Probability of pairs of numbers in random number selections

Can you attempt to write an algebraic formula for this?

The number of ways to draw exactly one pair in two of three drawings of K numbers from a set of N numbers is:

[C(N,K)] * [C(K,1) * C(N-K,K-1)] * [C(N-2K+1,K)] * C(3,2)

where C(N,K) = N! / [(N-K)! * K!]

Divide that by C(N,K)^3 to get the probability.

In your case, N=49, K=7.

It gets more complicated with more pairs.

markr, your equation only made me cross my eyes. It's been too long since I did anything like that if I ever did.


This may help to explain it a little better.

The birthday paradox
http://www.itsecurity.com/security.htm?s=239

parados, I've known about the classroom paradox, but I'd be interested in seeing how to formulate "six degrees of separation." I have seen from personal experience some mathematically slim chance incidents. The first one was some years ago when I used to work for Florsheim as a traveling auditor. I called a childhood friend of mine who now lives in Los Angeles, and we went to J-town for dinner. After sitting down, my older brother walked into the same restaurant, and he lives in Sacramento. Another incident was when my brother and his wife joined us for the Aegean Sea cruise from Istanbul to Athens. We took the Bospherous cruise that included lunch at a restaurant called the Garaj. After sitting at a table for six, another couple sat at our table, so I asked them where they were from. They guy said, Lodi, and that's where my brother is from. He even used my brother's ophthalmologic office, but this was the first time they met. On another occasion, I was on a transAtlantic cruise, and was talking to a couple in the Horizon lounge about a professor I met on my Budapest to the Black Sea cruise. A lady came up to me and asked if I knew the professors name. I said, yes, his name is Jerry Poe. She goes on to tell me she works with his wife all the time. What are the mathematical odds?

Although somewhat related, this is not the same problem. Each person has only one birthday (not seven), and the birthday problem solution doesn't address the probability of exactly one, two, three, etc. pairs.

markr, I'm sure paradose is aware that his classroom paradox isn't related, but since you provided the answer, he just included something also of interest. I think this is the first time I've seen it formula form, although I've known about the paradox most of my life.

markr,

I got the impression that Carmi felt the odds should preclude so many pairs from showing up. Your formula while nice and I am sure it works isn't something most people with nothing more than some high school algebra would understand. Since we get so many people wanting someone else to do their homework posting the forumula alone is just fine. I just wanted to show that the odds aren't as high as most people think it would be.


From the birthday paradox site I give a link to....


It also seems unreasonable that there are so many pairs when you get a lottery ticket with supposed random numbers.

I see. I misinterpreted "This may help to explain it a little better." to imply it was a solution to the problem.

No problem markr. I wasn't very clear.

The other thing Ci about people showing up in strange places is harder to quantify I am sure. Too many human interactions. Perhaps your brother suggested a restaurant to you because he knew you were going to a particular city then he suggested that same restaurant to his client. Humans rarely act in a random fashion. We are hardwired to respond to stimuli in the same ways.

I was always amazed (and somewhat disgruntled) with my father when I was a boy how he could always find someone that knew someone he knew no matter where he was.

I wish that were true, but he lives in Sacramento, while I live in Silicon Valley. We've not lived in the same city for many decades - about five. Besides, it's my friend that selected the restaurant.