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Probability of getting exactly one pair in poker

This probability is TWO PAIR This hand has the pattern AABBC where A, B, and C are from distinct kinds. The number of such hands is (choose-2)(4-choose-2)(4-choose-2)(choose-1)(4-choose-1). After dividing by (choose-5), the probability is A TRIPLE. Following this logic, I tried to calculate the probability of getting two pair. My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank. Sep 03,  · The probability of getting exactly one pair will be the number of hands with exactly one pair divided by the total number of 5-card hands. You have already found the denominator, (52 C 5). Now we must find the numerator, which, as you said, is a lot more tricky.

Poker probability

Math Library:. Multiplying all these factors together gives the total number of ways that we could choose exactly one pair. Its "Total" represents the Pascal's work on this problem began an important correspondence between him and fellow mathematician Pierre de Fermat Suppose the face values are to be 3, 7, K as suggested above. Some variants of poker, called lowball , use a low hand to determine the winning hand. You asked for "odds" in your question, but do you mean "probability?

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In poker , the probability of each type of 5-card hand can be computed by calculating the proportion of hands of that type among all possible hands. Probability and gambling have been an idea since long before the invention of poker. The development of probability theory in the late s was attributed to gambling; when playing a game with high stakes, players wanted to know what the chance of winning would be.

In , Fra Luca Paccioli released his work Summa de arithmetica, geometria, proportioni e proportionalita which was the first written text on probability. Motivated by Paccioli's work, Girolamo Cardano made further developments in probability theory.

His work from , titled Liber de Ludo Aleae , discussed the concepts of probability and how they were directly related to gambling. However, his work did not receive any immediate recognition since it was not published until after his death.

Blaise Pascal also contributed to probability theory. Determined to know why his strategy was unsuccessful, he consulted with Pascal. Pascal's work on this problem began an important correspondence between him and fellow mathematician Pierre de Fermat Communicating through letters, the two continued to exchange their ideas and thoughts. These interactions led to the conception of basic probability theory. To this day, many gamblers still rely on the basic concepts of probability theory in order to make informed decisions while gambling.

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Math [ Privacy Policy ] [ Terms of Use ]. I have calculated the total number of possible hands as 2,, 52 C 5 , but I get stuck trying to figure out the possibility of exact hands. We will explain the numerator, term by term, starting at the lefthand end. There are 13 face values 2, 3, Suppose the face values are to be 3, 7, K as suggested above. We have 4 ways of choosing the 3, since there are four 3's in the pack.

Multiplying all these factors together gives the total number of ways that we could choose exactly one pair. And this divided by 52C5 gives the required probability.

You asked for "odds" in your question, but do you mean "probability? Odds tell you how many times an event will happen and how many times it won't, e. I assume that you are asking for the probability of getting exactly one pair in a hand of five cards randomly dealt from a card deck. The probability of getting exactly one pair will be the number of hands with exactly one pair divided by the total number of 5-card hands. You have already found the denominator, 52 C 5.

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Now we must find the numerator, which, as you said, is a lot more tricky. The total number of 5-card hands with exactly one pair can be thought of as the following: number of different face values that the pair might have x number of ways to pick two cards of the same face value from the four available in the deck x number of ways three more cards can be chosen such that none of them have the same face value and none share the same face value as the one chosen for the pair.

Now we have to calculate these quantities. First you must find the number of different ways of choosing one face value for the pair e. There are 13 face values 2,3, Now that we have chosen a particular face value let's say we chose Jacks we need to calculate the number of ways to choose two Jacks from the four Jacks available in the deck, yielding 4 C 2. Now comes the hard part: calculating the number of ways three more cards can be chosen such that none of them share the same face value as the one chosen for the pair.

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