TITLE: Guessing the other dice number, given two dice were rolled, and one of them rolled a 3
QUESTION [5 upvotes]: An argument i had with a friend followed this question:
"Given two dice were rolled, and one of them rolled a 3. What would you bet the other dice rolled?"
The phrasing is just to enhance that there might be some number that has an higher probability of showing on the other dice. One of us said that since 7 is the most likely number to be rolled by two dices, then 4 should be the answer. The other said that once the 3 was set on one dice, we should not be looking at anything other than the other cube - so any number would be an equal guess.

REPLY [3 votes]: Ross has provided a nice answer that highlights the subtleties of deducing probabilities from what someone tells you, where you need to take into account how they decided what to tell you.
ConMan has provided an answer that treats the $11$ results containing a $3$ as equiprobable, but does so again in the paradigm of someone reporting the result, and in my view doesn't explain properly how the report is to be interpreted.
I note that your question doesn't actually mention anyone reporting the result; you merely write “and one of them rolled a $3$”.
That allows us to give an entirely objective interpretation to the question, free of the subtleties of the decisions of people reporting results: What is the conditional probability distribution for “the other die”, conditional on the event that at least one die shows a $3$? To make this interpretation well-defined, we need to define what “the other die” means, but it's intuitively clear enough: If there's a non-$3$, it's the non-$3$, and otherwise it's one of the $3$s (it doesn't matter which); alternatively, we could define its value as the sum of the dice minus $3$ (with the advantage that this defines a random variable in all cases, not just when there's a $3$).
Then, if we let $X$ denote the value of “the other die” and $T$ denote the event that at least one die shows a $3$, we have
$$
\mathsf P(X=3\mid T)=\frac{\mathsf P(X=3\cap T)}{\mathsf P(T)}=\frac{\frac1{36}}{\frac{11}{36}}=\frac1{11}\;,
$$
and for $x\ne3$ we have
$$
\mathsf P(X=x\mid T)=\frac{\mathsf P(X=x\cap T)}{\mathsf P(T)}=\frac{\frac2{36}}{\frac{11}{36}}=\frac2{11}\;.
$$
Thus, under this interpretation of the question, you should bet on any number other than $3$.
In the reporting paradigm, the corresponding scenario would be that you agree beforehand that the person rolling the dice will tell you whether or not there is at least one $3$, so the person has no decisions to make on what to report.