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?


1 Answer 1


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


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...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.