Product of Maximum and Minimum of Dice Rolls

Question

You roll three dice. What is the expected value of the product of the minimum and the maximum value of the three dice?

Is there a nicer way to do this besides going through all cases and summing up many different values?

Answer 1

I need some more clarification, but I think I know what you mean.

I can't think up of one, but I can think of something with not a lot of casework.

First, we deal with all three are the same number case.

The sum total for this case is $1^2+2^2+3^2+4^2+5^2+6^2=\frac{6(7)(13)}6=91.$

Second, we deal with two are the same number case.

This is also relatively simple. For the two numbers we pick, we have to multiply by $2$ to find out which one is used as the different number. We also multiply by $3$ since the number that shows once can be assigned to any number. This sum is $6\cdot((1+2+3+4+5+6)\cdot(1+2+3+4+5+6)-91)=6\cdot(441-91)=6\cdot(350)=2100.$

Third, we deal with when three numbers are all different.

We might be able to settle this quickly, but note that there are $6$ ways each combination of numbers that are all different can be assigned to dice.

The total for this case is $6\cdot(1\cdot(3+4+5+6)+2\cdot(4+5+6)+3\cdot(5+6)+4\cdot6)=6\cdot(18+30+33+24)=6\cdot(105)=630.$

The total for all cases are $91+2100+630=2821.$ Dividing this by $216,$ the total number of outcomes, gives the expected value is $\boxed{\frac{2821}{216}}.$

Answer 2

One way to solve this problem, is by distinguishing three cases:

1. All dice have the same value. In this case, the sum of the six equally likely values equals:

   $$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 91$$

2. Two dice have the same value. We can distinguish five values for the minimum and, depending on the minimum, a different number of maxima. We should multiply by 2, since either the minimum or the maximum value occurs twice, and multiply by 3, since there are three ways in which the minima/maxima can be selected. We thus find:

   $$3 \cdot 2 \cdot [1 \cdot (2 + 3 + 4 + 5 + 6) + 2 \cdot (3 + 4 + 5 + 6) + 3 \cdot (4 + 5 + 6) + 4 \cdot (5 + 6) + 5 \cdot 6] = 2 \cdot 3 \cdot 175 = 1050$$

3. All dice have a unique value. We can consider four cases for the distance between the minimum and the maximum value and, depending on this distance, a different number of intermediate values. We should multiply by 6, since the three values can be ordered in $3! = 6$ ways. We thus find:

   $$3! \cdot [1 \cdot (1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + 4 \cdot 6) + 2 \cdot (1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6) + 3 \cdot (1 \cdot 5 + 2 \cdot 6) + 4 \cdot (1 \cdot 6)] = 6 \cdot 189 = 1134$$

Adding all numbers together and dividing by the total number of possible throws, the expected value turns out to be:

$$\frac{91 + 1050 + 1134}{6^3} = \frac{2275}{216} \approx 10.53$$

A second way to solve this problem, is by distinguishing six cases for the distance $\delta$ between the minimum and the maximum value:

1. $\delta = 0,$ in which case all values are the same. We find:

   $$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 91$$

2. $\delta = 1.$ There are three permutations in which the minimum value occurs twice, and three permutations in which the maximum value occurs twice, and we thus find:

   $$(3 + 3) \cdot (1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + 5 \cdot 6) = 420$$

3. $d = 2,$ in which case there is one intermediate value which results in 6 unique permutations:

   $$(3 + 6 + 3) \cdot (1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + 4 \cdot 6) = 600$$

4. $d = 3,$ in which case there are two intermediate values:

   $$(3 + 2 \cdot 6 + 3) \cdot (1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6) = 576$$

5. $d = 4,$ in which case there are three intermediate values:

   $$(3 + 3 \cdot 6 + 3) \cdot (1 \cdot 5 + 2 \cdot 6) = 408$$

6. $d = 5,$ in which case there are four intermediate values:

   $$(3 + 4 \cdot 6 + 3) \cdot (1 \cdot 6) = 180$$

The expected value then equals:

$$\frac{91 + 420 + 600 + 576 + 408 + 180}{6^3} = \frac{2275}{216} \approx 10.53$$

This result can be confirmed by running the following Python snippet:

import itertools

s = 0
for a, b, c in itertools.product(range(1, 7), repeat = 3):
    s += min(a, b, c) * max(a, b, c)
print("Total sum: %i" % s)
print("Expected value: %f" % (float(s) / 216))

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Total sum: 2275
Expected value: 10.532407