I usually don’t have a pat answer for what philosophers do. I mean, there are a lot of puzzles that philosophers try to set up and then attack, and usually an inquisitive person has only a few minutes of attention in them to try and understand a conceptual entanglement. For example, the Hesperus and Phosphorous issue, as an issue is usually a hard one to motivate for people. It also has the added difficulty of requiring a discussion of what direct reference is and why modern philosophy treats names as directly referential.

Going forward though, I think I’m going to start using the following puzzle from philosophy of probability (of all places). I think I’ll explain this puzzle and then say that ordinary concepts are RIFE with such problems. Philosophers try to work out those problems.

Here it goes.

A machine makes squares. It can make squares with an area as large as 4 or as small as 0. The size of the square it makes on any given occasion is random. What is the probability that the square will have an area between 0 and 2? The answer is 1/2.

Now consider the length of a side of one of the squares that this machine makes. The lengths will range in size from 0 to 2. What is the probability that the square will have a side length between 0 and 1? 1/2 again right.

But now there is a problem. We said that there is a 1/2 probability that the machine will make a square with an area between 0-2, but since we also said that there is a 1/2 probability that the machine will make a square with a side length between 0-1. But since a side length of 1 will create a square with area 1 (1 squared is 1), then we have said that there is a 1/2 chance that the machine will make a square with an area between 0-1 and a 1/2 chance that the machine will create a square with an area between 0-2. This is a contradiction. Thus, our concept of probability needs to be revised.

Area

0 ——2——- 4 (.5 probability that the square will be to the left of the 2)

Side length

0 ——-1—— 2 (.5 probability that the square will be to the left of the 1)

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