Generalization of a boy or girl paradox with die!

Question

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.

Answer 1

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. $\frac{(\frac{1}{m})^n}{1-(\frac{m-1}{m})^n}$

Answer 2

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).