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Poker - Imagine - 2011-11-15
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?
Poker - Loose - 2011-11-15
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
Poker - Imagine - 2011-11-15
Well, I can't interpret it as probability @have to find the number of hands and number of hands only, instead of the probability.
Poker - OB3LISK - 2011-11-15
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.
Poker - Imagine - 2011-11-15
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.
Poker - Kalovale - 2011-11-15
OB3LISK Wrote: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.
Poker - OB3LISK - 2011-11-16
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.
Poker - InfiniteFusion - 2011-11-16
Here is a link that explains everything.
http://en.wikipedia.org/wiki/Poker_probability#Derivation_of_frequencies_of_5-card_poker_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