# 10 dice are rolled. What is the probability of getting 6 dice that are even numbers and the other four dice are 3's?

Ten dice are rolled. What is the probability of getting six dice that are even numbers and the other four dice are $3$'s?

My approach is that you have $6^{10}$ number of total outcomes, and

• $\Pr(\text{6 dice with even numbers}) = (1/2)^6$

• $\Pr(\text{the other 4 with 3's}) = (1/6)^4$

So, the probability of the question will be $\frac{(1/2)^6 \cdot (1/6)^4}{6^{10}}$

But I am not confident that this is correct.

## 2 Answers

You've forgotten the ordering of the dice. It could be the first four dice that come up as threes. It could be the first two and the last two. And so on.

Each one of those orderings has the same probability of happening, and that probability is the one you've calculated. That means you need to multiply what you have by the number of different orderings, which is $\binom{10}4=210$.

Oh, and now that I've gotten a closer look at your formulas (since they've become well formatted), I can see that you've mixed up two different (correct) ways of thinking: either you take the probability of each die being a success and multiply them to get $(1/2)^6(1/6)^4$, or you take the number of possible good dice throws, and divide by the possible number of throws to get $\frac{3^6\cdot 1^4}{6^{10}}$. Remember to keep straight in your head which one of these you're using. They give the same answer, but do not go together well.

• First of all, thank you for your help. So based on what you are telling me, i should have 210*(1/2)^6*(1/6)^4 divided by 6^10 correct? – michael colberg Oct 9 '16 at 21:59
• No. Not divided by $6^{10}$. Read the last paragraph of my answer. – Arthur Oct 9 '16 at 22:21

Thank you Arthur for correcting me, I got a bit carried away. It's night-time here where I'm at, after all. Out of the 10 dice you roll, 4 need to be 3's (the other 6 will be even and uniquely determined, then), but there are many ways in which this condition can be satisfied. For example, you can roll $(3, 3, 3, 3, 2, 2, 2, 2, 2, 2)$, or you can roll $(3, 2, 3, 2, 3, 2, 3, 2, 2, 2).$ All in all, there will be $\binom{10}{4}$ such combinations.

The final solution, therefore, if we denote $A$ as the set of all outcomes where the conditions are satisfied, is $$P(A)={\binom{10}{4}(\frac{1}{2})^6(\frac{1}{6})^4}.$$

Edit 2: Again, thank you. I think it should be correct now, finally.