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What is the probability of rolling exactly two sixes in 6 rolls of a die?

Solution by the Binomial Probability formula is

$$\binom{6}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^4 = \frac{15 \times 5^4}{6^6} = \frac{3125}{15552} \approx 0.200939.$$

But by basic probability understanding

Probability = successful outcome/ total possible outcomes

So that way the probability should be $15/ (6^6) \approx 0.0003215.$

I know that the second calculation is not giving the right answer, but somehow am not convinced on why it is wrong as I don’t see the approach being wrong.

Can someone help me understand why the second approach is wrong?

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To get to 15, you just count the possible "locations" where the double six can occur (like: a six in the first place can be combined with a six on either 5 of the other positions etc).

However, you are forgetting that each set of locations (e.g. a six on the first two rolls) occurs more than once in you 6^6 rolls: once for each of the combinations of 4 non-six rolls of the other dice.

Something like:

6 6 1 1 1 1
6 6 1 1 1 2
6 6 1 1 1 3

etc.

Of course, for each location (again, I mean 2 "positions of the sixrolled dice), there are 5^4 possible non-six rolls for the remaining 4 rolls...

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