Your odds of getting a 'three of a kind' poker result
In Texas holdem poker, from a single deck of 52 cards (13 'values' in every one of 4 suits) random cards are dealt as follows: 2 cards to each player and 5 on the table. The card values in ascending order are (ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, (ace). The Ace is special in the sense that it may either serve as a 1, or as the next higher value to a King. The 4 suits are spades, hearts, diamons and clubs, but these have no order in poker.
A player's result depends only the best combination of 5 cards that he can make out of the 7 cards that are either on the table or were dealt to him. The player may change the order of the cards.
The following combinations exist, in descending order:
- Royal flush: a special case of a straight flush, where the values are 10,J,Q,K,A.
- Straight flush: 5 cards of the same suit with consecutive values.
- Four of a kind: 4 cards of the same value (and any additional card).
- Full house: 3 cards of one value and 2 cards of another value
- Flush: 5 cards of the same suit
- Straight 5 cards of consecutive values
- Three of a kind: 3 cards of the same value (and two additional cards)
- Two pairs: 2 cards of the one value and 2 cards of another value (and one additional card)
- One pair: 2 cards of the one value (and three additional cards)
- Higest card: none of the above
Note: only the highest combination counts, so if from the 7 available cards, both a 'Flush' and 'Two pairs' could be formed, it will be classified as a Flush, not as a Two pairs, because a flush ranks higher than two pairs.
A priori, (before dealing any card) out of the 133,784,560 possible combinations of 7 cards that could be dealt, how many of these will be classified as a 'Three of a kind' result?
The answer is 6461620.
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1 solution
Pick 7 cards containing three cards of the same value as follows:
Note
So, avoiding Four of a kind and Full house there are 52×3×2÷3!×12!÷8!÷4!×4^4 = 6,589,440 combinations.
Now we also want to exclude combinations of 7 cards containing straights or flushes next to our 3 cards.
Straights
Depending on the value of the 3-of-a-kind cards, different numbers of straights can be formed containing one of that value. For example, if the value is 3, straights can be formed in the value ranges: A2345, 23456 or 34567. The counts are: A,2,K->2; 3,Q->3; 4,J->4; 5 through 10->5. Together that are 3×2+2×3+2×4+6×5=50 combinations: Each card in the straight can have any of 4 suits, so over all values there are 50×4^4 = 12800 straights to exclude; the suits of the triplet can be chosen in 4 over 3 = 4 ways. The overlap between Straights and Three of a kind is 51200.
Flushes
The 3 cards of the same 'kind' (value) are of different suits, so to also form a flush, all next 4 cards have to be of the same suit as one of the three suits. There are 52×3×2÷3!×3×12×11×10×9÷24 = 77,220 combinations that contain also a flush.
Straight flushes
We wanted to exclude hands that are either straights or flushes, but we want to do that only once for the hands that are both (straight/royal flush)! So we will again add the number of hands that are straight, flush and three of a kind at the same time.
It turns out that if within a deal of 7 cards, we may form three of a kind, and a flush, and a straight, we can always form a straight flush. Proof: The flush must have exactly 1 card in common with the 3 (same valued, different suited) cards that make up the three of a kind; the straight also uses exactly 1 of these 3, but any of them will do. Pick the same one that the flush uses, then the straight excludes the same 2 cards as the flush, and the straight and flush both consist of the same 5 cards, which then form a straight flush.
Straight flushes (including royal flushes) can occur in 4 suits, at 10 levels so 40 ways to have a straight flush, with 2 additional cards. In order to also form a three of a kind, the additional cards have to be of the same value as 1 of the 5 values of the straight. So pick one of these 5 values and then pick 2 out of 3 of the other cards of that value: 5×3×2/2=15 ways. Add 15×40=600
Result
The net total now is : 6,589,440 - 51,200 - 77,220 + 600 = 6,461,620 7-card combinations that are three of a kind, but nothing higher.
I had a program count the best combination for all different deals of 7 cards; for comparison, I also counted deals of just 5 cards
Note that for a 7-card deal (as opposed to a 5-card deal) it is rarer to have no combination at all than one or even two pairs!