Compute the probability that a five-card poker hand is dealt to you that contains all hearts
Video Transcript13 hearts, and we're going to choose five of them and divide this by all 52 cards, choose five to get the probability. So if we have 13, choose five, Divided by 52, Choose five, you get approximately, and I'm going to write this as a fraction 33 Out of 6,066,640. Show
(most recent edit: January 2, 2005)A SINGLE PAIRThis the hand with the pattern AABCD, where A, B, C and D are from the distinct "kinds" of cards: aces, twos, threes, tens, jacks, queens, and kings (there are 13 kinds, and four of each kind, in the standard 52 card deck). The number of such hands is (13-choose-1)*(4-choose-2)*(12-choose-3)*[(4-choose-1)]^3. If all hands are equally likely, the probability of a single pair is obtained by dividing by (52-choose-5). This probability is 0.422569.TWO PAIRThis hand has the pattern AABBC where A, B, and C are from distinct kinds. The number of such hands is (13-choose-2)(4-choose-2)(4-choose-2)(11-choose-1)(4-choose-1). After dividing by (52-choose-5), the probability is 0.047539.A TRIPLEThis hand has the pattern AAABC where A, B, and C are from distinct kinds. The number of such hands is (13-choose-1)(4-choose-3)(12-choose-2)[4-choose-1]^2. The probability is 0.021128.A FULL HOUSEThis hand has the pattern AAABB where A and B are from distinct kinds. The number of such hands is (13-choose-1)(4-choose-3)(12-choose-1)(4-choose-2). The probability is 0.001441.FOUR OF A KINDThis hand has the pattern AAAAB where A and B are from distinct kinds. The number of such hands is (13-choose-1)(4-choose-4)(12-choose-1)(4-choose-1). The probability is 0.000240.A STRAIGHTThis is five cards in a sequence (e.g., 4,5,6,7,8), with aces allowed to be either 1 or 13 (low or high) and with the cards allowed to be of the same suit (e.g., all hearts) or from some different suits. The number of such hands is 10*[4-choose-1]^5. The probability is 0.003940. IF YOU MEAN TO EXCLUDE STRAIGHT FLUSHES AND ROYAL FLUSHES (SEE BELOW), the number of such hands is 10*[4-choose-1]^5 - 36 - 4 = 10200, with probability 0.00392465A FLUSHHere all 5 cards are from the same suit (they may also be a straight). The number of such hands is (4-choose-1)* (13-choose-5). The probability is approximately 0.00198079. IF YOU MEAN TO EXCLUDE STRAIGHT FLUSHES, SUBTRACT 4*10 (SEE THE NEXT TYPE OF HAND): the number of hands would then be (4-choose-1)*(13-choose-5)-4*10, with probability approximately 0.0019654.A STRAIGHT FLUSHAll 5 cards are from the same suit and they form a straight (they may also be a royal flush). The number of such hands is 4*10, and the probability is 0.0000153908. IF YOU MEAN TO EXCLUDE ROYAL FLUSHES, SUBTRACT 4 (SEE THE NEXT TYPE OF HAND): the number of hands would then be 4*10-4 = 36, with probability approximately 0.0000138517.A ROYAL FLUSHThis consists of the ten, jack, queen, king, and ace of one suit. There are four such hands. The probability is 0.00000153908.NONE OF THE ABOVEWe have to choose 5 distinct kinds (13-choose-5) but exclude any straights (subtract 10). We can have any pattern of suits except the 4 patterns where all 5 cards have the same suit: 4^5-4. The total number of such hands is [(13-choose-5)-10]* (4^5-4). The probability is 0.501177.
Chris P. What is the probability that all are of the same suit (all hearts, all clubs, all spades, or all diamonds)? 2 Answers By Expert Tutors
Quan P. answered • 10/14/13 Middle school, High school, College Math courses You can answer this question by the following steps: 1. Understanding the all possible outputs: 2. Calculate the probability for each output and then find the total probability. P(all hearts) = P(1st card is heart) * P(2nd card is heart) * P(3rd card is heart) * P(4th card is heart) * P(5th card is heart) = (13/52)*(12/51)*(11/50)*(10/49)*(9/48) The processes to find P(all clubs), P(all spades), P(all diamonds) are similar to find P(all hearts) But, notice that the number of hearts, clubs, spades, and diamonds are equal to each other, 13 cards. Therefore, you just multiply the P(all hearts) by 4, you will get the P(5 cards are same suit). Hope it helps. Quan.
Michael F. answered • 10/14/13 Mathematics Tutor The answer is the number of ways of selecting five cards from any single suit, namely 13C5, the number of favorable cases, divided by the number of ways of dealing 5 cards from a poker deck, namely 52C5, the result is (13×12×11×10×9)/(52×51×50×49×48) or about .000495198 for any individual suit. That any of the 4 suits be the suit of the flush is 4 times that. Thank you Quan P. Still looking for help? Get the right answer, fast.ORFind an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. What is the probability that a five card poker hand contains exactly one ace?The probability that a 5 card poker hand contains exactly one ace is ≈0.2995.
How many 5First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this previously, and found that there are 2,598,960 distinct poker hands.
What is the probability of getting 4 aces in a hand of poker with 5 cards?Chances of getting four Aces in 5-Card Draw: 1 in 54,145 hands on average. Chances of getting any four of a kind in Hold 'Em: 1 in 4,165 hands on average.
How many possible 5This means that if there are 52 cards, how many combinations of 5 cards can be drawn (answer 2,598,960 combinations).
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