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# Probability of Poker Hands

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Introduction Table of Content

 Subject Page # Introduction 1 Probability of getting no pairs (five different face values, not in sequence, not all cards in the same suit) 2 Probability of getting one pair (two cards of one face value and three cards of different face values, none matching the pair) 6 Probability of getting two pairs (one pair of each two different face values and a card of a third face value) 8 Probability of getting three of a Kind (exactly 3 cards of one face value and 2 different cards) 10 Probability of getting a straight (five cards in sequence, but not all the same suit) 12 Probability of getting a flush (five cards of the same suit but not in sequence) 14 Probability of getting a full house (three cards of one face value and two cards of another face value) 16 Probability of getting four of a kind (four cards of one face value and one other card) Probability of getting a straight flush (five cards in sequence of the same suit) Conclusion Work Sited

INTRODUCTION  Probability of getting no pairs

Five different face values, not in sequence, not all cards in the same suit A no pair, also known as a high card, occurs when a player has a set of five cards that are not in sequence, have five different face values, and that they are not of the same suit. No pair is the lowest ranked poker hand, and if all players have no pair, the player with the highest card wins (Ace being the highest, two being the lowest). Incase of a tie with the highest card, the second highest card is looked at. If a tie still occurs, the pot is split between the players. Following are the examples of a poker hands with no pairs.

Middle

4576

-----------

10829

Explanation:

Choosing two cards with the same face value:

In a standard deck of 52 cards, there are 13 different face values with each having 4 different suits, for example, there 9♣, 9♠, 9♦, and 9♥. 4C2 represents the number of ways of choosing two cards from the particular face value. For example, there are six ways to choose two cards from the nines. In addition, this could be possible with any of the thirteen face values and therefore we multiply it by 13. Therefore, there are 78 different ways to get a pair.

Choosing three cards of different face values, none matching the pair:

Assuming that the pair is already dealt, we must now choose three other cards with none of them matching the first pair. Since we have already dealt one face value of out thirteen possible face value, our sample space has now reduced to twelve. In other words we cannot have the same face value as the pair for any of the remaining three cards. For example, if we dealt a pair of nine in the first two flips, we can have any face value accept the nine for the remaining three cards. The remaining cards can have the face values of A, 2, 3, 4, 5, 6, 7, 8, 10, J, Q, and K. Therefore there are 12C3 ways to choose the three remaining cards with three different face values. Moreover, there are four possible suits for each face value; we must multiply it by 4^3 since there is no restriction on the suits. For example, if the three different cards are 3, 6, and 8, it is possible that 3 and 8 may share the same suit. Therefore, there are 12C3 x 4^3 ways to choose three different cards with different face values and none matching the pair.

Number of ways to choose cards without restriction:

Conclusion

♥, J♥, Q♥, K♥ and A♥.  Following is an example of the flush. As we can notice, all the five cards are from the same suit Measuring the probability of getting a flush (including the royal flush):

Pflush =       4C1 X 13C5 – (10 X 4)

-----------------------------

52C5

=         4 X 1287 - 40

---------------------

2598960

=           5108

---------------

2598960

=            1277

--------------

649740

Explanation:

To begin with, we must select 5 cards with each having the different face values. There are thirteen different face values and we are only choosing five of them. So the total number of combinations of five different face values are given by, 13C5. Furthermore, we must times it by 4C1 because there are 4 different suites for each. For example, we can have a flush with spades, diamonds, clubs and hearts.

Getting a straight flush:

Moreover, we must eliminate the possibility of getting a straight flush, in other words, possibility of them being in a sequence. There are 10 ways the five cards could be in the sequence. This was proven previously by listing all the possible straights. Furthermore, there are 4 different suits and this could occur with any suits and therefore, we must multiply 10 X 4. As a result, there are 40 ways to get a straight flush and therefore, we must subtract it from the total number of ways of getting a flush.

Number of ways to choose five cards without restriction:

In order to find the probability, we must use the formula P(A)= N(A)/N(S), where N(A) is the number of outcomes in which even A can occur and N(S) is the total number of possible outcomes. Therefore, N(S) in this case there are 52C5 ways to choose the 5 cards from the standard deck of 52 cards without any restrictions.

Probability of getting a full house

Three cards of one face value and two cards of another face value   Conclusion Chintan Patel and Ricky Bwaja                                      Probability of Poker Hands

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