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Here is the link to the boy or girl paradox Wikipedia page. The question I have is: Let's say you have two fair 6-sided dice and you roll them simultaneously. If at least one is 6, what's the probability that both of them are six?

My understanding is that it's similar to the boy or girl problem in which the sentence "If at least one is 6", change outcome space to $11$ instead of $36$. The number $11$ comes from the fact that there are $12$ cases with one $6$ but we counted (6,6) twice, so $12-1 = 11$. This has been answered on this website (and others) before. The answer is $\frac{1}{11}$.

My question is how we can generalize this to N fair 6-sided dice. Assume I simultaneously roll N fair 6-sided dice. At least one of them is 6. What is the probability that all of them are 6? My approach was to count all possible outcomes. We basically have $N$ slots we need to fill with numbers $1$ to $6$. We know that one of them is 6 (giving us $N$ possibilities since any of the empty slots can be $6$). Among the $N-1$ empty slots, there are $6^{N-1}$ combinations, making the total $N*6^{N-1}$ cases. However, we're overcounting some cases and subtract $N-1$ from the above total to make it $N*6^{N-1} - (N-1)$. Any suggestions? Am I missing something? Please let me know if anything is unclear.

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    $\begingroup$ You need to be more explicit about the experiment you are repeating. If you roll two dice and always cover one of them randomly, then the answer to your initial question is 1/6 (the probability of the unseen dice being a 6 is 1/6, conditional on the seen dice). To get a different answer you need to design the experiment so that if at-least-one six is rolled you then either cover the non-six or else randomly cover one of them if both are a six. The second experiment, I would argue, is very unnatural. $\endgroup$
    – guy
    Oct 29 '20 at 20:02
  • $\begingroup$ Isn't "If we can see that one of them is 6" equivalent to "at least one is 6"??? Please have a look at this: math.stackexchange.com/questions/884364/…. $\endgroup$
    – Peyman
    Oct 29 '20 at 20:24
  • $\begingroup$ No, it isn’t equivalent, because you can identify one die as “the one I saw.” And the probability of two sixes given at least one is six is different from the probability of two sixes given that “the one I saw” is six. The difference is that you can talk about “the other die” in the latter scenario but not the former. $\endgroup$
    – guy
    Oct 29 '20 at 22:38
  • $\begingroup$ Also, I’m stating this informally, but you can make this distinction precise mathematically. $\endgroup$
    – guy
    Oct 29 '20 at 22:39
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    $\begingroup$ I've updated the question. Do you think it's clear now? Thanks. $\endgroup$
    – Peyman
    Nov 1 '20 at 0:55
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We roll $n$ dice, each with $m$ sides The probability of getting all ones is: $(\frac{1}{m})^n$

The probability of at least a single one result is: $1-(\frac{m-1}{m})^n$

and the ratio between these is your result, the conditional probability. $\frac{(\frac{1}{m})^n}{1-(\frac{m-1}{m})^n}$

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Let $m$ be the size of the die, and let $n$ be the number of rolls.

Let $A$ be the event of rolling only 'ones' and let $B$ be the event of at least one 'one'.

You can use the following to get to a solution

  • Since these events are nested (A only occurs if B occurs) you can use $$P(A\vert B) =\frac{ P(A \text{ and } B)}{ P(B) }= \frac{P(A)}{P(B)}$$

  • The total number of possibilities to roll the dice is $m^n$

  • There is only one possibility for event $A$ rolling only 'ones'

  • To compute the number of possibilities for event $B$ it is easier to compute this indirectly by the event $\neg B$ (ie. the negation, when there are zero 'ones' rolled).

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  • $\begingroup$ Would you mind to elaborate? How to calculate $P(B)$ is the generalized version? $\endgroup$
    – Peyman
    Nov 8 '20 at 1:20
  • $\begingroup$ @Peyman, I have added an update. $\endgroup$ Nov 8 '20 at 8:06

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