Starting hand probabilities

[MathBabe]

We all know there are 169 starting hands in Hold'Em. But the odds of getting each hand are NOT 168:1. I've been meaning to write this up for a while, and I think I've worked it out in a way that isn't too confusing.

(Executive summary: odds of getting a specific pocket pair are 220:1, two specific suited cards 330:1, and two specific unsuited cards 110:1. Nice easy numbers to remember).

The number of starting hands is easily calculated by 52*51/2 (52 possible cards for the first card, times 51 possible cards for the second, divided by 2 since we don't care which order we get them in). That's 1326. BUT, in practice we don't care if we have AsAc or AsAh - they're both AA. That's what cuts the number down, and understanding how gives us the probability of each type of hand.

For each pocket pair, there are 6 versions you could have based on suits. That's odds of 1320:6, or 220:1 (a probability of .45%). The odds of getting dealt a pocket pair on any hand are 13 times that, 1238:78, or 16:1 (5.9%).

For each two suited cards, like JTs, there are only 4 versions you could have (one for each suit). So, the odds are 1322:4 or 331:1 (.3%). There are 13*12/2 combinations of two unpaired cards, which is 78, so the total odds of getting any two suited cards are 1014:312 or about 3:1 (23.5%).

For each two unsuited cards, like AKo, there are 12 versions you could have (4 possible suits for the first card, times 3 for the second - we don't divide by 2 because we DO care about order in this case). So, for each particular unsuited cards, you have odds of 1314:12 or 110:1 (.9%). There are 13*12/2 =78 combinations of suited cards, so the total odds are 300:936 or about 1:3 FOR it happening. (70.6%).

We can check by adding them all up. 78 pocket pairs + 312 suited cards + 936 unsuited cards = 1326, which matches the original number. Adding up the probabilities, we get 5.9% + 23.5% + 70.6% = 100%.