The average outcome of the sum of a pair of dice and the higher sum of two pairs

Question

I'm helping my son to design a game using dice. He wants to figure out the expected outcome of the sum of a pair of $6$-sided dice, if $2$ pairs of dice are thrown and the higher pair is kept.

I get as far as the expected outcome of a single pair of dice is $7$ by calculating the weighted average of all $36$ outcomes.

Where do we go from here, to account for taking the higher of $2$ pairs of $6$-sided dice?

Answer 1

I can see several ways to handle this.

1. Look at all $6^4=1296$ possible values of the four dice and average these values. No weights are necessary since the outcomes are equally likely.
2. Look at all $11^2=121$ possible sums of the two dice. Each pair of dice can sum from $2$ through $12$, which is $11$ possible values. You would then calculate the weighted average of these outcomes. Weights are necessary here.
3. Look at all $11$ possible final outcomes, $2$ through $12$, and take the weighted average of these outcomes. For example, the final total $2$ happens only when both pair of dice sum to $2$, with probability $\frac 1{36}\cdot\frac 1{36}=\frac 1{1296}$. The final total $7$ happens when the first pair sums to $7$ and the second pair sums to $7$ or less, with probability $\frac 16\cdot\frac{21}{36}=\frac 7{72}$, or when the first pair sums to less than $7$ and the second pair sums to $7$, with probability $\frac{15}{36}\cdot\frac 16=\frac 5{72}$. These add up to probability $\frac 16$. You could also calculate this as twice the probability that one pair sums to $7$ and the other sums to at most $7$ less the probability that both pair sum to $7$, which is $2\cdot\frac 16\cdot\frac{21}{36}-\left(\frac 16\right)^2=\frac 16$. Clearly weights are needed here and finding them is the hard part.

Option three seems the easiest here if you are doing this by hand, option one is easiest if you use a computer program or spreadsheet.

Doing option 1 on a spreadsheet, I get the value $\frac{10850}{1296}=\frac{5425}{648}=8+\frac{241}{648}\approx 8.37191358$. I get the same answer doing option 2 and option 3 on a spreadsheet. Option 3 was easier than I expected: let me know if you want a copy of the quick-and-dirty Excel spreadsheet I made that does all three options.

Answer 2

He wants to figure out the expected outcome of the sum of a pair of dice(6 sided) if two pair of dice are thrown and the higher pair is kept.

$$\begin{align}\mathsf E(\max(X+Y, U+V)) & = \frac 1 {6^4} \sum_{x=1}^6 \sum_{y=1}^6 \sum_{u=1}^6 \sum_{v=1}^6 \max(x+y, u+v) \\ & = \dfrac{5425}{648} \\ & \approx 8.37191\ldots \end{align}$$

Answer 3

Use fact that cdf of max is product of cdfs.

In R:

> die = c(1,1,1,1,1,1)
> dice = convolve(die,die,type="o")
> dice
 [1] 1 2 3 4 5 6 5 4 3 2 1
> dice.pdf = dice/36
> dice.pdf
 [1] 0.02777778 0.05555556 0.08333333 0.11111111 0.13888889 0.16666667
 [7] 0.13888889 0.11111111 0.08333333 0.05555556 0.02777778
> dice.cdf = cumsum(dice.pdf)
> dice.cdf
 [1] 0.02777778 0.08333333 0.16666667 0.27777778 0.41666667 0.58333333
 [7] 0.72222222 0.83333333 0.91666667 0.97222222 1.00000000
> max.cdf = dice.cdf^2
> max.cdf
 [1] 0.0007716049 0.0069444444 0.0277777778 0.0771604938 0.1736111111
 [6] 0.3402777778 0.5216049383 0.6944444444 0.8402777778 0.9452160494
[11] 1.0000000000
> ev = 2 + sum(1-max.cdf)
> ev
[1] 8.371914

You can do that with pencil and paper by keeping those as integers until the end:

1 2 3 4 5 6 5 4 3 2 1

Do cumulative sum:

1 3 6 10 15 21 26 30 33 35 36

Square that:

1 9 36 100 225 441 676 900 1089 1225 1296

Subtract from 1296:

1295 1287 1260 1196 1071 855 620 396 207 71 0

Add those to get 8258. Then

2 + 8258/1296 = 5425/648 ≈ 8.371914

Answer 4

A short Python script might be useful here:

sum = 0

for i in [1,2,3,4,5,6]:
    for j in [1,2,3,4,5,6]:
        for m in [1,2,3,4,5,6]:
            for n in [1,2,3,4,5,6]:
                sum += i+j if i+j > m+n else m+n

print float(sum) / 6**4

The result is $8.37191358025$.