A little probability question in DC

Aug 10, 2020

Here's a little math question. But wait! Before you go, "Blah, blah, math" and click away, take a quick look at the question, and then the story behind it.

Suppose you have an urn with eleven marbles in it. There are two green marbles, seven yellow marbles, and two agates. You take out the agates because this problem is only about the solid colors, leaving you with nine.

If you pick three marbles out of the urn at random, what's the probability that you'll get both green ones?

You might be thinking, "No one cares about urns with marbles in them." Or, maybe "Probability problem! Awesome!". Or, "When would I care about that?"

The answer to the last question might be "last Friday" (that is, August 7, 2020). On that day, the US Court of Appeals for the District of Columbia Circuit ruled that the House of Representatives can pursue its lawsuit to compel former White House Counsel Don McGahn to testify before the Judiciary Committee. I have no idea about what will happen with that, but here's how we get to the math part.

1. A district court judge ruled that the House suit could proceed.

2. A three-judge panel at the DC Circuit ruled 2-1 that it could not.

3. The House appealed for an en banc hearing of the whole court.

4. After two of the eleven judges recused themselves, the remaining nine heard the case.

5. The full court ruled 7-2, with the two original judges who voted to dismiss in the minority.

Arguably, any three-judge panel that did not include both of those judges would have ruled differently in the initial appeal.

I learned about this from the Cafe Insider Podcast, hosted by Preet Bharara <>and Anne Milgram, who wondered about the odds of this.

So: What's the probability that such a panel would be been selected?

Well, it's exactly the same problem as our urns and marbles!

So let's figure it out.

We can list the possible combinations of marbles (or judges) that would include both the green ones: GGY, GYG, and YGG. It seems intuitive that the probability of any of those should be the same. That intuition is correct, so we won't prove that here.

So let's focus on the first possibility. If we think about picking the marbles out of the urn one at a time, the probability of getting a green one on the first draw is 2/9 or two out of nine, because there are two green ones out of a total of nine.

Now there are eight left, and the probability of getting the other green marble is 1/8, for the same reason.

So 2/9 of the time we get a first green one, and 1/8 of the time after that we get the other green one. These are what are called "independent events", because they don't have anything to do with each other, and when you've got independent events you can multiply the probabilities. So 2/9 * 1/8 = 2/72 (As Tom Lehrer might say, "'Fractions? I didn't know there would be fractions!' I hear you cry.")

And we've got three ways that can work out: GGY, GYG, YGG. That's three possibilities with the same probability, and 3 * 2/72 is 6/72, which reduces to 1/12.

If we want to say this as odds, we give it as a ratio of the number of times a thing happens vs. the number of times it doesn't. A probability of 1/12 has one time where it happens and eleven where it doesn't (on average), so the odds would be 1:11 in favor, or 11:1 against.

Thanks to my brother-in-law for this approach to it. I solved it a different way at first, but his approach is easier to follow, I think.