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## #1 2010-05-21 00:08:56

nevinsmith
Member
Registered: 2010-05-19
Posts: 39

### probability and circles!!!!

a circle has it's circumference evenly distributed with 2010 dots, 4 dots chosen at random are a,b,c,d

find the probability of line segments a-b  intersecting with c-d!

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## #2 2010-05-21 12:43:42

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: probability and circles!!!!

Hi nevinsmith;

I am working on my own solution to this. I am going to assume the the points can be picked with replacement. What do you think? Or are they to be picked without replacement?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2010-05-22 12:44:32

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: probability and circles!!!!

Hi nevinsmith;

In order to solve your problem which sounds like a contest problem rather than a high school problem. If it is I encourage you to submit your own work, rather than mine. It is unfair to utilize a forum to help you win a contest. It is unfair to the other contestants.

To solve the problem I made this assumption.

1) The 2 lines are chosen without replacement. Means they will be 2 distinct lines. It seemed to make the most sense.

Solution: For two lines to never meet they must be parallel. Your question then reduces down to a complete graph of degree 2010. How many parallel lines does such a graph have including the edges? This stumped me for a while. My thanks to JBL who pointed out the answer to this subproblem. The number of parallel line segments of a complete graph of degree n is a bijection with the Orchard crossing number of the complete bipartite graph K_{1,n}.

The Orchard crossing number is well understood. For any nth degree complete graph the number of edges is (n(n-1)) / 2 this is the total number of lines. In your case the total number of unique lines is

There are:

Ways to pick two distinct lines or edges in a complete graph. We hold this number.

The number of parallel edges in a n degree complete graph is given by the orchard crossing problem. It is:

For n = 2010 we have 1 013 056 080. So:

Probabilty that they intersect is the complement of that:

P(2 lines intersecting) = 1 - 0002485 = .999751. Quite rare.

Soon I will hopefully have a computer simulation answer that will back this up.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #4 2010-05-22 12:56:49

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: probability and circles!!!!

I think this question is actually a lot easier than that. You're stumbling because the line segments not being parallel doesn't necessarily mean that they'll intersect.

For lines AB and CD to cross, their order around the circle needs to be ACBD or ADBC.

Overall, the possibilities for orders are:

ABCD
ABDC
ACBD
ACDB

They're all equally likely, so the probability is 2/6 = 1/3.

Why did the vector cross the road?
It wanted to be normal.

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## #5 2010-05-22 13:51:43

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: probability and circles!!!!

That's very interesting.

nevinsmith wrote:

a circle has it's circumference evenly distributed with 2010 dots, 4 dots chosen at random are a,b,c,d

find the probability of line segments a-b  intersecting with c-d!

I went with the idea that he didn't want them to intersect even outside the circle. He does not say in his problem. For that you need to find parallel lines.  I felt that this condition made the problem more difficult and more interesting.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #6 2010-05-22 20:44:05

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: probability and circles!!!!

He calls them line segments. For me, that makes things fairly unambiguous.

Why did the vector cross the road?
It wanted to be normal.

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## #7 2010-05-22 20:46:23

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: probability and circles!!!!

Lines that are not parallel must intersect in some finite distance. Therefore whether they intersect inside the circle or not they are still line segments.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #8 2010-05-22 20:55:19

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: probability and circles!!!!

Well, I still think your leap of logic is bigger than mine.
But anyway, both versions of the question have been answered, so the poster should be happy.

Why did the vector cross the road?
It wanted to be normal.

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## #9 2010-05-22 21:15:47

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: probability and circles!!!!

Hope so, anyway I learned some things doing mine and looking at yours. So it was a good day.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #10 2010-05-24 00:06:41

ZHero
Real Member
Registered: 2008-06-08
Posts: 1,889

### Re: probability and circles!!!!

mathsyperson wrote:

. . . For lines AB and CD to cross, their order around the circle needs to be ACBD or ADBC.

Overall, the possibilities for orders are:

ABCD
ABDC
ACBD
ACDB

They're all equally likely, so the probability is 2/6 = 1/3.

I worked upon a Long Hand approach just trying to refute this.

Consider that there are P[sub]1[/sub], P[sub]2[/sub], P[sub]3[/sub], . . . , P[sub]n+2[/sub] points on a circle in that order and n≥2 is even.

I take a fixed point P[sub]a[/sub] and another point P[sub]a+k[/sub].

The idea is that for P[sub]a[/sub] and P[sub]a+1[/sub], no line segment formed with the remaining n points as their ends will intersect P[sub]a[/sub]P[sub]a+1[/sub].
(coz there is no point in minor arc P[sub]a[/sub]P[sub]a+1[/sub])

Similalry, For P[sub]a[/sub] and P[sub]a+2[/sub], (n-1) line segments formed out of the remaining n points will intersect P[sub]a[/sub]P[sub]a+2[/sub]!!
(there is only one point, P[sub]a+1[/sub], in minor arc P[sub]a[/sub]P[sub]a+2[/sub])

For P[sub]a[/sub] and P[sub]a+3[/sub], 2(n-2) line segments formed out of the remaining n points will intersect P[sub]a[/sub]P[sub]a+3[/sub]!!
(there are two points, P[sub]a+1[/sub] and P[sub]a+2[/sub], in minor arc P[sub]a[/sub]P[sub]a+3[/sub])
and so on.......

In each of the cases, the total number of line segments that can be formed out of the remaining n points is n(n-1)/2.

Now, the probability can be found out as..
(total number of intersecting lines for each of the line segments P[sub]a[/sub]P[sub]a+1[/sub], P[sub]a[/sub]P[sub]a+2[/sub], etc) ÷ (total number of line segments that can be drawn from the remaining n points in every case.)

The numerator part is a little twisted !!!!

The probability for all even values of n is

Last edited by ZHero (2010-05-24 00:52:27)

If two or more thoughts intersect, there has to be a point!

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## #11 2010-05-24 18:11:35

ZHero
Real Member
Registered: 2008-06-08
Posts: 1,889

### Re: probability and circles!!!!

If two or more thoughts intersect, there has to be a point!

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## #12 2010-05-25 18:47:28

nevinsmith
Member
Registered: 2010-05-19
Posts: 39

### Re: probability and circles!!!!

mathsyperson- thnx for the solution! ya your right it means intersect only inside the circle. i kinda thought it was something like that but didn't think it was equal in probability haha. thnx everyone

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