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Poker Probabilities Mathematical Calculations
52-card
deck
poker
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. Royal FlushThe number of different royal flushes are four (one for each suit). Straight FlushThe 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 KindThere 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 HouseThere 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. FlushThere 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. StraightThe 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 KindThere 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 PairThere 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 PairThere 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. NothingFirst 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.
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. 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 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
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
Summary with fractions
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…> 32-card deck pokerI’ve played 32 card 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! 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
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