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QA - Poker Probabilities

 

 

 

 

 

 

 


Log............................................................................................................................................................................................................ 3

Poker Probabilities..................................................................................................................................................................... 4

Poker (Encyclopædia Britannica).................................................................................................................................... 4

Poker probability in 52-card deck poker..................................................................................................................... 6

Probability..................................................................................................................................................................................... 6

Q: What is the probability getting Straight Flush, Four of a Kind, House etc. when given five Cards?...................... 6

Conditional probabilities.................................................................................................................................................. 11

Q: What is the probability to get a House or Four of a kind if you have Three of a kind, and choose to change two or one of the remaining cards?......................................................................................................................................................................... 11

Q: What is the probability to get a house if you have Two Pair, and choose to change the remaining card?.......... 12

Q: What is the probability to get a house, four of a kind, three of a kind or Two Pair if you have a Pair, and choose to change 2 or 3 of the remaining Cards?............................................................................................................................................................. 13

Q: What is the probability that the other players get at least one pair, two pair etc. when the 5 cards are dealt?. 17

Q: What is the probability that the other players get at least one pair, two pair etc. when all have changed cards? 17

Poker probability in 32-card deck poker................................................................................................................... 18

Introduction............................................................................................................................................................................... 18

Probabilities................................................................................................................................................................................ 18

Q: What is the probability getting Straight Flush, Four of a Kind, House etc. when given five Cards?.................... 18

References........................................................................................................................................................................................ 20

 


Log

Date

Description

2000-01-27

First version published at http://www.pvv.org/~nsaa/poker.html

2000-10-08

Corrected errors and added description in the following calculations :

P(Two pair | one pair AND change 3) = 2718/16215 = 2592/16215

P(Three of a kind | one pair AND change 3) = 1917/16215 = 1854/16215

P(House | one pair AND change 3) = 102/16215  = 165/16215

 

Thanks to jerome.sheahan@nuigalway.ie for these corrections.

2000-11-08

Added explanaition to P(3 of a kind) and P(House)

2004-08-03

Corrected error in the following conditional calculation

P(House | A Pair AND Change 2) = (Combin(2 3;2) + Combin(2;1)*Combin(3;1))/Combin(47;2) =

(1 3+6)/1081 =7 9/1081

Thanks to pabloposada@puebla.megared.net.mx for this correction

 

Changed the decimal separator to decimal sign(,) instead of  decimal point(.) since this is in accordance with ISO (ISO 31-0:1992 (E), Note 17).

2005-08-24

Added links to Wikipedia.org and made some further comments to quoted text from Encyclopædia Britannica.


Poker Probabilities

This document was originally created after some hefty discussion between me and some of my friends. What is the best suit: Flush or Strait? If you play Standard five-card draw and you have a pair and an Ace. Should you keep the pair or should you also keep the Ace? What about playing with 32-card deck? This document does these calculations, and many more. For the moment this document has not dealt with the Texas Hold’em poker variant.

 

I’ve also done some quotations from other sources where I have corrected miscalculations. This was done before the first version of this document was published at 2000-01-17 when good poker calculations were not commonly available. Today setup like the Wikipedia article Poker probability explains and state the standard probabilities very well.  

 

Poker (Encyclopædia Britannica)

I will in this section give some commets from an online article about poker published at   http://www.britannica.com/bcom/eb/article/6/0,5716,62116+1,00.html, but now removed.

 

The text in italic is from this article in Encyclopædia Britannica.

 

 Poker

 

 a family of card games, almost invariably played as gambling games. Although played

 internationally, Poker is most popular in North America.

 

Today Poker is played all over the world and is extremely popular everywhere, mainly because of internet, and the possibility to play online outside state control which’s very restrictive in most countries.

 

 A Poker hand usually consists of five cards. Players try for combinations of two or more cards

 of a kind, five-card sequences, or five cards of the same suit. (See below Rank of Hands.)

 

Today Texas Hold’em is the most popular mainstream variant, mainly because it’s played on the television aired World Series of Poker (abbreviated WSOP), widely recognized as the world championship of the game. Most online casinos also offer variants of Texas Hold’em as the standard game in online poker. Texas Hold’em consists of two cards down and five cards up, i.e. seven cards to use for each player at the final round.

 

 Poker is played with a standard 52-card deck in which all suits are of equal value, the cards

 ranking from the ace high, downward through king, queen, jack, and the numbered cards 10 to

 the deuce. The ace may also be considered low to form a straight (sequence) ace through five

 as well as high with king-queen-jack-10.

 

I’ve played poker with 32-card deck in Germany, and as you can see below the ranking of the hand changes!

 

Rank of hands.

 

 The traditional ranking is (1) straight flush (five cards of the same suit in sequence, the highest

 ace-king-queen-jack-ten being called a royal flush; (2) four of a kind, plus any fifth card; (3)

 full house; (4) flush; (5) straight; (6) three of a kind; (7) two pair; (8) one pair; (9) no pair,

 highest card determining the winner.

 

Below I’ve calculated all the possibilities, not just give the rank. As I stated above, 32-card deck, changes this rank.

 

 To determine the winner in hands in which there are hands of the same rank, the one

 containing the highest card wins. If the high cards are identical, the second highest wins, and so

 on. With full houses, the higher three of a kind wins; with two pairs, the highest pair wins, or if

 the pairs are identical, the odd high card wins, as in the case of identical pairs. For the occasion

 when none of the above applies, e.g., two flushes with identical cards in different suits, house

 rules may apply (the winning hand being the one in the higher bridge suit, or the two hands

 may split the pot). There is no universally accepted code of Poker rules. A code prepared by

 Oswald Jacoby in 1940 and a set of rules in the United States Playing Card Company's Official

 Rules of Card Games, published from 1945, are the ones usually adopted subject to house rules

 in the United States

.

Many people playing poker start arguing what to do when two people have the same flush with identical cards in different suits.  Make a rule before the game have started or split the pot 50/50 as default, if nothing else has been agreed upon.

 

 Wild cards. Most serious poker players decry the use of wild cards, but among the less serious,

 a wild card can be declared by the dealer (the deuce is most popular, but any rank can be

 used, or a distinguishable face card: the one-eyed jack). When there are wild cards in the

 game, the highest hand becomes five of a kind, though some house rules preserve the sanctity

 of the royal flush

 

Be aware that wild card can seriously change the probability and I will really recommend not using Wild cards in any serious game. You should at least be aware of the probabilities and have the correct rank calculated (if it changes it – I’ve not calculated this so I can’t really say). But to site a passage in Wikipedia’s article Wild card (poker)

Another issue with wild cards is that they distort the hand frequencies. In 5-card stud, the stronger hands are less frequent than the weaker hands; i.e., no pair is most common, followed by one pair, two pair, three of a kind, etc. When you add wild cards, the stronger hands gain frequency while the weaker hands lose frequency. For example, if you have a pair and a wild card, you will always choose three of a kind rather than two pair. This causes three of a kind to be more common than two pair.


Poker probability in 52-card deck poker

 

About the rank.

 

Probability

Q: What is the probability getting Straight Flush, Four of a Kind, House etc. when given five Cards?

 

A: Probability getting different hands when dealt five Cards

 

(Rank of Hands)

Probability getting this hand:

Out of comb(52;5) = 2598960 ways to draw five cards you will get the following hands these number of times

Exact probability

Approx.

Probability 1/

Approx. Probability num.

Royal Flush

4

1/649740

1/649740

0,00000154

Straight Flush

36

9/649740

1/72193

0,0000139

Four of a Kind

624 [5]

1/4165

1/4165

0,000240

House

3744[11]

6/4165

1/694

0,00144

Flush

5108 [8]

1277/649740

1/509

0,00197

Straight

10200

5/1274

1/255

0,00392

Three of a Kind

54912

88/4165

1/47

0,0211

Two Pair

123552

198/4165

1/21

0,0475

One Pair

1098240[11]

352/833

1/2.4

0,423

None

1302540

1302540/2598960

1/2

0,501

 

 

Probability getting at least this hand:

Out of comb(52;5) = 2598960 ways to draw five cards you will get the following hands these number of times

Exact probability

Approx.

Probability 1/

Approx. Probability num.

Royal Flush

4

 

1/649740

0,00000154

Straight Flush

40

 

1/64974

0,0000154

Four of a Kind

664

 

1/3914

0,000256

House

4408

 

1/590 [1]

0,00170

Flush

9516

 

1/273

0,00366

Straight

19716

 

1/132

0,00759

Three of a Kind

74628

 

1/35

0,0287

Two Pair

198180

 

1/13

0,0763

One Pair

1296420

 

1/2 [1]

0,499

None

2598960

 

1

1

 

1 in 5 for at least a pair of Jacks (11) [1]

 

Why:

 

P(House) = 13*12*COMBIN(4;3)*COMBIN(4;2) = 3744

 

 Why?

 13*12 because: 13 Columns horizontal and 12 rows vertical

   1-1-1-2-2        2-2-2-1-1        ...   13-13-13-1-1

   1-1-1-3-3        2-2-2-3-3              13-13-13-2-2

   1-1-1-4-4        2-2-2-4-4              13-13-13-3-3

   ...

   1-1-1-13-13-13   2-2-2-13-13-13         13-13-13-12-12

 

Lets look at one of these 1-1-1-2-2

 You have Combin(4;3) to draw 1-1-1 and Combin(4;2) to draw 2-2

 

 

P(3 of a kind) = 13*COMBIN(48;2)*(COMBIN(4;3) - P(House)

               = 13*1128*4 - 3744

               = 54912

 

 Why?

 13*COMBIN(48;2) 13 Columns horizontal and Combin(48;2) rows vertical as  above

   1-1-1-2-2        2-2-2-1-1        ...   13-13-13-1-1

   1-1-1-2-3        2-2-2-1-3              13-13-13-1-2

   1-1-1-2-4

   1-1-1-2-5

   ...

   1-1-1-3-3        2-2-2-3-3              13-13-13-2-2

   1-1-1-3-4        2-2-2-3-4              13-13-13-2-3

   1-1-1-3-5

   ...

   1-1-1-4-4        2-2-2-4-4              13-13-13-3-3

   ...

   1-1-1-13-13-13   2-2-2-13-13-13         13-13-13-12-12

 

 Lets look at 1-1-1-2-3. You have Combin(4;3) ways to draw 1-1-1.

 Since we now have Counted all the 1-1-1-2-2,1-1-1-3-3 etc. we must

 substract the house calculation.

 

-Another way to look at it is like this

 You have 13 Columns.

 Lets look at 1-1-1-y-x. You have Combin(4;3) ways to draw 1-1-1 and you

 have Combin(48;2) to draw y-x

 Since we now have Counted all the 1-1-1-2-2,1-1-1-3-3 etc. we must

 substract the house calculation.

 

From [14] : http://www.thewizardofodds.com/game/pokerodd.html

Royal Flush

The number of different royal flushes are four (one for each suit).

Straight Flush

The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen, or King. Thus there are 9 possible high cards, and 4 possible suits, creating 9 * 4 = 36 different possible straight flushes.

Four of a Kind

There are 13 different possible ranks of the 4 of a kind. The fifth card could be anything of the remaining 48. Thus there are 13 * 48 = 624 different four of a kinds.

Full House

There are 13 different possible ranks for the three of a kind, and 12 left for the two of a kind. There are 4 ways to arrange three cards of one rank (4 different cards to leave out), and combin(4,2) = 6 ways to arrange two cards of one rank. Thus there are 13 * 12 * 4 * 6 = 3,744 ways to create a full house.

Flush

There are 4 suits to choose from and combin(13,5) = 1,287 ways to arrange five cards in the same suit. From 1,287 subtract 10 for the ten high cards that can lead a straight, resulting in a straight flush, leaving 1,277. Then multiply for 4 for the four suits, resulting in 5,108 ways to form a flush.

Straight

The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen,King, or Ace. Thus there are 10 possible high cards. Each card may be of four different suits. The number of ways to arrange five cards of four different suits is 45 = 1024. Next subtract 4 from 1024 for the four ways to form a flush, resulting in a straight flush, leaving 1020. The total number of ways to form a straight is 10*1020=10,200.

Three of a Kind

There are 13 ranks to choose from for the three of a kind and 4 ways to arrange 3 cards among the four to choose from. There are combin(12,2) = 66 ways to arrange the other two ranks to choose from for the other two cards. In each of the two ranks there are four cards to choose from. Thus the number of ways to arrange a three of a kind is 13 * 4 * 66 * 42 = 54,912.

Two Pair

There are (13:2) = 78 ways to arrange the two ranks represented. In both ranks there are (4:2) = 6 ways to arrange two cards. There are 44 cards left for the fifth card. Thus there are 78 * 62 * 44 = 123,552 ways to arrange a two pair.

One Pair

There are 13 ranks to choose from for the pair and combin(4,2) = 6 ways to arrange the two cards in the pair. There are combin(12,3) = 220 ways to arrange the other three ranks of the singletons, and four cards to choose from in each rank. Thus there are 13 * 6 * 220 * 43 = 1,098,240 ways to arrange a pair.

Nothing

First find the number of ways to choose five different ranks out of 13 which is combin(13,5) = 1287. Then subtract 10 for the 10 different high cards that can lead a straight, to be left with 1277. Each card can be of 1 of 4 suits so there are 45=1024 different ways to arrange the suits in each of the 1277 combinations. However we must subtract 4 from the 1024 for the four ways to form a flush, leaving 1020. So the final number of ways to arrange a high card hand is 1277*1020=1,302,540.

 

Specific High Card Lets find the probability of drawing a jack high, for example. There must be four different cards in the hand all less than a jack, of which there are 9 to choose from. The number of ways to arrange 4 ranks out of 9 is combin(9,4) = 126. We must then subtract 1 for the 9-8-7-6-5 combination which would form a straight, leaving 125. From above we know there are 1020 ways to arrange the suits. Multiplying 125 by 1020 yields 127,500 which the number of ways to form a jack high hand. For ace high remember to subtract 2 rather than 1 from the total number of ways to arrange the ranks since A-K-Q-J-10 and 5-4-3-2-A are both valid straights.

Five Card Draw High Card Hands

Hand

Combinations

Probability

Ace high

502,860

0.19341583

King high

335,580

0.12912088

Queen high

213,180

0.08202512

Jack high

127,500

0.04905808

10 high

70,380

0.02708006

9 high

34,680

0.01334380

8 high

14,280

0.00549451

7 high

4,080

0.00156986

Total

1,302,540

0.501177394

 

 

 

From [11] http://www.sscnet.ucla.edu:80/soc/faculty/campbell/210a_Fall1997/210a_notes_10_14_97.htm

 

The first thing we need to know is how many elementary events there are that can occur. We just calculated it, it's 52!/(47!*5!)=2598960. Now all we have to do is work out how many hands correspond to each of the above three situations, and divide by this number.

 

PAIR:

To dealing with the probability of the pairs first, the first thing is to work out how many possible pairs there are. Well for any given value, there are (4 2) pairs that can be drawn, and there are 13 possible values, so there are 13 (4 2) ways of having a pair. How many combinations of the remaining 12 values are there that do not result in a pair among the remaining three cards? (12 3) Thus given 12 remaining values, there are (12 3) of picking three distinct ones from them, for example, 2 3 4, 2 3 5, 2 3 6, .... K Q A. Of course, for each of the three cards any suit is OK, we can have any combination of the 4 suits, so we have to multiply by 4^3. Thus the probability of having one pair, and three distinct remaining cards, is 13 (4 2) (12 3) 4^3 / (52 5). If we work it out, it's about 0.40.

 

FULL HOUSE:

A similar approach can be taken for the full house. There are 13 (4 3) 12 (4 2) ways of having a full house, so the total probability

of a full house is 13 (4 3) 12 ( 4 2) / (52 5) = 0.0014.

 

FLUSH

A flush is 4 (13 5)/(52 5)

 

What about Royal Straight flush Need to substract  40!!!

 

FOUR OF A KIND

What about four of a kind? There are 13 ways of four of a kind, 12 choices for the remaining card, so 13*12 / (52 5). Pretty

unlikely!

 

WRONG !!! Not 12 choices for the remaining Cards BUT 48 See [5] who agrees with me.

 

Flush: (From http://www.schoolnet.ca/vp-pv/amof/e_combI.htm)

We give now a simple question that can be answered with a knowledge combinations and binomial coefficients. What is the probability of getting a flush in a five card poker hand on the initial deal? (A flush means that all five cards are in the same suit.) First, we have to recognize that a five card poker hand is a combination of 5 cards chosen from 52 cards. Thus the total number of possible hands is the binomial coefficient C(52,5) = 2,598,960. The ranks of the cards making up the flush is a combination of 5 ranks chosen from 13 rank. The suit of the cards making up the flush is a combination of 1 suit chosen from 4 suits. Multiplying,

there are thus C(13,5)*C(4,1) = 1287*4 = 5148 ways of getting a flush. The probability of getting a flush is the ratio of the number of ways of getting a flush divided by the total number of hands; it is 5148/2598960 = 33/16660 = .001980792317. Not very high

odds --- about 2 in every 1000 hands!

 

Need to subtract 4 (royal flush) and 36 Straight flush

 

 

 

 


Conditional probabilities

Q: What is the probability to get a House or Four of a kind if you have Three of a kind, and choose to change two or one of the remaining cards?

 

 

 

Summary

You have Three of a kind and Change Cards

Gets House

Gets Four of a kind

Total

2

0,0611

0,0425

0,1036 (approx.. 1/9)

1

0,0638

0,021

0,0851 (approx. 1/12)

You should always change two Cards. Then you will get a house or Four of a kind in 1 out of 9 times.

 

Summary with fractions

You have a Three of a kind and

Change 2 Cards – Probability

Change 1 Cards - Probability

Same (Three of a kind)

969/1081 (approx. 1/1,1)

43/47 (approx. 1/1.09)

House or better

112/1081 (approx. 1/9)

4/47 (approx. 1/12)

 

House

 

66/1081 (approx. 1/16)

 

3/47 (approx. 1/16)

 

Four of a kind

 

46/1081 (1/23.5)

 

1/47

 

WHY:

Change 2 Cards

 

There are Comb(47;2) = 1081 possible ways to draw 2 Cards from the remaining (52-5=) 47 unknown Cards.

 

P(House | Three of a kind AND Change 2) =

(2*Combin(3;2) +

10*Combin(4;2))/Comb (47;2) =

66/1081 (approx. 1/6)

 

Why? You got f.ex. this hand 7-7-7-6-8, you throw away six and eight

2*Combin(3;2)  : (7-7-7-6-6 or 7-7-7-8-8) It’s 3 six’ or 3 eight’s in the remaining 47 Cards

10*Combin(4;2)  : (7-7-7-1-1 or 7-7-7-2-2 or etc) It’s 4 one’s, 4 two’s, etc in the remaining 47 Cards

 

P(Four of a kind | Three of a kind AND Change 2) =

(Combin(1;1)*(47-1))/Comb(47;2) =

46/1081 (approx. 1/23.5)

 

Why? You got f.ex. this hand 7-7-7-6-8 and you throw away 6-8.

You have 1 seven among the remaining 47 unknown Cards. It’s possible to combine this seven with the all the other 46 unknown Cards.

 

P(House OR Four of a kind | Three of a kind AND change 2) =

(66+46)/Comb(47;2) =

112/1081 (approx. 1/9)

 

Change 1 Card

 

There are Comb(47;1) = 47 possible ways to draw 1 Cards from the remaining (52-5=) 47 unknown Cards.

 

P(House | Three of a kind AND Change 1) =

Combin(3;1)/Comb(47;1) =

3/47 (approx.. 1/15)

 

Why? You got f.ex. this hand 7-7-7-6-8 and you throw away the eight.

You have 3 six’s among the remaining 47 unknown Cards.

 

P(Four of a kind | Three of a kind AND Change 1) =

1/Comb(47;1) =

1/47

 

P(House OR Four of a kind | Three of a kind AND change 1) =

(3+1)/Comb(47;1) =

4/47 (approx.. 1/12)

 

 

 

Q: What is the probability to get a house if you have Two Pair, and choose to change the remaining card?

 

A:

P(House | Two Pair) = (2+2)/Comb(47;1) = 4/47 (approx. 1/12)

Why? For example you have these Two Pair (12, 12) and (3, 3) and you discard the fifth card (a 5). Then you have 47 remaining (52-5) cards where 2 of them are 12’s and two of them are 3’s. Then it is 4 out of 47 to get either a third 12 or a third 3.

 


Q: What is the probability to get a house, four of a kind, three of a kind or Two Pair if you have a Pair, and choose to change 2 or 3 of the remaining Cards?

 

A: There are four strategies. Keep all Cards, Change one, two or three of the remaining Cards. It’s quit obvious that you always should either change two or three Cards if you will maximize your probability to get better Cards (Except when you are “bluffing”).

 

 

Summary

You have a Pair and Change x Cards

Get Two Pair

Get Three of a kind

Gets House

Gets Four of a kind

Total

3

0,160

0,11

0,01

0,0028

0,29

2

0,172

0,078

0,0083

0,00093

0,26

 

 

Summary with fractions

You have a Pair and

Change 3 Cards – Probability

Change 2 Cards - Probability

Same (a Pair)

11559/16215 (approx. 1/1,4)

801/1081 (approx. 1/1,3)

Two Pair or better

4656/16215 (approx. 1/3,5)

280/1081 (approx. ¼)

 

Two Pair

 

2592/16215 (approx. 1/6)

 

186/1081 (approx. 1/6)

 

Three of a kind

 

1854/16215 (approx. 1/9)

 

84/1081 (approx. 1/13)

 

House

 

165/16215 (approx. 1/98)

 

9/1081

 

Four of a kind

 

45/16215 (approx. 1/360)

 

1/1081[1]

 

 

WHY:

Change 3 Cards

 

There are Comb(47;3) = 16215 possible ways to draw 3 Cards from the remaining (52-5=) 47 unknown Cards.

 

P(Two pair | one pair AND change 3) =

(Combin(3;2)*3*(47-2-1-2) +

Combin(4;2)*9*(47-2-2-2))/Combin(47;3)=

(378+2214)/16215=

2592/16215 (approx. 1/6)

 

Why? You have 5 known Cards where two of them are a pair, and the rest is different (ex. 7-7-5-6-8). You have 47 remaining unknown Cards. This 47 unknown Cards contains a pair (7-7), 3 three of a kind (5-5-5, 6-6-6, 8-8-8) and 9 Four of a kind (1-1-1-1,2-2-2-2,3-3-3-3,4-4-4-4,9-9-9-9,…,13-13-13-13).

You are not interested in the other pair. This card will give you three or four of a kind.

 

The 3 Three of a kind can be combined in 3*Combin(3;2) ways. This again can be combined with 47 (all unknown) – 3 unknown cards used in Combin(3;2) – 2 other cards belonging to the pair (other two 7’s).

 

The 9 Four of a kind can be combined in 9*Combin(4;2) ways. This again can be combined 47 (all unknown) – 4 unknown cards used in Combin(4;2) ) – 2 other cards belonging to the pair (other two 7’s).

 

 

You still don’t believe me?

 

You got this hand 7-7-5-6-8 and you throw away 5-6-8. Then the possibilities to get two pair with either 5-5, 6-6 or 8-8 combined with 7-7 is:

5-5-x

5-x-5

x-5-5

In these 3 combinations the last single Card can be substituted with all remaining 47 Cards except the three 5’s and the two other 7’s.

=3*(47-2-1-2)

added by

6-6-x

6-x-6

x-6-6

In these 3 combinations the last single Card can be substituted with all remaining 47 Cards except the three 6’s and the two other 7’s.

=3*(47-2-1-2)

added by

8-8-x

8-x-8

x-8-8

In these 3 combinations the last single Card can be substituted with all remaining 47 Cards except the three 6’s and the two other 7’s.

=3*(47-2-1-2)

=3*3*(47-2-1-2)

=3*Combin(3;2)*(47-2-1-2)

=378

 

In the same manner

 

You got this hand 7-7-5-6-8 and you throw away 5-6-8. Then the possibilities to get two pair with either 1-1,2-2,3-3,4-4, 9-9,10-10,11-11,12-12 or 13-13 combined with 7-7 is:

1-1-x-x

1-x-1-x

1-x-x-1

x-1-1-x

x-1-x-1

x-x-1-1

In these 6 combinations the last Card can be substituted with all remaining 47 Cards except the four 1’s and the two other 7’s.

=6*(47-2-2-2)

added by

2-2-x-x

….

etc..

=9*6*(47-4)

=9*Combin(4;2)*(47-2-2)

=2214

Q.E.D.

 

 

P(Three of a kind | one pair AND change 3) =

[Combin(2;1)*Combin((47-2);2)

-Combin(2;1)*3*Combin(3;2)

-Combin(2;1)*9*Combin(4;2)]/16215=

2*(990-9-54)/16215=

1854/16215 (approx. 1/9)

 

Why?  You have 5 known Cards where two of them are a pair, and the rest is different(f.ex. 7-7-5-6-8). You have 47 remaining unknown Cards. This 47 unknown Cards contains a pair (7-7), 3 three of a kind (5-5-5, 6-6-6, 8-8-8) and 9 Four of a kind (1-1-1-1,2-2-2-2,3-3-3-3,4-4-4-4,9-9-9-9,…,13-13-13-13).

 

You have two 7’s that will give you the Third 7 (Combin(2;1))and 47-2 other cards to fill the Combin(45;2) remaining hand.

 

You need to subtract the possible house you can get with either 5-5, 6-6 or 8-8. F.ex. 7-7-7-5-5. You have 3 pair like this, and each pair can be drawn out of three 5’s, 6’s or 8’s (Combin(3;2)) . The two remaining 7’s Combin(2;1).

 

You also need to subtract the house you can get with either 1-1, 2-2, 3-3,4-4,9-9,…or 13-13. F.ex. 7-7-7-1-1You have 9 pair like this, and each pair can be drawn out of four 1’s, 2’s, 3’s etc.(Combin(4;2)) . The two remaining 7’s Combin(2;1).

 

P(House | one pair AND change 3) =

(3*Combin(3;3) +

9*Combin(4;3) +

Combin(2;1)*3*Combin(3;2) +

Combin(2;1)*9*Combin(4;2) )/16215=

165/16215 (approx. 1/98)

 

Why? You got this hand 7-7-5-6-8 and you throw away 5-6-8.

 

Add the bullet points:

House with the pair (7-7)

·         You have three 5-5-5, 6-6-6, 8-8-8 (3*Combin(3;3)) 

·         and nine 1-1-1-1,2-2-2-2, etc (9*Combin(4;3))

 

House with an extra card in the pair (7-7-7)

You can draw the extra 7 in Combin(2;1) ways.

·         You have three 5-5-5, 6-6-6, 8-8-8. You can draw 2 out of 3 of these (Combin(2;1)*3*Combin(3;2)). 

·         And nine 1-1-1-1,2-2-2-2, … , etc You can draw 2 out of 4 of these (Combin(2;1)*9*Combin(4;2)).

 

P(Four of a kind | one pair AND change 3) =

(Combin(2;2)*(47-2))/16215=

45/16215 (approx. 1/360)

 

Why? You got this hand 7-7-5-6-8 and you throw away 5-6-8. Then the possibilities to get the two other 7’s is all the 47 remaining unknown Cards except the two last 7’s.

 


Change 2 Card

 

There are Comb(47;2) = 1081 possible ways to draw 2 Cards from the remaining (52-5=) 47 Cards.

 

P(Two pair | one pair AND change 2) =

(Combin(3;1)*(47-3-2) +

9*Combin(4;2) +

2*Combin(3;2))/Combin(47;2)=

186/1081 (approx. 1/6)

 

Why? You got this hand 7-7-5-6-8 and you throw away 6-8

Combin(3;1)*(47-3-2)  : (7-7-5-5-*) One out of three 5’s multiplied by the 47 remaining - three 5’s - two 7’s

 

P(Three of a kind | A Pair AND Change 2) =

(Combin(2;1)*(47-3-2))/Combin(47;2) =

84/1081 (approx. 1/13)

 

Why? You got this hand 7-7-5-6-8 and you throw away 6-8

Combin(2;1)*(47-3-2)  : (7-7-5-7-*) One out of two 7’s multiplied by the 47 remaining - three 5’s - two 7’s.

 

P(House | A Pair AND Change 2) =

(Combin(3;2)+

Combin(2;1)*Combin(3;1))/Combin(47;2) =

(3+6)/1081

9/1081

 

Why? You got this hand 7-7-5-6-8 and you throw away 6-8

Combin(3;2) : 7-7-5-5-5 – You can draw two 5 out of tree remaining.

Combin(2;1)*Combin(3;1) : 7-7-5-7-5 One out of two 7’s and one out of three 5’s

 

P(Four of a kind | A Pair AND Change 2) =

Combin(2;2)/Combin(47;2) =

1/1081

 

 

Q: What is the probability that the other players get at least one pair, two pair etc. when the 5 cards are dealt? 

A: <…missing for the moment…>

Q: What is the probability that the other players get at least one pair, two pair etc. when all have changed cards?

 

-Suppose that people just throw away cards that don’t destroy any (one pair, two pair etc.)

A: <…missing for the moment…>


Poker probability in 32-card deck poker

 

Introduction

I’ve played 32 card deck poker in Germany and I discussed a lot with people what’s the correct rank of hands. As you see below it’s more difficult to get Flush than Four of a kind when you’re playing with 32 cards!

 

Probabilities

Q: What is the probability getting Straight Flush, Four of a Kind, House etc. when given five Cards?

 

A: Probability getting different hands when dealt five Cards

 

(Rank of Hands)

Probability getting this hand:

Out of comb(32;5) = 201376 ways to draw five cards you will get the following hands these number of times

Exact probability

Approx.

Probability 1/

Approx. Probability num.

Royal flush

4

1/50344

1/50344

0.0000199

Straight flush

12

3/50344

1/16781

0.0000596

Flush

208

 

1/968

0.00103

Four of a Kind

224

 

1/899

0.00111

House

1344

 

1/150

0.00667

Straight

4080

 

1/49

0.0203

Three of a Kind

10752

 

1/19

0.0533

Two Pair

24192

 

1/8

0.120

One Pair

107520

 

1/1.9

0.534

None

53040

 

1/3.8

0.263

 

 

Probability getting at least this hand:

Out of comb(32;5) = 201376  ways to draw five cards you will get the following hands these number of times

Exact probability

Approx.

Probability 1/

Approx. Probability num.

Royal flush

4

 

1/50344

0.0000199

Straight Flush

16

 

1/12586

0.0000795

Flush

224

 

1/899

0.00111

Four of a Kind

448

 

1/449

0.00222

House

1792

 

1/112

0.00900

Straight

5872

 

1/34

0.0292

Three of a Kind

16624

 

1/12

0.0826

Two Pair

40816

 

1/5

0.203

One Pair

148336

 

1/1.36

0.737

None

201376

 

1

1

 

Why:

 

You have

 

Royal Straight Flush

<…missing for the moment…>

References

 

[1]

http://www.britannica.com/bcom/eb/article/6/0,5716,62116+3,00.html

 

[2]

http://search.britannica.com/bcom/search/results/advanced/1,5844,4,00.html?p_query0=Poker&p_phraseT0=1

 

[3]

“Profitable Things to Watch in a Poker Game” http://www.cardplayer.com:80/caro.htm

1.      First question: Is this game worth my time? I need to see mistakes made by others that I wouldn't make myself. If I can't spot them, I'm probably in a bad game.

2.      Second question: What is my fantasy seat? By applying the criteria we've talked about in previous lessons (sit to the left of the loose players so that they act before you, also sit to the left of knowledgeable, aggressive players, and sit to the right of tight nonentity players)

3.      Try to reconstruct hands. Focus on just one opponent and – after seeing the showdown and while the next deal is being prepared - go back mentally and try to equate that player's hand with how he played at each stage of the action

4.      When looking for tells, focus on just one player.

5.      When you're out of a hand and don't feel like observing, don't.

6.      A simple, accurate way to rate your table. For 20 hands that you're not involved in: (a) Add one point for each call; (b) Subtract one point for each raise; and (c) Subtract one extra point for each check-raise (minus two points total). First bets are ignored in the count. Reraises count as a single raise (minus one point). All players' actions count, even when they act more than once on a single betting round. The higher the score, the better. You'll have to compare your results to other games of the same size, type, and number of players. But soon, you'll know with surprising accuracy how profitable today's game is compared to yesterday's.

 

[4]

“A Glossary of Poker Terms” http://www.conjelco.com/pokglossary.html

 

[5]

“what is the probability of getting a Royal Flush or Four of A Kind” http://www.sit.wisc.edu/~smwise/

 

[6]

http://www.cs.cornell.edu/cs100-sp99/ProgramDocs/P3_sol.htm

P3B Output:

 

P is the probability of having exactly one pair in

an n-card hand.

 

  n           P

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

  2         0.0588

  3         0.1694

  4         0.3042

  5         0.4226

  6         0.4855

  7         0.4728

  8         0.3923

  9         0.2751

 10         0.1599

 11         0.0745

 12         0.0262

 13         0.0062

14            0.0007

 

[7]

“Mathematical Probability” http://www.ms.uky.edu:80/~viele/sta281/mathprob/mathprob.html

 

[8]

C(13,5)*C(4,1) = 1287*4 = 5148 ways of getting a flush. (including Straight flush 40)”

 

[9]

“The Chances Of Winning The UK National Lottery” http://lottery.merseyworld.com/Info/Chances.html

 

[10]

“Link to nsaahttp://www.schoolnet.ca/vp-pv/amof/e_combI.htm

 

[11]

“We will consider three types of hands, following the book. A) One pair, with three different remaining cards, B) full house (3 of a kind, plus one pair), and C) flush (all cards of the same suit)”

http://www.sscnet.ucla.edu:80/soc/faculty/campbell/210a_Fall1997/210a_notes_10_14_97.htm

 

[12]

“5 cards selected at random from an ordinary deck.” http://www.math.iupui.edu/~momran/m118/chall.htm

 

[13]

“Ref. Book with questions, Answers” http://www.cs.colostate.edu:80/~anderson/cs201/exercises-4.4.html

 

[14]

http://www.thewizardofodds.com/game/pokerodd.html

 


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