# How to count the number of starting hands in Texas Hold 'em I'm revisiting the excellent book by Tony Guerrera, Killer Poker by the Numbers. In the first chapter he goes over how to count the number of possible starting hands but leaves, as an exercise for the reader, counting the number of starting hands that are unpaired and unsuited. He does show the probability of being dealt 2 unsuited unpaired cards as 1/110.5.

My calculation seems to be wrong and I can't figure out where I'm missing.

If you select any one of 52 cards, then there remains 36 cards left that are not of that suit or that rank. And there are 52*51 permutations of 2-card hands available.

This gives me 52*36/52*51 = 1872 / 2652 = 70.5%

I'm obviously way off here but I can't tell what I'm doing wrong.

Probability of a pair is `3/51`, because the first card dealt does not matter. Any card can pair up. The second card needs to be of the same rank as the first card. There are 3 cards left that satisfy this condition, but because one card is already dealt there are 51 cards left. Therefore the answer is `52/52 * 3/51` which is just `3/51`.

Probability of no pair is `48/51`. Similarly the first card does not matter, but the second card needs to not pair, which are all remaining card minus three, which is `48/51`.

Probability of a suited hand is `12/51`, since there are 13 cards of each suit in a deck of cards. The first card you get does not matter, but the second card needs to be of the same suit, which are 12 cards.

Probability of a non suited hand is `39/51`. Same as the previous example, but now you need to essentially dodge 12 cards from the remaining 51 cards. `51 - 12 = 39.`

Probability of a non suited non pair hand is `36/51`, which indeed rougly equals `70.5%`. Just like the previous example, but now you also need to not pair. Since these three cards are all different suits than the original cards we are not counting any cards twice, just substract three more.

Now we understand this logic we can tackle your problem. So we need to get dealt J9o. This means our first card has to be a jack or a nine, which are 8 cards total. Since there are 52 cards in a deck odds of this happening is `8/52`. Now after that we need to hit the other card. There are 3 cards left that will get the job done, because suits need to be different. Odds of this happening therefore is `3/51`. To calculate the odds of both these events occuring we need to multiply both numbers. We get `8/52 * 3/51 = 24/2652.` We can simplify this to `1/110.5`.

• I uploaded a photo of the text. Maybe I'm misinterpreting the question.
– Ramy
Dec 1, 2017 at 17:34
• can you expand your counting a bit? I see you're using 51 as the denominator every time. Can you explain why this is the case?
– Ramy
Dec 1, 2017 at 17:41
• Ah I see, I understand now. I will have my updated answer up in about 30 minutes. Dec 1, 2017 at 18:03
• Done. Try answer 1 and 3 yourself to see if you fully understand it now! GL. Dec 1, 2017 at 18:34
• @Ramy exactly. Sorry for not pointing that out. Dec 1, 2017 at 18:49