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**phatus****Member**- Registered: 2007-10-15
- Posts: 11

Hi

I'm after some info on how to work out probability using poker hands.

I've been told that if i hold two different cards in my hand the chance of one of them coming down in the next five cards is about a 55% chance. Worked out by doing this:

44/50 0.8

43/49 0.87

42/48 0.87

41/47 0.87

40/46 0.87

x these numbers by each other and get 0.45 (roughly).

1 - 0.45 = 0.55

What i wanted to know now is if i hold two suited cards in my hand like two spades what is the chance of 3 spades coming down in the next 5 cards and how do i work this out?

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

Sounds like you're playing some Texas Hold'em!

You have 2 cards in your hand, leaving 50 cards that the other 5 will come from. There are 2118760 different sets of 5 cards out of those 50. (That's "50 choose 5" = 50! / (5! * 45*)).

There are 11 spades left in the deck. The number of ways exactly 3 spades can show up in the next 5 cards is:

(11 choose 3) * (39 choose 2) ! you want 3 of the the remaining spades plus 2 non-spades

(11! / (8! * 3!)) * (39! / (37! * 2!))

165 * 741 = 122265

So the odds of exactly 3 spades being flopped is the number of "good" hands divided by the total number of possible hands: 122265/2118760 =~ 5.7%

If you want the odds of 3 or more spades, you need to calculate the the number of possible hands with 4 spades and 5 spades:

4 spades = (11 choose 4) * (39 choose 1) = 330 * 39 = 12870

5 spades = (11 chosoe 5) * (39 choose 0) = 462 * 1 = 462

(122265 + 12870 + 462) / 2118760 =~ 6.4%

For the example you provided, it's really more complicated than that. The odds of getting making a pair is different than the odds of making a pair or better (2 pair, 3 of a kind, full house, etc.). I'm sure the odds of all these events are already out there on the internet. If you interested in how to compute them, you've come to the right place!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi phatus;

Using the hypergeometric distribution which is designed for these type urn problems without replacement is the easiest way to go. It yields

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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