Two dice are rolled.

Event A is all combinations for which the sum is greater than 9.

Event B is all combinations for which at least a face is 6. What are P(A|B) and P(B|A)?

Maybe you will find it very easy, but I couldn't come up with an answer based on using Bayes' theorem and not just writing down the possibilities.

  • $\begingroup$ What does "at least a face is B" mean? $\endgroup$
    – Peter Flom
    Jun 22, 2013 at 19:30
  • $\begingroup$ Sorry, I meant 6. I've updated the question $\endgroup$
    – Ryan
    Jun 22, 2013 at 19:40

4 Answers 4


To extend Peter Flom's counting answer, assume we know (by counting or some other source), the following things:

  • $P(A) = \frac{1}{6}$,
  • $P(B) = \frac{11}{36}$, and
  • $P(A|B) = \frac{5}{11}$.

Bayes' theorem lets us calculate P(B|A) from those values in a fairly straightforward way. Namely, it says that: $$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}$$ so, you'd plug in those values and get $$P(B|A) = \frac{\frac{5}{11} \cdot \frac{11}{36}}{\frac{1}{6}}= \frac{5}{6}$$ This matches what you'd get by counting (see Peter Flom's answer for the enumeration). If you knew $P(B|A)$ instead, you could use a similar procedure to calculate $P(A|B)$, but you still need to do some counting to find $P(A)$, $P(B)$ and one of $P(A|B)$ or $P(B|A)$.

Your example is a good homework exercise, but we often find ourselves in a situation where $P(B|A)$ would be useful to know, but incredibly difficult to calculate, while $P(A|B)$ is less useful but easier to estimate. Bayes' rule lets convert the former situations into the latter. For example, we'd very much like to know P(email is spam | its contents), which is hard to recover directly. However, given a lot of labelled emails, we can easily estimate P(email contains a certain word | the source email is spam), just by counting how often a word occurs in spam and non-spam emails and the overall rate of spam.


Based on the enumeration {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}, isn't P(B) = 11/36?

Alternatively, take the number of ways a single die will NOT show a 6 when rolled (five) and multiply this by the number of ways the second die will NOT show a 6 when rolled (five). Subtract the product from the total number of ways two dice can appear (36) and you get 11.

P(A|B) = P(A and B) / P(B) = (5/36) * (36/11) = 5/11

P(B|A) = P(A and B) / P(A) = (5/36) * (36/6) = 5/6


The other answers are not complete and/or ignoring the part of the question that says:

and not just writing down the possibilities

This may be important in the case of a large number of faces of the die $N$, in which case writing down all possibilities becomes too tedious.

A die $k$ has an outcome space $X_{k}=\{1,2,\ldots,N\}_{k}$.

Assuming fair dice in the following.

It follows, that for throwing dice 1 and 2, $P(B)=\frac{2N-1}{N^{2}}$. (It may help to imagine that the possible outcomes form a matrix: the outer product of $X_{1}$ and $X_{2}$). Substituting $N=6$ gives


We can now evaluate $P(A|B)$ without enumerating all possibilities. Because, since one die already shows $6$ and the sum $X_{1}+X_{2}>9$, it follows for the other die $X>2N-9$, or $X>3$. So:


To evaluate $P(A$), I don't see how to avoid enumerating at least some possibilities. However, in any case, the enumeration is over a finite number of possibilities, since it necessarily includes no more than $9^{2}$ possibilities, regardless of $N$. This leads to:


Substituting into Bayes formula, gives:



P(A) = P(10, 11, 12). There are 3 rolls that total 10, 2 for 11 and 1 for 1 so that's 1/6.

P(B) = 1/6 + 1/6 -1/36 = 11/36

P(A|B): rolls are (4,6), (5,6), (6,4), (6,5), (6,6); 5 rolls out of 12 = 5/11
P(B|A): rolls are (4,6), (5,6), (6,4), (6,5), (6,6): 5 rolls out of 6 = 5/6 (all but (5,5))

  • $\begingroup$ The OP wanted a solution without counting possibilities. $\endgroup$ Jan 27, 2017 at 13:19

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