What's the "pool of 35"? Does not compute.

If you're asking something like: you have a company of 1473 people. Two people are picked at random to be fired. What are the odds that Fred AND Sue get fired? In that case, you're almost there. The probability is 2/1473 that is one of them first, then it's 1/1472 that it's the other one next out of the remaining people. Multiply those two odds together and you get 8.92994441 × 10-8, or about 1 in 11.2 million.

I can't do math, but I do think I understand the "35" - dg is saying that out of 1473, a group of 35 is pulled. So...of the 1473, what are the odds of any single individual being in that group of 35, right?

Almost. Say a company has 1473 employees and 35 are going to be sacked, chosen completely at random. What are the odds of both Fred and Sue getting sacked?

In a dream many years ago I took statistics. I still have the book on my shelf. It's the first time I've cracked it open in 20+ years. Gibberish possibly follows.

probability fred and sue are sacked = (probability fred is sacked) * (probability sue is sacked given fred is sacked).

a = probability fred is sacked = 1 - (1472/1473 * 1471/1472 * [33 more terms skipped])

b = probability sue is sacked given fred has been sacked = 1 - (1471/1472 * 1470/1471 * [32 more terms skipped])

probability both fred and sue sacked = a * b;

now waiting for a person who really knows statistics to come along and show me how wrong i am...

Hrm.

I *think* you would do it like...

Each person has a 35/1473 chance of getting sacked. (2.37%) If it's completely random, then the events are independent, so the probability of both of them getting sacked is 35/1473 X 35/1473 = 0.056 %

I think.

Oh, I wasn't thinking about Sue being sacked, given Fred is. Just both at the same time.

I once taught this stuff, but not at 8 in the morning. You have t balls, t-2 are yellow and 2 are red. You choose p at random (without replacement.) What's the probability of getting 2 red ones?

Writing a little code, i calculate:

probability fred sacked : 0.023761
probability sue sacked given fred sacked:0.023098
probability fred and sue both sacked : 0.000549

Who writes their statistics code in C? Just me.

So t(t-1)/p(p-1). That is, choosing the first is 35/1473 and the second is 34/1472. And then you multiply them. There's something non-intuitive about that since it looks like they have to be the first and second fired, but we're ignoring order so they might as well be.

Or, equivalently, there a tCp ways of choosing a pool of p out of t. "favorable" is that we choose fred & sue and any 33 others: t-2Cp-2 and then take the ratio of favorable to total, which gives (again) t(t-1)/p(p-1).

choosing the first is 35/1473

Wouldn't that be true only if you replaced each sample you took? Essentially 35 tries in the pot of 1473?

No. But that's why I included the second explanation that avoids that point of view but gives the same answer.

Y'all are so cool.

I took finite math to fulfill my stuff-with-numbers requirement in college. I thought it was pretty awesome at the time, but I didn't really get it and have just about no memory of it.

Rock on with your bad math selves.

I need to go back for a little refresher course in probability :D

Your answer simplifies to 35/1473*34/1472, DarkForest.. you got it right.

Holy shit you people are awesome!

For the sake of sanity, I need to ignore the issue that either Fred or Sue could be selected at any point along the 35 selections made and assume that they would be the first and second picked, I think.

If I have this right, the probability of both being picked is 0.0549%, correct?

Yes. I also ran a simulation of this, and it came out 0.0548%. You could try it out too, but then you'll have to fire them three hundred million times.

You don't have to be worried about order of picking; you can think of it as order of discovery. If you can tolerate little more babbling:

Imagine you're the boss. You call all the employees out to the parking lot, and randomly hand each person a folded scrap of paper. Inside, 1438 are blank while the other 35 each have a drawing of a sad bunny.

After they're all handed out, you walk up to Fred. You know there are 35 bunnies and 1438 blanks out there, so "clearly" Fred has a 35/1438 chance of having a bunny. Fred opens up the paper, and lo, it's a bunny. Fred looks downcast.

You turn to Sue, sitting next to Fred. She hasn't opened her paper yet, but you know there are 1472 other scraps of paper floating around the remaining employees and 34 of them have bunnies on them. They were all given out randomly, independently---the fact that Fred has a bunny tells you nothing about where the other 34 bunnies are. So the chance of Sue having a bunny is 34/1472. She opens it up, and she has a bunny.

Even if you stop right there and don't look at any other papers, the odds of both Fred and Sue having had those bunnies would be the two odds multiplied together, ~0.0549%. It's also obvious that going to Fred before Sue had nothing to do with whether or not they had bunnies.

Note that Fred and Sue might have each talked beforehand, and Fred might have said, "I have a 35/1473 chance of bunny, and therefore so must you, and therefore we have a (35/1473)^2 chance of both having bunnies, right?" But Fred is wrong; he and Sue do not have magical pieces of paper that when opened, 35/1473 of the time materialize a bunny drawing independent of all other papers; the way the papers were handed out matters.

Again, awesome work!

Thanks for your help, everyone - I learned something today!

DarkForest, as you saying there are three hundred million possible outcomes? Just curious ...

no, i was just joking (lame). i ran my simulation through 300 million trials to get a probability...