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I was just going back and refreshing my knowledge on binomial coefficients/probability and I was trying to calculate some poker odds since I would be able to look them up and verify my attempts.

I have run into an issue when trying to calculate how many off-suit hands there are for an opening deal in Texas Hold'em.

The way I'm attempting it is that I pick 2 suits of the 4 available $(\binom{4}{2} = 6)$ and then pick 2 ranks of the 13 ($\binom{13}{2} = 78$) to ensure that I don't get any pairs or any suited cards.

I multiply the results together $\binom{4}{2} \times \binom{13}{2} = 6 \times 78 = 468$

Only problem is I'm missing half since the number of possible unsuited hands you can get is apparently 936 according to https://en.wikipedia.org/wiki/Texas_hold_%27em_starting_hands

Where am I going wrong with my logic here?

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When you pick two suits and two ranks you have missed matching the suits with the ranks. If you pick Spades and Hearts, then $2$ and $3$, you could have $2S, 3H$ or $2H, 3S$. That is another factor of $2$.

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  • $\begingroup$ Isn't that the point of multiplying the number of combinations for picking two suits with the number of combinations of picking 2 ranks? $\endgroup$ – Adam Ritter May 20 '20 at 4:31
  • $\begingroup$ No, that gets you the number of combinations like $2+3, H+S$ but that is still not enough to specify the hand as in my example. You still need to decide whether the $2$ is $H$ or $S$. Then the $3$ is the other one. $\endgroup$ – Ross Millikan May 20 '20 at 4:38
  • $\begingroup$ Ah, I see, so you're really getting back sets consisting of 2 + 3, H + S. Are there any math rules/formulas that discuss how the distribution is supposed to happen when it comes to combinatorics? It feels intuitive in this case but what happens when I'm dealing with something along the lines of ($\binom{13}{4} \cdot \binom{12}{5} \cdot \binom{5}{3}$) ? $\endgroup$ – Adam Ritter May 20 '20 at 23:12
  • $\begingroup$ Careful counting is an important part of combinatorics. You need to count things once and once only, or correct for any errors. If I were doing this problem I would say the first card to draw has $52$ possibilities, then the second has $52-13-4+1=36$ possibilities, where I subtract the cards of the same suit and same rank, then add back the one I subtracted twice. This gives $1872$, but we have counted every hand twice because either card could come first. Dividing by $2$ gives the correct $936$ $\endgroup$ – Ross Millikan May 20 '20 at 23:40

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