# Top 3 of 4 Dice Rolls

I'm trying to prove why the mean of the distribution of sums of the top 3 out of 4 fair 6 sided dice is rolls 12.25. Anybody who's rolled a D&D character knows the idea.

$r_n = Rand([1,6])$

$x = \frac{\sum_{i=1}^4{r_i} - min(r_i)}{3}$

Pardon the notation, I wasn't sure how to properly define the problem.

So, I came to derive 12.25 with a computer program that just does several million iterations and comes up with something that's approaching 12.25. I just don't know why or how to prove it. I thought of splitting the interval [1,6] into 4 equal subsets and add the midpoint of the top 3. But that didn't work. Can someone explain why it's 12.25 and how to prove it?

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If you want to use a computer, you can enumerate all $6^4=1296$ rolls and calculate it exactly. –  Ross Millikan Nov 8 '11 at 1:34

We use the same idea as Ross Millikan. In order to bring out the structure, there will be as little calculation as possible.

Let the random variable $X$ denote the smallest roll, and let $Y$ denote the sum of the three larger rolls. Then $$E(X)+E(Y)=E(X+Y)=14.$$ We calculate $E(X)$. From this, $E(Y)$ is easily found.

Let $q_1$ be the probability that $X\ge 1$, let $q_2$ be the probability that $X\ge 2$, and so on. Then $P(X=1)=q_1-q_2$, $P(X=2)=q_2-q_3$, and so on until $P(X=6)=q_6$. Thus $$E(X)=1\cdot(q_1-q_2)+2\cdot(q_2-q_3)+3\cdot(q_3-q_4)+4\cdot (q_4-q_5)+5\cdot (q_5-q_6) +6q_6.$$ This simplifies to $$q_1+q_2+q_3+q_4+q_5+q_6.$$ But $$q_i=\frac{(7-i)^4}{6^4},\qquad\text{and therefore}\qquad E(X)=\frac{1^4+2^4+3^4+4^4+5^4+6^4}{6^4}.$$

Comment: Let $W$ be a random variable that only takes on integer values. For any positive integer $n$, let $q_n=P(W\ge n)$. Then $$E(W)=\sum_{n=1}^\infty q_n,$$ provided the sum is defined.

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Ahh, yes, this look about right. Except how are you defining $q_i$? –  dharga Nov 8 '11 at 22:09
@dharga: In the main post, and in the appendix, $q_i$ is the probability that our random variable is $\ge i$. That's explained at the beginning of the third paragraph. "Let $q_1$ be the probability that $X\ge 1$, let $q_2$ be the probability that $X\ge 2$, and so on." –  André Nicolas Nov 8 '11 at 22:20
Sorry, I got that part. You define it as $q_i = \frac{(7-i)^4}{6^4}$. My probability is quite rusty, how did you get to that? –  dharga Nov 9 '11 at 14:06
For concreteness, pick $i=5$. The probability the minimum is $\ge 5$ is the same as the probability that all tosses are $\ge 5$. The probability of $\ge 5$ on one toss is $2/6$. For all $4$ tosses $\ge 5$, probability is $(2/6)^4$. –  André Nicolas Nov 9 '11 at 14:43

One approach is to find the number of rolls with each number as the lowest. There is only one roll with $6$ the minimum. There are $2^4-1=15$ with $5$ the minimum. This continues to the fact that there are $6^4-5^4=1296-625$ rolls with $1$ the minimum. The total of all the dice is $1296*4*3.5=18144$ The thrown out dice are $6*1+5*15+4*65+3*175+2*369+1*671=2275$ so the total score after throwouts is $15869$, giving an average of about $12.2446$

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There are $6^4$ rolls for which the minimum is at least $1$, $5^4$ rolls for which the minimum is at least $2$, and so on. Thus the sum over all the minima is $1^4+2^4+3^4+4^4+5^4+6^4=2275$. It's a bit of an overkill, but if you like you can also calculate this sum using the general formula for the sum of the first $n$ fourth powers:

$$\sum_{k=1}^6k^4=\left.\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\right|_{n=6}=\frac{6\cdot7\cdot13\cdot125}{30}=2275\;.$$

As Ross shows in his answer, this leads to an average of about $12.2446$.

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