# probability

• Aug 21st 2008, 11:18 AM
perash
probability
What is the probability that n points chosen randomly on a circle lie on a
semicircle?
• Sep 18th 2013, 12:28 AM
brosnan123
Re: probability
Solution:

Given n points chosen randomly on a circle lie on a semicircle
then the probability =n/2^(n-1)
• Sep 18th 2013, 04:43 AM
Hartlw
Re: probability
It's the same as the probability of placing n objects in one of two boxes, which is 1/2^n
• Sep 18th 2013, 06:59 AM
BobP
Re: probability
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 !)
• Sep 18th 2013, 07:13 AM
Hartlw
Re: probability
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)
• Sep 18th 2013, 07:19 AM
HallsofIvy
Re: probability
No, there is yet another possibility, the one I would interpret the problem as meaning: n points lie on a circle and the largest angular distance between them is less than $\pi$. There is no semicircle necessarily "determined".
• Sep 18th 2013, 08:11 AM
Hartlw
Re: probability
Quote:

Originally Posted by HallsofIvy
No, there is yet another possibility, the one I would interpret the problem as meaning: n points lie on a circle and the largest angular distance between them is less than $\pi$. There is no semicircle necessarily "determined".

A semi-circle is specified in the OP.

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
4) etc.

For this interpretation, the probability of n points being in a semi-circle is 1/ (2^(n-2))
• Sep 18th 2013, 09:00 AM
BobP
Re: probability
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.
• Sep 18th 2013, 09:38 AM
zzephod
Re: probability
Quote:

Originally Posted by Hartlw
A semi-circle is specified in the OP.

The interpretation of the problem is dictated by commom sense.

You cannot say that without insulting everyone who has already posted a different interpretation. If common sense dictated what you say all those who disagree with you have no common sense. Things are even worse if what you post is wrong (specifically your point two; two points uniformly distributed on a circle do not determine a semi circle - they determine a semi-circle with probability 0)

.
• Sep 18th 2013, 09:40 AM
zzephod
Re: probability
Quote:

Originally Posted by HallsofIvy
No, there is yet another possibility, the one I would interpret the problem as meaning: n points lie on a circle and the largest angular distance between them is less than $\pi$. There is no semicircle necessarily "determined".

That looks right, the points all lie on some semi-circle, no one said it had to be unique.

.
• Sep 18th 2013, 01:47 PM
Hartlw
Re: probability
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 don’t. 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 don’t. 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 don’t. 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.
• Sep 18th 2013, 06:46 PM
zzephod
Re: probability
Quote:

Originally Posted by Hartlw
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 don’t. p=1/2

Suppose the first two points are at \theta=0 and \theta=\pi/2. Then the three points are in a common semi-circle if the third point is placed between \theta=-\pi/2 and \pi. Which has probability 3/4.

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• Sep 19th 2013, 05:18 AM
Hartlw
Re: probability
Quote:

Originally Posted by zzephod
Suppose the first two points are at \theta=0 and \theta=\pi/2. Then the three points are in a common semi-circle if the third point is placed between \theta=-\pi/2 and \pi. Which has probability 3/4.

.

Draw a circle and arbitrarily place two points on it. Place the third point anywhere. Then either it is in a semi-circle with the other 2 or it isn't. There are no other possibilities. Then the probability of three random points being in a semi-circle is p=1x1x1/2.

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.
• Sep 20th 2013, 08:31 AM
Hartlw
Re: probability
Quote:

Originally Posted by zzephod
Suppose the first two points are at \theta=0 and \theta=\pi/2. Then the three points are in a common semi-circle if the third point is placed between \theta=-\pi/2 and \pi. Which has probability 3/4.

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zzephod is absoluteley right, as was BobP earlier. I wrote zzephod to tell him why he was wrong and he showed me he was right and I was wrong. My apologies to all.

Their point is: If instead of 0 and 90deg with a probability of 270/360 = ¾ 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.

It’s an entireley new ball game.