All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
The odds stay unchanged (Posted on 2004-06-17) Difficulty: 4 of 5
On a certain island each of the inhabitants is a member of one of the two existing clubs.

The membership distribution is such that when two random people meet, the probability of those two belonging to the same club is equal to the probability of them belonging to distinct clubs.

When 100 newcomers arrive on the island and each enrolls in one of the two clubs, the distribution still retains this feature. How many people belong to either club?

See The Solution Submitted by Ady TZIDON    
Rating: 4.2727 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Another Approach | Comment 15 of 21 |

I would also like to understand how the "triangle approach" would work for the general sense. It is a shorter way than the way I found, and very interesting to me :)

If there were A people in club alpha and (X-A) people in club beta (meaning there were X people on the island) then you can figure out there was a (A^2 + (X-A)^2 - X) / (X^2 - X) chance for people from the same club would meet and a (2A(X-A))/(X^2-X) chance two people from different clubs would meet. (Drawing a square with A rows and columns shaded would help this. Remember to cross off the squares in the diagonal since you can't pick the same person twice!)

Since the number of people can be switched with no harm, assume club Alpha has fewer people.

A^2 + (X-A)^2 - X = 2A(X-A) since both are equal.

A^2 + X^2 - 2AX + A^2  - X = 2AX - 2A^2

4A^2-4AX+X^2 = X

(2A-X)^2 = X

2A-X = sqrt(X)

2A = X +- sqrt(X)

A = (X +- sqrt(X))/2

A = (X - sqrt(X))/2 (Since A is less than B)

From just this beginning we can see that X must be a perfect square, and so the number of people on the island later must be a perfect square. Since 100 is a perfect square we can go to our friend Pythagoras for the answer.

Sure enough, 10, 24, 26 is a pythagorean triple. So 24^2 or 576 is the number before and 26^2 or 676 is the number after. (A can be calculated in both cases from the above equation.)

Through later investigation, we find that this is the ONLY number since 100 is such a nice number to work with; it only has 1 even factor less than its square root other than 4. (4 doesn't work since 100/4 is odd)

I will get to posting the general case soon; my fingers need a rest though :)

Edit: The HTML codes didn't work :( I fixed them here.

Edited on June 19, 2004, 4:09 pm
  Posted by Gamer on 2004-06-18 23:22:00

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information