A 6-sided cube has faces numbered 1 through 6. If the cube is rolled

twice, what is the probability that the sum of the two rolls is 8?

Why is the solution 5/36 and not 6/36? Aka, why isn’t (4,4) counted twice?

(2,6) (3,5) (4,4) (5,3) (6,2)

A 6-sided cube has faces numbered 1 through 6. If the cube is rolled

twice, what is the probability that the sum of the two rolls is 8?

Why is the solution 5/36 and not 6/36? Aka, why isn’t (4,4) counted twice?

(2,6) (3,5) (4,4) (5,3) (6,2)

when calculating probabilities, you count outcomes, not pairs. So, you count (4,4) as a single outcome where the first and second die both show 4. You don’t count it twice because you’re not looking at the order in which they were rolled; you’re just interested in the result of the two rolls.

Therefore, there are 5 favorable outcomes (pairs that sum to 8) out of the 36 possible outcomes, giving you a probability of 5/36.

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Curious does the 36 possible outcomes include (4,4) twice?

Let’s break down your statement:

Aka out of 6 children, they have a 1/6 probability of being chosen first

Correct. And if that *does* happen, we are done, as no one cares who is picked the second time - our aim has been satisfied. This itself means that the answer *must* be greater than \frac{1}{6} - which automatically invalidates your answer of \frac{1}{30}.

Can you complete the problem? As a hint, the remaining case for us to consider is when Jan is chosen second.