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!

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    $\begingroup$ Shouldn't $P[B]$ be $$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$ ? $\endgroup$ – Mohamad Misto Jun 4 '15 at 14:45
  • $\begingroup$ A curiosity is that in this case $P(B \mid A)=P(A)$ so $P(A \cap B)=P(A^2)$ and $P(A \mid B)=\dfrac{P(A)^2}{P(B)}$ $\endgroup$ – Henry Jun 4 '15 at 14:59
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    $\begingroup$ Well laid out. Apart from a minor slip, all correct. You will notice near total collapse when you simplify. Now can you simplify the argument? $\endgroup$ – André Nicolas Jun 4 '15 at 14:59
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    $\begingroup$ You should get $10/19$. $\endgroup$ – André Nicolas Jun 4 '15 at 15:26
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    $\begingroup$ There are $\binom{20}{2}$ equally likely ways to choose the locations of the two $6$'s. There are $(10)^2$ ways to choose one in each half. So the probability is $100/\binom{20}{2}$. $\endgroup$ – André Nicolas Jun 4 '15 at 15:29

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

  • $\begingroup$ Your reasoning seems sound to me. It doesn't agree with all of the other answers, though. I am not yet sure why... $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 15:24

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 $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.

  • $\begingroup$ Surely, when considering the number of ways in which we can obtain exactly one $6$ in the first ten rolls, we must account for the fact that we obtain a $6$ in the second set of ten rolls, too. In that case, the number of ways of rolling exactly one $6$ in the first ten rolls AND exactly one $6$ in the second ten rolls is $10^2 = 100.$ Is my reasoning sound? $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 15:20
  • $\begingroup$ @CKKOY there are 100 ways you can roll exacly one 6 in the first 10 rolls and exacly 1 six in the next 10 rolls but that's not what the question is asking. The question is asking how many ways you get exactly 1 six in the first 10 rolls given all the ways you can get 2 6's in 20 rolls including one where both 6's occur in the first 10 rolls or both 6's occur in the last 10 rolls. $\endgroup$ – Warren Hill Jun 4 '15 at 15:26
  • $\begingroup$ Is your interpretation of the question equivalent to the following? 'The question is asking how many ways you can roll exactly one $6$ in the first ten rolls (out of twenty), given that you rolled two $6$s in the total twenty rolls.' Please see the other answers and comments, which imply that the correct answer is $10/19$. $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 15:30
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    $\begingroup$ @CKKOY sorry you are correct. I've corrected my answer. $\endgroup$ – Warren Hill Jun 4 '15 at 15:35
  • $\begingroup$ Apology accepted. $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 15:40

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.

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    $\begingroup$ Your answeris wrong, and $10\over 19$ is the correct one. According to mathematics. $\endgroup$ – user228113 Jun 4 '15 at 16:43
  • $\begingroup$ @Laszlo: Your calculation of $0.104336$ is the rounded value of $P[A \cap B].$ However, the final answer, $10/19$, is roughly equal to $0.104336/{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$ $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 16:48
  • $\begingroup$ @G.Sassatelli Mathematics is not a matter of enouncing something. You must argument your opinion, as I did. What is wrong with "my mathematics"? $\endgroup$ – Laszlo Jun 4 '15 at 18:01
  • $\begingroup$ @CKKOY Exactly, as I said. You DO NOT need the denominator. $\endgroup$ – Laszlo Jun 4 '15 at 18:05
  • $\begingroup$ To avoid any other useless dispute: I made a simulation for one million of 20 times rolling sets (it is only two lines in Mathematica!). The number of cases with one six in the first 10 and the second 10 rolls was: 104,524. Repeated this 100 times, got a normal distribution with the mean value exactly at 104336 (it took 20 minutes). Still sustain that my solution is wrong? $\endgroup$ – Laszlo Jun 4 '15 at 18:34


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    $\begingroup$ It would be helpful if you wrote a brief explanation of the answer that you have written. $\endgroup$ – Caleb Owusu-Yianoma Jun 4 '15 at 15:13

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