# Flawed Logic: The Problem Of Blue-spotted Monks

I recently ran across the Problem of the Blue-spotted Monks again at https://richardwiseman.wordpress.com/2011/04/04/answer-to-the-friday-puzzle-98/.  Actually, this is a slight variation of the well-known Blue-Eyed Monk problem that can be found at https://en.wikipedia.org/wiki/Common_knowledge_%28logic%29, among other places. One site even called this problem “The Hardest Logic Puzzle in the World”A, but that was probably just a case of self-promotion. For today’s discussion, I chose the first version of the problem because it is simpler (we don’t have to worry about the case in which one person has red eyes). If you haven’t done so, go ahead and read the problem. We will discuss the solution in the next section.

The classic answer uses mathematical inductionD to first consider the (trivial) case in which only one monk has the disease. Then they move on to the case of two monks. Their error begins in the case of three infected monks, assuming that the monks were required to start back at one in making their individual analysis.  That is wrong; in the comments section of some of the references, several people take issue with this assumption.  I believe this to be a misapplication of the induction process.  The monks have only to consider two possibilities; either they are infected or they are not. For the sake of argument, let’s say there are ten infected monks. Most of the monks will see ten fellow monks with blue spots. Ten of the monks will see nine monks with blue spots. None of the monks will see only one monk with blue spots, or just two monks with blue spots, etcetera. Those monks seeing ten spotted monks have only two possibilities – either there are ten infected monks or there are eleven. Those monks seeing nine spots need only consider the possibility that there are either nine or ten monks infected. Most of the monks, in considering their first possibility (that there are only ten infections) realize that those ten, in considering the possibility that there are only nine infections, must allow for the possibility that there are nine monks who see only eight infected monks. With the information available, nobody sees any reason to consider any other lesser possibility. Just like in the conventional solution, each monk, having two possibilities, must allow the lesser possibility (which means they are not infected) to resolve itself first before concluding that they are infected and turn themselves in. All of the monks know that the least number of infections that any of them must consider is eight, not one. The first day nothing would happen. The second day, the monks that saw only nine other spotted monks will conclude that the possibility of anyone sighting eight spotted monks was groundless, so they will all turn themselves in. The third day, those who saw ten spotted monks will all breathe a sigh of relief.

### Considering The Second Possibility

Now we need to look at the big picture.  This isn’t rocket science.  Why hasn’t anybody seen the error in conventional wisdom before today?  I, like the monks, must consider the possibility that I am the one infected.  If nobody else comes forward with a confirmation of the correct answer soon, I guess I’ll be forced to turn myself in.  Please hurry.

### Silent

An old liberal of unspecified race, gender, size, and sexual orientation that believes in both God and science and is not the least bit intimidated by numbers.

## 2 thoughts on “Flawed Logic: The Problem Of Blue-spotted Monks”

1. If you are a monk seeing 9 spots you know that if you have a spot they each also see 9 but if you do not they each see 8. If you have no spot then each of those others only see 8 but do not know if they have a spot so, in such a scenario, as far as they are aware there would be 8 or 7. So it is not logical to pack before the ninth day when all 10 can deduce they have a spot. See: http://puzzling.stackexchange.com/questions/37218/about-the-100-blue-eyes-problem for a detail overview of the incorrect “skip ahead” failure.

1. I didn’t understand this response. The first two sentences seem to repeat my argument. The third sentence is his conclusion, but is clearly missing any new steps I would need to get his point.

Fortunately, I received another response from some person who didn’t want to be identified (yes, that IS VERY strange considering – hmm, never mind, I’m just going to humor him) – We can call him B.M. (maybe B. for short) – that did make sense to me. Reading the explanations at the website Jonathan references, they all seem to need to make an allowance for someone who only sees five or six spotted monks, when in my example of ten infected monks I make it clear that none of the monks could actually believe that, or believe another monk could believe that. As B. pointed out to me, that doesn’t matter. In my example (as in all cases) there are two groups – in this case those who see nine infected monks and those who see ten. In examining the possibilities, the two groups are in neighboring parallel universes (those are B’s terms). The problem is NOT that some monk could actually believe that there were only three infected monks; the problem is that in doing your analysis (as I did in the example) you don’t know whether the other group is in the universe one blue spot higher than yours (meaning you are infected) or one spot lower, so the two universes are always just one step out of sync and unable to agree to a closer starting point for the logic countdown. In my example the group with ten sightings made an allowance for the group with nine sightings so everyone could be on the same page, so to speak, but the nine-spotting group made no allowance for their other neighboring universe, which it would be impossible for them to know not to do. That was my mistake. Had they made that allowance, then the first day they would be waiting for their possible seven-spot-sighters to leave while the other group was waiting for the eight-spot-sighters to leave. On the second day each group would wait for the next non-existent group so nothing would happen that day either. The third day, both groups would conclude that it must be them and everybody would leave the island. The only way to get the two groups in sync would be to start at the only absolute reference point, which would be zero, to start the countdown (or count-up, actually). If that makes any sense, thank B. If this logic is flawed also, let me know as soon as possible; I think I’ve already made a big enough fool of myself this week. Thanks in advance.

This site uses Akismet to reduce spam. Learn how your comment data is processed.