Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls

Question

I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.

My thoughts

Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find $$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$

By the binomial distribution formula, we get that $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$

Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$

Am I correct in thinking this?
If so, then it follows that the required probability is $$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$ which, I know, can be simplified further!

Answer 1

((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5

Answer 2

I think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't care what number it is otherwise.

One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.

There are thus $\dfrac{20 \cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.

In how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making

The answer is simply $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.

Answer 3

I took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls:

• $B_1$: Both 6's come in the first 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
• $B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\begin{pmatrix} 10 \\ 1\end{pmatrix} \begin{pmatrix} 10 \\ 1\end{pmatrix} = 100$ ways for this to happen.
• $B_3$: Both 6's come in the second lot of 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.

Now $$P(A|B) = P(B_2|B_1 \cup B_2 \cup B_3) = \frac{100}{45+100+45} = \frac{10}{19}.$$

Answer 4

Since all drawings are independent, is confusing to use the conditional probability formula and it is not necessary at all. Think as you are rolling simultaneously two dices 10 times each. Having in both groups exactly one 6 has the same probability p. Because these two are independent the simultaneous event (one 6 in each group) has the p^2 squared probability. The answer is what you thought A and B was: [(10C1)⋅(1/6)⋅(5/6)^9]^2 = 95,367,431,640,625/914,039,610,015,744 = 0.104336. (far to be 10/19)

This is what Adelafif said too, but he made an error on the coefficient n-k=9 (not 5).

What Paul Wright said is very funny but totally wrong. Having one 6 in ("the first") 10 rollings has the same probability independently how many time you are rolling the dice after!. This probability is (10C1)⋅(1/6)^2⋅(5/6)^8 (much less than Paul calculated). With Paul's theory, we should win a lottery pot considerably easier.