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So, I know that the probability that three randomly choosen points on a circle will be on a semi-circle is 3/4 (as is discussed here: Probability the three points on a circle will be on the same semi-circle). However, I'm not sure what is wrong with the following logic:

After the first point is chosen, we can rotate the circle so that the first point is at the circle's zenith, like so:First point chosen.

Now, the second point breaks the symmetry, so that one side is preferred over the other (if the second point falls on the left hand side of the circle, the third must also fall on the left; if it falls on the right hand side of the circle, the third must fall on the right). This is unless that the second point happens to fall exactly on the first point, or exactly opposite the first point (both have probability measure zero).

In any case, the probability that the point falls somewhere on the circle is 1 (Let's assume that the second point has fallen on the left-hand side):

Two randomly chosen points on a circle.

Now, since the location of the third point is an independent event, we see that it can fall either on the side with the second point (in which case all three are on the same semi-circle), or it can fall on the other side (in which case they are not). Since the second point breaks the symmetry into two distinct sides, the probability that the third point will be on one side or another is 1/2.

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    $\begingroup$ There are choices of the third point on the right side that make all three points lie on a semicircle. (Right side, up towards the top...) Not the semicircle you drew, another one. $\endgroup$ – David C. Ullrich Dec 20 '15 at 20:20
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    $\begingroup$ You've added an additional constraint by requiring it to be that semicircle. It doesn't have to be that one to have three points on the same semicircle. For instance 2 could be on the end of the semicircle. You are right that without loss of generality the first point can be taken to be at the top, but once you have chosen that, the circle is now rigid, you can't rotate it again. $\endgroup$ – Ian Dec 20 '15 at 20:21
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Community wiki answer so that the question can be marked as answered:

As pointed out in the comments, you get a lower probability because you've sharpened the constraint to particular semicircles.

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