Here is a epistemic logic puzzle that I wrote for my students in the undergraduate logic course I have been teaching this semester at the College of Staten Island at CUNY.  We had covered some similar puzzles in lecture, including Cheryl’s Birthday and the blue-eyed islanders.

Bonus Question. Suppose that Alice and Bob are each given a different fraction, of the form $\frac{1}{n}$, where $n$ is a positive integer, and it is commonly known to them that they each know only their own number and that it is different from the other one. The following conversation ensues.

 

JDH: I privately gave you each a different rational number of the form $\frac{1}{n}$. Who has the larger number?

Alice: I don’t know.

Bob: I don’t know either.

Alice: I still don’t know.

Bob: Suddenly, now I know who has the larger number.

Alice: In that case, I know both numbers.

What numbers were they given?

Give the problem a try! See the solution posted below.

Meanwhile, for a transfinite epistemic logic challenge — considerably more difficult — see my puzzle Cheryl’s rational gifts.

Solution.
When Alice says she doesn’t know, in her first remark, the meaning is exactly that she doesn’t have $\frac{1}{1}$, since that is only way she could have known who had the larger number.  When Bob replies after this that he doesn’t know, then it must be that he also doesn’t have $\frac{1}{1}$, but also that he doesn’t have $\frac{1}{2}$, since in either of these cases he would know that he had the largest number, but with any other number, he couldn’t be sure. Alice replies to this that she still doesn’t know, and the content of this remark is that Alice has neither $\frac{1}{2}$ nor $\frac{1}{3}$, since in either of these cases, and only in these cases, she would have known who has the larger number. Bob replies that suddenly, he now knows who has the larger number. The only way this could happen is if he had either $\frac{1}{3}$ or $\frac{1}{4}$, since in either of these cases he would have the larger number, but otherwise he wouldn’t know whether Alice had $\frac{1}{4}$ or not. But we can’t be sure yet whether Bob has $\frac{1}{3}$ or $\frac{1}{4}$. When Alice says that now she knows both numbers, however, then it must be because the information that she has allows her to deduce between the two possibilities for Bob. If she had $\frac{1}{5}$ or smaller, she wouldn’t be able to distinguish the two possibilities for Bob. Since we already ruled out $\frac{1}{3}$ for her, she must have $\frac{1}{4}$. So Alice has $\frac{1}{4}$ and Bob has $\frac{1}{3}$.

Many of the commentators came to the same conclusion. Congratulations to all who solved the problem! See also the answers posted on my math.stackexchange question and on Twitter:

Epistemic logic puzzle: Still Don’t Know.

Another question from the logic final exam. Epistemic logic. https://t.co/UBpQPTyVPc

— Joel David Hamkins (@JDHamkins) December 20, 2017