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Two dice (with numbers 1 to 6 on the faces) are rolled.

One die rolls a 6.

What is the probability of rolling a double 6?

One solution is to say that P(2 sixes) = $\frac{1}{6}$ since the first die gives a 6, so the only way to get a double six is by rolling a six on the other die (which has a 1 in 6 chance).

Another solution is to say that there are 11 possible combinations if one die rolls a six i.e. (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) and (6, 1). So the probability of rolling a double six if one six has already been rolled is $\frac{1}{11}$.

Which answer is correct and why?

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    $\begingroup$ The second one is correct. The exercise says that one rolls a 6. This roll can be the first one or the second one. Given that someone roll $a$ six, what is the probability to roll a double 6? $\endgroup$ – callculus Oct 27 '17 at 22:39
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    $\begingroup$ This is related to the Two Child Problem or Boy-or-Girl Paradox en.wikipedia.org/wiki/Boy_or_Girl_paradox $\endgroup$ – Jay Oct 28 '17 at 1:32
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There is confusion between two questions.

  1. You have rolled a six. Now for a double-six, you need to get a 6 on the second roll. The second roll is independent of the first. So the probability of a 6 again on the second roll is $1/6.$

  2. If both dice have already been rolled out of your sight, and you are told that there is at least one 6, then conditional on that information, what is the probability that the dice actually show a double-6. Then the analysis of @Shayne2020 (+1) leading to the answer $1/11$ is correct.

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  • $\begingroup$ Dear Teacher, Dear Professor, I'm so sorry for this comment..I asked mathoverflow for help. But the question was downvoted. I had to delete the question. I really need help. But unfortunately, no one helped. If I ask you for help, can you take a look? If you have a few minutes. Then, I will delete this comment..Thank you very much.. mathoverflow.net/q/298997/123863 $\endgroup$ – Mathematics is Life Apr 29 '18 at 17:40
  • $\begingroup$ @MathematicsIsLife: This seems to be an entirely different question. If you could explain what you are doing and summarize your data, someone might be able to help. For example, your first vector $P_1$ when multiplied by 42 can be summarized as: values 0, 1, 2, 3, 4, 5 with respective frequencies 25, 4, 5, 5, 2, 1 . $\endgroup$ – BruceET Apr 29 '18 at 18:01
  • $\begingroup$ Teacher, the sum of all values are equal to $1$. For any $P(x).$ Because, this is probabilistic distribution...(English is my second language..Sorry for wrong words) $\endgroup$ – Mathematics is Life Apr 29 '18 at 18:11
  • $\begingroup$ Teacher, I can not do anything because I do not have a computer and I can not use mathematical softwares..:( $\endgroup$ – Mathematics is Life Apr 29 '18 at 18:42
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This can be phrased as

Given that at least one die rolls a 6, what is the probability of rolling a double 6?

We can use the conditional probability formula: $$P(A|B) = \frac{P(A\ \mathrm{and}\ B)}{P(B)}$$ This means "the probability of event A given event B, is the probability of A and B divided by the probability of B".

\begin{align} P(\mathrm{double\ 6}|\mathrm{at\ least\ one\ 6}) &= \frac{P(\mathrm{double\ 6}\ \mathrm{and}\ \mathrm{at\ least\ one\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{P(\mathrm{double\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{1/36}{11/36}\\ &= \frac{1}{11} \end{align}

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One die rolls a 6.

This doesn't clarify which one. Was it the first die or the second one? (Note that the dice are distinguishable by the turn of their throws.)

So, the latter logic is correct. The conditional probability reduces the sample space $S$ into ${(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) (6, 1)}$

Now, $|S|= 11$

And the only favorable outcome (rolling two 6's ) is $(6,6)$.

Thus, the probability is $\frac{1}{11}$

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The odds are $1$ in $6$ in both cases (seeing one die vs not seeing the dice) because the arguments given for all the $11$ permutations neglect to address doubles for the $5$ (which equal $5$ combinations) that are not $(6,6)$. Independently, if one die is known to be six, there is a $1$ in $6$ chance of the other being six.

The answer is $1$ in $6$.

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