We are choosing 3 ranks out of 11 for the 3 board cards that aren't paired, and then there are 4 suits for each of those ranks, so 4^3. The first term in the denominator is the same as the numerator for a 2-pair.
* The second term is for matching only 1 hole card which we can do 2 ways times 3 suits each, and then there are C(11,4) ways to choose the 4 ranks of the other 4 board cards, and 4^4 ways to choose their suits. The third term is for matching 0 hole cards, so there are C(11,5) ways to choose those ranks, and 4^5 ways to choose those suits.
David's method might be easier to understand.
*Hey, I'm only counting 3 things, and David's counting 4. His things are a little easier to count because the C(47,2) doesn't require considering suit, but the others do, with 3^2 in one, while the others just involve C(4,2) and C(3,2). Though his certainly seems subjectively simpler, I'm finding it hard to quantify why because the difference isn't as great as it first appears. I think it's mostly because the subtraction of the pairs can be regarded as something of an afterthought, with the rest being really simple. But you still have to do the pairs for the exact answer, and there are 2 types. They are simple, but so is what I did. Plus you have to think about the problem backwards, i.e.,they announce the keno numbers, and then we draw a random keno ticket. David likes simple solutions, and I think it's important to be able to think that way. It's also important to understand general principles like Bayes' theorem, and if you think about the problem in the forward direction, the Bayes' theorem solution is obvious. And if you wanted the probability of 1 pair or 0 pair, with mine you only have to move a different term from the denominator to the numerator. If you hit on that first, you can stop thinking. You generally stop running once you've already caught the bus. But it's always best to have a second solution as a check for anything this complex. Had I done David's solution as a check, I would have found my missing factor of 2. But David's solution is so conceptually simple that I might not even feel a need to check it. So maybe that proves it's simpler.
Last edited by BruceZ; 04-16-2014 at 02:20 PM.