This puzzle requiring inductive reasoning about other people's knowledge exists in several forms with the same basic pattern. Here is one:

In a certain monastery, ten of the monks have a disease which causes them to have blue spots on their foreheads but has no other symptoms. All the monks have taken a vow of silence, and there are no mirrors in the monastery, so nobody knows whether he has a blue spot on his forehead or not. If a monk discovers that he has a blue spot on his forehead, he will have to leave the monastery by the end of the day. All the monks are perfect logicians - that is, they can instantly deduce all the logical consequences of any statement made to them - and they all know that all the other monks are perfect logicians. One day, the Guru, who is known to be truthful, gathers all the monks together and announces "At least one monk in this monastery has a blue spot on his forehead." Nothing happens for nine days, but on the tenth day, all the monks with blue spots leave. Why?

Another version of this problem involves a small town some of whose inhabitants are commiting adultery, and whenever this happens everyone except the wronged spouse hears about it within an hour. The law in this town is that if your spouse can prove that you have been unfaithful you will be put to death that day. The town's priest wishes to stop the problem without accusing anyone, so he makes a proclamation which everyone can hear: "There is adultery in this town." Note that if the proclamation is made through the unreliable postal service rather than being broadcast for everyone's ears simultaneously, nothing will happen.

Why is there a delay before the monks leave? Well, start by considering the case of one monk with a blue spot. When the guru makes his statement, the monk will look around, notice that no-one else has a blue spot, deduce that he is the one, and leave that day.

With two monks with blue spots, A and B, A will think that B is the only monk with a blue spot, and will expect him to leave. When this hasn't happened by the end of the first day, A will realise that B saw someone else with a blue spot, and that this someone else must be him. B will be making the same deductions, and they will both leave on the second day.

With three monks with blue spots, each monk with a blue spot will think that the other two are in the situation described in the previous paragraph, will expect them to leave on the second day for the reasons described above, and when they haven't left by the third day, will realise that he also has a blue spot. And so on - if N monks have blue spots, they will all leave on the Nth day.

One puzzling aspect of this problem is what extra information the guru's statement provides. The monks already knew that at least one monk has a blue spot on his forehead, because they could see each other. Again, this is easier to understand if there are fewer monks with spots. If we go back to the case of two monks with spots, A knows that at least one monk (namely, B) has a blue spot, but he doesn't know that B knows this. With three monks with spots (A, B and C):

  • A knows that at least 2 monks (B and C) have blue spots
  • A knows that B knows that at least one monk (C) has a blue spot
  • A does not know that B knows that C knows that at least one monk has a blue spot.
The effect of the guru's statement is that afterwards, everyone knows that everyone knows that ... at least one monk has a blue spot, for any number of repetitions of "everyone knows that".

An interesting thought experiment. There are some details missing in the writeup above, however. In the simplest scenario, all the monks must be present at the guru's announcement, all monks must know before or at the end of the day if any other monks are leaving (or have left), no monks can be cured, and no monks can catch a new case of the disease.

Exploring the limits of how far logic can take you is fun (for some of us, anyway), so consider the case where a monk can catch a new case of blue-spot disease. Let day 0 be the day of the guru, and number the days sequentially from then for notational convenience. Assume that every monk sees every other monk at the end of every day, and that every monk knows that this happens. If, at the end of day N, a monk sees only N others with spots, that monk must leave (and all the others with spots will leave with him); otherwise, everyone stays. (A monk will never see fewer than N (other) spotty monks on day N, unless one or more gets healed.) This way, even if a new case occurs, so long as the spots appear by the end of the day, the system will still work.

Things become a lot more intricate if a monk can be healed. There are several distinct classes of problem; if a monk can heal spontaneously, or if the only way to get healed is in an infirmary (where perhaps records can be kept, even if it's just a tally of the number of monks who got cured that day), or if either can happen. The problem can be divided further on whether the disease is a once-in-a-lifetime type (like mumps) or not (like a cold). If spontaneous healing can occur, a solution like above probably cannot be found. If there is an infirmary where monks can heal, and it's a once-in-a-lifetime type, and the monks not only remember how many monks had spots but which ones had, there may be a way using the tally described, but there would probably have to be a special case for healing the last monk. I, for one, would like to read more if anyone has thought further on the problem.

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