Given n points chosen randomly on a circle lie on a semicircle
then the probability =n/2^(n-1)
No it's not, it's rather more like saying that you have, say, 2m boxes with each of the n objects being assigned randomly to a box, and then asking for the probability that at least m of the boxes, (any m of the 2m boxes), contain no object at all.
(And I think that you need to take this in the limit as 2m goes to infinity since in the original problem there are an infinite number of possible semi-circles !)
The first post is a little vaque:
The possibilities are:
a) You are given the two semi-circles and one is specified. Then p=1/2^n
b) You are given the two semi-circles and the first choice specifies one. Then p=1/2^(n-1)
c) The first point determines a semi-circle, then use reasoning of a), b)
d) The first two points determine the semi-circle, p=1/2^(n-2)
The interpretation of the problem is dictated by commom sense.
For ex, option d) in post #5 above can be visualized as follows:
1) place (pick) a point on the circle. probability it is in a semi-circle is p=1.
2) place a second point on the circle. Probability they are in a semi-circle is p=1. This determines a semi-circle within which all the rest of the points have to lie.
3) place a third point on the circle. Probability it is in the semi-circle is p=1/2
For this interpretation, the probability of n points being in a semi-circle is 1/ (2^(n-2))
Sorry Hartlw, but you're wrong at 2), (though you may argue about your interpretation).
For ease of reference, picture the points on a clock face.
Suppose for example that your first point is at 12 and the second at 3.
'The' semi-circle is not determined, it could run from 9 through 12 to 3 or from 12 through 3 to 6, (or any between those extremes), that's 3/4 of the face within which the third point could lie.
If you continue to add points in such a way that it's still possible to 'cover them' with a semi-circle, 'the' semi-circle may still (possibly) not be defined.
Taking a somewhat extreme case, suppose that the first thousand points all happen to lie within the quadrant from 12 through to 3, then there is still a 75% chance that the next point is OK, anywhere from 9 round to 6.
After placing two points arbitrarily (p=1 each time), there are only 2 options for placing the third point: either the three points lie in a semi-circle or they donít. p=1/2
After placing three points, there are only 2 options for placing the fourth point: either the four points lie in a semi-circle or they donít. p=1/2
After placing four points, there are only 2 options for placing the fifth point: either the five points form a semi-circle or they donít. p=1/2
Probability of n points being in a semi-circle is then the product of the n probabilities, or p=1/(2^(n-2)), as done previously.
Draw a circle and try it.
The probability of the fourth point being in a semi-circle with the other three is the probability of the first three being in a semi-circle times the probability of the fourth point being in the semi-circle if they are. If the three are in a semi-circle, then no matter where I place the fourth point, there are only one of two possible outcomes: it is in a semi-circle with the other three or it isn't. So for the fourth, p=1x1x1/2x1/2.
Their point is: If instead of 0 and 90deg with a probability of 270/360 = ĺ for the third point, the first two points are at 0 and 150, then any point on an arc of 210 deg is within a semi-circle with the first two points, so the probability of creating a semi-circle with the third point is 210/360.
And if the first two points are at 0 and 180deg, the probability of a third point being in a semicircle with the first two is 360/360=1.
Itís an entireley new ball game.