# One Dice or Two?

My son rolls two (fair) dice and sums them. I roll a single die. What is the probability that my dice roll will beat his total?

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## Solution

There are lots of ways to solve this, including walking through all possible combinations by hand; the solution space is not that large. There are only 216 (6 × 6 × 6) possible ways that three dice can be rolled.

A little thinking, however, can reduce our workload. We know the criterion is that my die should

*beat*my son's total, so that means it must be at least one greater. The lowest possible hand my son can roll is {1,1} for a total of two. This means that the only possible chance I can beat him is if I roll {3,4,5,6}. Also, if his total is six or greater, he is guaranteed a win. We only need to look at ways my son can roll five or less.Here are the different ways my son can roll totals 2–5. There are are 36 possible ways his two dice can be rolled.

Total: 2 {1,1}

Total: 3 {1,2}, {2,1}

Total: 4 {1,3}, {3,1}, {2,2}

Total: 5 {1,4}, {2,3}, {3,2}, {4,1}

*Pr*= 1/36Total: 3 {1,2}, {2,1}

*Pr*= 2/36Total: 4 {1,3}, {3,1}, {2,2}

*Pr*= 3/36Total: 5 {1,4}, {2,3}, {3,2}, {4,1}

*Pr*= 4/36## Let me count the ways …

- If I roll a three, which happens 1/6 of the time, the only total I can beat is two, which happens 1/36, so the chance of this happening is 1/6 × 1/36 = 1/216.
- If I roll a four, which happens 1/6 of the time, the two totals I can beat are two and three, which is (1/36 + 2/36). The chance of this happening is 1/6 × 3/36 = 3/216.
- If I roll a five, which happens 1/6 of the time, I can beat a two, three, or four which is (1/36 + 2/36 + 3/36). The chance of this happening is 1/6 × 6/36 = 6/216.
- If I roll a six, which happens 1/6 of the time, I can beat a two, three, four, or five which is (1/36 + 2/36 + 3/36 + 4/36). The chance of this happening is 1/6 × 10/36 = 10/216.

Adding these all together, we get the total probability as (1+3+6+10)/216 = 20/216 = 5/54 ≈ 9.26%

*The chance of a single roll beating a double roll is ≈ 9.26%*