Let's calculate the odds that you have a given hand. Here are the assumptions, 52 card deck (no wilds) and you are dealt exactly 6 cards. Also straights and flushes require all cards to be in it; e.g., not calculating five card straight flush/straight as a new separate hand.
To simplify the math, I'm calculating the odds you have this hand or better (in almost all cases). For example, when calculating the odds of say a flush, I'm finding the odds that all 6 cards have the same suit -- not necessarily the odds that all 6 cards have the same suit and also don't make a higher hand like a straight flush. Similarly, if you are dealt 2/2/3/3/3/K that counts as a high card, pair, two pair, three-of-a-kind, full house.
Note there are nine distinct straights with the high card being A,K,Q,J,10,9,8,7,6 (that is A-K-Q-J-10-9 to 6-5-4-3-2-A).
Please familiarize yourself with the notion of combination in math; e.g., choosing six distinct cards from a 52 card deck has nCr(52, 6)=20358520=52!/(6!*46!) different combinations of cards, which we use to find the odds of a hand. (You had 52*51*50*49*48*47=14658134400 ways of selecting the cards, but 6*5*4*3*2*1=720 possible orderings of these cards that you divide by when you don't care about order - in poker we treat the two-card hand A♠ K♥ is the same as K♥ A♠).
To summarize:
Hand | Combinations | Odds |
---|---|---|
Straight Flush | 36 | 1 in 565514.44 |
4-kind + Pair | 936 | 1 in 21750.55 |
3-kind + Three-of-kind | 2496 | 1 in 8156.46 |
Flush | 6864 | 1 in 2965.98 |
4-kind | 14664 | 1 in 1388.33 |
Straight | 36864 | 1 in 552.26 |
Full House (3-kind + pair) | 175968 | 1 in 115.694 |
Three Pairs (pair+pair+pair) | 370656 | 1 in 54.93 |
3-kind | 958048 | 1 in 21.25 |
Two Pairs (pair+pair) | 6334848 | 1 in 3.21 hands |
Pair | 17963400 | 1 in 1.13 |
High Card | 20358520 | 1 in 1 |
Please note in step 8 (three pair), I assume the three pairs have different ranks. Thus something like A♠A♥K♠K♦A♦A♣ which is technically both (4-kind+pair) as well as as three pairs, I didn't include as being three pairs (as it is natural to exclude this possibility). Similarly for two pairs I don't include the possibility of 4-kind (like A♠A♥K♠Q♦A♦A♣) being considered a two pair.