Number of ways to choose 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 one pair
Two cards of one face value and three cards of different face values, none matching the pair
A pair occurs when a player obtains two cards of the same value from his set of five cards. The other three cards do not match the pair and do not have a pair among themselves. The pair with the higher value defeats the pair with the lower value. If two hands have the same pair, than the kicker or the highest non-paired cards are compared in order to determine the winner. It is ranked higher than the no pair hand because the probability if getting one pair hand is lower than the probability of getting no pairs. Following is an example of a one pair hand. As we can observe, this hand has a pair of Jacks and the remaining three cards are all different with none matching the pair.
Measuring the probability of getting a pair:
Pone pair = 4C2 x 13 x 4^3 x 12C3
--------------------------
52C5
= 13 x 6 x 220 x 64
----------------------
2598960
= 1098240
---------------
2598960
= 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:
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 two pairs
Exactly 3 cards of one face value and 2 different cards
A poker hand which has a pair of two different face values and a card of a third face value is known as a two pair hand. If the two hands have the two pair hand than the player with the higher face value pair wins. If two hands have the same highest pair, than the second pairs are compared to determine the winner. The two paired hand is ranked higher than the one paired hand since it is less likely to obtain than the one pair hand. Following is an example of the hand which contains two pairs. As we can observe, this hand has a pair of 7 and a 9 and the fifth card has a different face value that the two pairs.
Measuring the probability of getting two pairs:
PTwo pairs = 4C2 x 4C2 x 13C2 x 11 x 4
---------------------------------
52C5
= 6 x 6 x 78 x 11 x4
-------------------------
2598960
= 123552
----------------
2598960
= 10269
--------------
216580
Explanation:
Choosing the two pairs with the different face value:
As we already know, there are thirteen different face values in the standard deck of 52 cards. In the two paired hand, we must select two face values from the thirteen different face values. 13C2 represents the number of ways of selecting the two different face value cards for the 13 different face values. Furthermore, now we need to take the suites in consideration. Both pairs can have the any suites from the hearts, spades, clubs and diamonds. Therefore, the 4C2 x 4C2 represents the number of ways of selecting the suites for the both pairs. For example, 10♠, 10♣, 4♠, 4♥ and 8♥ have two pairs, a pair of 10s and the pair of 4s. In this hand, we have chosen spades and clubs for 10s and hearts and spades for the 4s.
Choosing the 5th card with the different face values:
Secondly, we must select the fifth card. The fifth card cannot share the same face value with either pairs. As a result, our sample space has reduced to 11 since we must eliminate the face values of the pairs. For example, if our two pairs are As and 2s, our fifth card cannot be an As or the 2s. In addition, there is no restriction on the suites and we have 4 options, hearts, spades, diamonds and clubs.
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 three of a kind
One pair of each two different face values and a card of a third face value
This hand occurs when we get three cards from the same face value, and the other two unmatched cards. Three of a kind hand is also known as trips. If two hands have the trips, the hand with the higher trips face value wins. The probability of getting a three of a kind is less likely than no-pairs, one pair and two pairs and therefore, it is ranked higher. Following is an example of a three of a kind hand. As we can see, there are three cards form the same face value, 10, and the other two cards are unmatched.
Measuring the probability of getting three of a kind:
PThree of a Kind = 4C3 x 13C1 x 12C2 x 4 x 4
----------------------------------
52C5
= 4 x 13 x 66 x 16
----------------------
2598960
= 54912
--------------
2598960
= 1144
------------
54145
Explanation:
Choosing three of a kind:
As we already know, there are thirteen different face values in the standard deck of 52 cards. When choosing the first card, we have thirteen different choices for the face value, for example, it could be any form A to K. Since we are only choosing one face value for all three cards, we can represent as 13C1. However, we must also take the suits in consideration. Even though the face values for all three cards are same, the suits differ. We can have the combination of any three suits from the total of four suits. As a result, we can use 4C3 to show that we can have a combination of 3 suits. Some of the combinations of the three suits are as follow:
♣, ♠, ♦
♥,♣, ♠
Choosing the remaining two unmatched cards:
Secondly, we must select the remaining two face cards. One important thing that we have to keep in mind while selecting the cards in that our sample space has now changed. Since the other two cards are not allowed to have the same face value as the three of a kind, our sample space has now reduced to twelve. Since the order in not important, we used combinations instead of the permutations. In other words, we can have the combination of any 2 face values from the remaining 12 face values. Therefore, we used 12C2 to show that we can have a combination of any two face values from the reduced sample space. Unlike the previous scenario where the suits were restricted, there are absolutely no restrictions in this case. 4 X 4 shows that for the two unmatched cards, any suits are possible. For example, if we have trips of 10s and the other two cards have the face values of 2 and 3, we can have four choices of suits for the 2 (2♣, 2♠, 2♦, or 2♥) and four choices for the 3(3♣, 3♠, 3♦, or 3♥).
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 straight
Five cards in sequence, but not all the same suit
A straight is a poker hand with five face values in sequence. However, the suits for the five different cards are different. The straight with the highest card is ranked higher than the other straight hands. The difference between the straight and the straight flush is that the straight flush have same suit for all the cards. Straight is less likely to occur than the previous hands and therefore it is ranked higher them in the game of poker. Following is an example of the poker hand with five cards in sequence. As we can observe, the face values for all the cards are in sequence, but they do not share the same suits.
Measuring the probability of getting a straight:
Pstraight = 4^5 x10 - 10 x 4
-------------------------
52C5
= 1024 x 10 - 40
---------------------
2598960
= 10200
---------------
2598960
= 255
--------------
64974
Explanation:
Getting a Straight:
There are ten total ways to get a straight while considering the face values only. The ten possible straights are as follow:
- A, 2, 3, 4, 5
- 2, 3, 4, 5, 6
- 3, 4, 5, 6, 7
- 4, 5, 6, 7, 8
- 5, 6, 7, 8, 9
- 6, 7, 8, 9, 10
- 7, 8, 9, 10, J
- 8, 9, 10, J, Q
- 9, 10, J, Q, K
- 10, J, Q, K, A
Secondly, since there are no restrictions for the suits, we can have 4 possible suits for each card. 4C1 represents the number of ways to choose a suit out of four total suits. Thus making it 4^5 for the five cards of the hand.
Getting a straight flush:
Furthermore, one of the given conditions for this hand is that, all the cards cannot share the same suit. A straight flush occurs when we have five cards in a sequence and at the same time they all share the same suit. An example of the straight flush is 2♣, 3♣, 4♣, and 5♣. The number of ways of getting a straight flush can be represented by 10 x 4, where 10 is the number of ways of selecting the first card. The first card could any from the following; A, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Moreover, 4 indicate that the straight flush could be from any of the four suites, ♣, ♠, ♦, or ♥. As a result we must subtract the possibility of getting a straight flush from the total possibility of getting a straight.
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 flush
Five cards of the same suit but not in sequence
Flush occurs when the hand gets five cards with the different face values but, they all share the same suit. The probability of getting this hand is lower than the probability of getting the previous hands and therefore it is ranked higher in the rank of poker. In the game of poker, a flush and the royal flush are two different hands but since it is not specified in this assignment, we have decided to include the royal flush in this category. Royal flush is when all they five cards are share the same suit and at the same time, they are all face cards. An example of the royal flush would be 10♥, 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
