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 Question from Jeremy, a teacher: If rolling two six-sided dice, what is the probability that a sum of 5 will be rolled before a sum of 11?

We have two responses for you

Hi Jeremy.

The chances of rolling an 11 on any particular casting of the dice is 2 in 36 (2 because you could roll 6 and 5 or 5 and 6; 36 because that is 6 possibilities of the first die and 6 for the second die).

Similarly, the chances of rolling a 5 on any particular casting is 4 in 36 (1,4; 4,1; 2,3; 3,2).

So on a given roll, it is twice as likely to be a 5 than an 11.

Now consider many rolls. We stop when a 5 or an 11 appear. The second roll also has the same behaviour. The third roll, as well. In fact, it is always twice as likely.

Thus, we'd expect to get a 5 first twice as often, so that the probability that a sum of 5 appears before a sum of 11 is 2/3.

Cheers,
Stephen La Rocque.

Hi Jeremy,

Here is a more formal argument with the same result.

As Steve says the probability of rolling a 5 is 4/36 and the probability of rolling ian 11 is 2/36 so the probability of rolling neither is 1 - (2/36 + 4/36) = 30/36. So how can you roll a 5 before an 11?

1. You could roll a 5 on the first roll: probability 4/36.

2. You could neither on the first roll and then a 5 on the second roll: probability 30/36 × 4/36.

3. You could roll neither on the first two rolls and then a 5 on the third: probability (30/36)2 × 4/36.

4. You could roll neither on the first three rolls and then a 5 on the fourth: probability (30/36)3 × 4/36.

The pattern continues so the probability you roll a 5 before an 11 is

4/36 + 30/36 × 4/36 + (30/36)2 × 4/36 + (30/36)3 × 4/36 + ···
= 4/36(1 + 30/36 + (30/36)2 + (30/36)2 + (30/36)3 + ···)

This is 4/36 times a geometric series with a = 1 and r = 30/36. Thus the probability you roll a 5 before an 11 is

4/36 × (1/( 1 - r)) = (4/36) × (36/6) = 4/6 = 2/3.

Harley

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