If the cards are all different then there are 13C7 = 1716 different hands.
Poker hands are combinations of cards (when the order does not matter, but each object can be chosen only once.)The number 52C5 of combinations of 52 cards taken 5 at a time is (52x51x50x49x48) / (5x4x3x2x1) = 2,598,960.The number of hands which contain 4 aces is 48 (the fifth card can be any of 48 other cards.)So there is 1 chance in (2,598,960 / 48) = 54,145 of being dealt 4 aces in a 5 card hand.The odds are 54,144 to 1 against. The probabilityis 1/54145 = (approx.) 0.000018469 or 0.0018469%.
This is a combinations question. There are (52 C 13) possible hands. This is 52!/((13!)((52-13)!)) = 635013559600
Well the short answer is that your chances are pretty slim. The long answer: The total number of 13-card hands is 52-choose-13, which equals 52! / (13! * 39!). Any hand without any pairs must have exactly one of each value from 2 to ace. All there is to choose is the suit of each card, for 413 possible pairless 13-card hands. The probability of choosing one of these hands is the quotient of these two numbers, or 413 * 13! * 39! / 52!, which equals 4194304/39688347475, approximately 0.00010568 or 1 in 9462. Another explanation: Say the first card in your hand is an ace. For the second card, you have to avoid the other three aces in the deck. There are 51 cards, and 48 of them are OK, so the probability that the second card doesn't pair the first card is 48/51. Suppose your second card was a king. For the third card, you now have to avoid all remaining aces and kings. There are 50 cards left in the deck and you have to dodge six of them. So the probability that the third card doesn't pair either of the first two cards is 44/50. Each time you pick a card, there are three more cards you have to avoid, and one fewer card in the deck. So for the 4th card, you have to avoid 9 cards out of 49, giving a probability of 40/49. This continues until the 13th card, when the probability is 4/40. All these things have to happen one after the other, so you have to multiply the probabilities together: 48/51 x 44/50 x 40/49 x 36/48 x 32/47 x 28/46 x 24/45 x 20/44 x 16/43 x 12/42 x 8/41 x 4/40. This gives the same probability as above (about 1 in 9462).
That means your hands are not small, but large. You're welcome.
If the cards are all different then there are 13C7 = 1716 different hands.
2
three
Assuming the 52 cards are all different, the first card can be any of the 52, the second card can be any of the remaining 51, and the third card can be any of the remaining 50, so there are 52x51x50 different three card hands possible.
Asssuming you mean in a 13 card hand, to get exactly nine cards in a suit, there are 4 x C(13,9) x C(39,4) possible hands out of C(52,13) total hands. That is 4 x 58,809,465 / 635,013,559,600 = 0.000370445412454, or about 1 in 2700.
There are 15,820,024,220 ways.
You can draw C(52,13) = 52! /13! 39! = 635 013559 600 different 13-card hands from a deck of 52 cards.
Shuffling a deck of cards creates new combinations of hands . Unless you're playing dishonestly, all the cards in a game will be the same. Only after they're dealt will the hands be different. In genetics, crossing over creates new combinations of genes from a set of existing genes.
If the hand must contain three 8's and to cards that are not 8's - the total number of possibilities is 2801.
52*51*50*49*48------------------------120answer is 2598960
You can make 2,598,960 different 5 card hands (not counting permutations) with a standard 52 card deck.
There are 26 red cards and 26 black cards. 3 red cards can be chosen in 26C3 ways 2 black cards can be chosen in 26C2 ways The required answer is 26C3 X 26C2 ways. Answer: 1067742 S Suneja