# Conditional Probability Question relating to two dice.

I'm taking a beginner Statistics class, and I'm stumped by a conditional probability example. Here is the question:

Roll two dice. If one die is a 6, what is the probability that the total of the dice is 10?

P(A) = Rolled a 6

P(B) = Total is 10

P(B|A) = ???

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Intuitive answer: If one die is a 6, the only way they add up to 10 is if the other die is 4. There is a 1/6 chance of rolling a 4. Therefore: P(B|A) = 1/6

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Formulaic answer:

$P(B|A) = P(B \cap A)/P(A)$

$P(B \cap A) = 2/36$ (out of 36 combinations of dice, there are only 2 that have a 6 and add up to 10: 6 and 4 or 4 and 6).

$P(A) = 11/36$

$P(B|A) = P(B \cap A)/P(A) = (2/36)/(11/36) = 2/11$

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The intuitive answer (1/6) and the formulaic answer (2/11) are very close, but not the same. Can you help me understand why?

We haven't yet gotten to Baye's Theorem. Maybe Bayes helps explain it?

## 1 Answer

If one die is a 6, the only way they add up to 10 is if the other die is 4.

You don't know which die is the six and there is a chance that they both are.   It is this common outcome that is throwing off your intuition, since it means there are not six equally probable results for the "other die".

There are eleven equally probable ways two die can roll such that "at least one is a '6'" and two of these outcomes have the other dice a '4'.

• Ah, thanks! I inferred "the first die is a 6" when that wasn't implied by the question. I see how that could very slightly affect the odds. – Michael Cornn Oct 6 '17 at 1:24