Poker

How can I find the number of outcomes of each of the following hands in poker?
straight flush
4-of-a-kind
full house
flush
straight
3-of-a-kind
two pairs
pair
high card
I have to be able to deduct the number of hands possible for each of these hands for Finite (preferably with combinations/permutations, but whenever I try to think about it, I end up with the wrong answers. Can anyone tell me how to calculate the number of outcomes for each hand and what to consider for these calculations?

Permutation isn't important, since the order doesn't matter. First, you'll need to know how many hands are possible using a standard deck of 52 cards. So you'll need to do Combination(52 x 5) = 2,598,960.

Then you need to look at each specific case and how many possible ways you can get that hand. For example, a Straight Flush can be:

- A, 2, 3, 4, 5
- 2, 3, 4, 5, 6
- 3, 4, 5, 6, 7
...
- 9, 10, J, Q, K
- 10, J, Q, K, A (Royal Flush)

So, you have 10 possible ways of getting a Straight Flush (including a Royal Flush) and there's 4 suits, so there are 40 possible Straight Flushes in a deck of cards. Divide it by the number of hands and you get the probability of getting a Straight Flush.

40 / 2,598,960 = 0.00001539...

For the Four of a Kind, it would be the Four of a Kind + any card.

- A, A, A, A, 2 (Clubs)
- A, A, A, A, 2 (Diamonds)
...
- K, K, K, K, Q (Hearts)
- K, K, K, K, Q (Spades)

You'd have the 13 possible Four of a Kinds + any of the other 48 cards (52 - 4 cards), which is 624. There are 624 way of getting a Four of a Kind.

624 / 2,598,960 = 0.00024...

At least, I think that's how it's done. I really hated Probability & Statistics class. : P

Well, I can't interpret it as probability @have to find the number of hands and number of hands only, instead of the probability.

This is very simple lol.

straight flush
4-of-a-kind = 1 hand of course since you need all 4 of a card.
full house
flush
straight
3-of-a-kind
two pairs
pair
high card

Honestly this isn't statistics if you're looking at probability, it's more like logic. I don't remember specifically what half of those hands are again...

Look at pair. Let's label Spades as A, Clubs as B, Hearts as C, and Diamonds as D.

Let's look at a single card, let's look at 9.

9A and 9B is a pair, 9A and 9C is a pair, 9A and 9D is a pair, 9B and 9C is a pair, 9B and 9D are a pair, and 9C and 9D are a pair. Six distinct sets that are possible involving 9.

Now that is for one number, we have 13 numbers (2-10 + A + K + Q + J). So multiply our 6 sets x 13 different cards = 78 different possible hands for a pair.

That's not really statistics.

My teacher had 1,098,240 for pairs; I think he factored in the number of hand combinations possible that have a pair and only a pair.

Look at pair. Let's label Spades as A, Clubs as B, Hearts as C, and Diamonds as D.

Let's look at a single card, let's look at 9.

9A and 9B is a pair, 9A and 9C is a pair, 9A and 9D is a pair, 9B and 9C is a pair, 9B and 9D are a pair, and 9C and 9D are a pair. Six distinct sets that are possible involving 9.

Now that is for one number, we have 13 numbers (2-10 + A + K + Q + J). So multiply our 6 sets x 13 different cards = 78 different possible hands for a pair.

That's not really statistics.

That's only 78 different combinations of a pair, not the hands with the pair.

So...you have to count the five cards on the table....

This class sounds really fun. But it seems likema peach to work this out. I'll finish pair off later I just woke up.

Here is a link that explains everything.

http://en.wikipedia.org/wiki/Poker_proba...oker_hands

For the question involving pairs, the answer is 4C2*13*12C3*4^3 = 1,098,240

4C2 tells number of combinations of pairs of a certain rank.
13 is the number of ranks.
12C3 is the number of combinations of ranks that don't share the rank of the pair.
4^3 relates to the number of suits per card (the cards that aren't in pairs).

Alternative (better) explanation for poker pairs problem:

http://mathforum.org/library/drmath/view/56500.html