes for which the sum of the numbers that come up from all 6 dice is 35.

By inspection, we can only get a sum of 35 by combining five 6’s and one 5 {6, 6, 6, 6, 6, 5} (Nothing more, nothing less!)

And,

# of ways to arrange $\{6, 6, 6, 6, 6, 5\} = \displaystyle\frac{6!}{5!}$ ways

Thus, the desired probability is $\displaystyle\frac{6}{46656}$ or $\displaystyle\frac{1}{7776}$

Solution II: (Applicable to all )

First, we need to find the number of ways for which the sum of the numbers that come up from all 6 dice is 35.

The number ways of obtaining a sum of 35 is equal to the number of positive integer solutions of

$a+b+c+d+e+f=35$  where $1\le a,b,c,d,e,f\le 6$

{can take integer values from 1-6 only since a conventional six-sided die contains of numbers from 1-6 only)

Using Combination with repetition,

Let n be a positive integer such that

$x_1+x_2+x_3+\ldots+x_r=n$  where $1\le x_1,x_2,x_3,\ldots,x_r\le m$

The number of positive integer solutions is,

${n-1\choose r-1}-{r\choose 1}{n-1-m\choose r-1}+{r\choose 2}{n-1-2m\choose r-1}-{r\choose 3}{n-1-3m\choose r-1}+\ldots$

For which $n-1-mk$>$r-1$ for some positive integer k.

Back in our problem,

The number of positive integer solutions of

$a+b+c+d+e+f=35$ where $1\le a,b,c,d,e,f \le 6$

Is ${34\choose 5}-{6\choose 1}{28\choose5}+{6\choose 2}{22\choose 5}-{6\choose 3}{16\choose 5}+{6\choose 4}{10\choose 5}=6$, it means that there are 6 ways for which the sum of the numbers that come up from all 6 dice is 35.

And since the total outcome is $6^6$ or $46656$, thus, the probability of obtaining a sum of 35 from a throwing 6 dice is $\displaystyle\frac{6}{6^6}=\displaystyle\frac{1}{7776}$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

### 3 Responses

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