Puzzling Probability Part 2

This is part 2 of a series, in which I am exploring some interesting probability puzzles.

Imagine a typical dice game. You roll 2 dice and compute their sum. For example, the first die is a 4 and the second die is a 3. The sum is 4 + 3 = 7. From our everyday experience with rolling dice, we know that 7 comes up most often, and numbers like 3 or 11 are rarer.

Now imagine a pair of strange dice, with the following numbers:

Die 1: 1, 3, 4, 5, 6, 8

Die 2: 1, 2, 2, 3, 3, 4

Is rolling these two dice is the same as rolling 2 standard dice? Or better yet, how could we determine if the strange dice are equivalent to standard dice?

The first idea I had was to add up all the numbers on both dice:

(1 + 3 + 4 + 5 + 6 + 8) + (1 + 2 + 2 + 3 + 3 + 4) = 42

If you add up all the numbers on two standard dice, you also get a sum of 42. At first, I thought this might be enough to confirm that the strange dice were equivalent. However, consider the following two dice:

Die 3: 6, 6, 6, 6, 6, 6

Die 4: 1, 1, 1, 1, 1, 1

The sum of the all the numbers on the above two dice is clearly 42. But the only number that comes up when rolling both is 7. For an alternate pair of dice to be equivalent to the standard dice, there should be a way of rolling a 2, or a 12, or the other numbers typically rolled from standard dice.

Creating a table is very helpful for eliciting the answer in this particular situation. Here are all the possible sums for two standard dice:

dice-1

The numbers along the outside are the standard dice numbers and the numbers on the inside of the grid are the sums. If we count up the occurrences of each sum, we have the following:

dice-2

Now, we can repeat the same process with the strange dice:

dice-3

dice-4

How interesting! The strange dice have the same sums with the same frequency as the standard dice. However, I doubt I could get away with using these dice at a casino, even if they are mathematically equivalent.

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