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Log............................................................................................................................................................................................................ 3
Poker Probabilities..................................................................................................................................................................... 4
Poker (Encyclopædia Britannica).................................................................................................................................... 4
Poker probability in 52card 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 32card 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
Date 
Description 
20000127 
First version published at http://www.pvv.org/~nsaa/poker.html 
20001008 
Corrected errors and added description in the
following calculations : P(Two pair  one pair AND change 3) = P(Three of a kind  one pair AND change 3) = P(House  one pair AND change 3) = Thanks to jerome.sheahan@nuigalway.ie for these corrections. 
20001108 
Added explanaition
to P(3 of a kind) and P(House) 
20040803 
Corrected error in the following conditional
calculation P(House  A Pair AND Change
2) = (Combin( ( 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
310:1992 (E), Note 17). 
20050824 
Added
links to Wikipedia.org and made some
further comments to quoted text from Encyclopædia
Britannica. 
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
fivecard 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 32card 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 20000117 when
good poker calculations were not commonly available. Today setup
like the Wikipedia article Poker probability
explains and state the standard probabilities very well.
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
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, fivecard 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 52card 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 kingqueenjack10.
I’ve played poker
with 32card 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
acekingqueenjackten 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,
32card 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
.
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 oneeyed 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 5card 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.
About the rank.
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
11122 22211 ...
13131311
11133 22233 13131322
11144 22244 13131333
...
111131313 222131313 1313131212
Lets look at one of these 11122
You have Combin(4;3)
to draw
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
11122 22211 ...
13131311
11123 22213 13131312
11124
11125
...
11133 22233 13131322
11134 22234 13131323
11135
...
11144 22244 13131333
...
111131313 222131313 1313131212
Lets
look at 11123. You have Combin(4;3) ways to draw
Since we now have Counted all the 11122,11133 etc. we must
substract the house calculation.
Another way to look at it is like this
You have 13 Columns.
Lets
look at 111yx. You have Combin(4;3) ways to draw
have Combin(48;2) to draw yx
Since we now have Counted all the 11122,11133 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 4^{5} = 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 * 4^{2} =
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 * 6^{2} * 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 * 4^{3} = 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 4^{5}=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 98765 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 AKQJ10 and 5432A 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/vppv/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
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 (525=) 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 77768, you throw away six and eight
2*Combin(3;2) : (77766 or 77788) It’s 3 six’ or 3 eight’s in the remaining 47 Cards
10*Combin(4;2) : (77711 or 77722 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)*(471))/Comb(47;2) =
46/1081 (approx. 1/23.5)
Why? You got f.ex. this hand 77768 and you throw away 68.
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 (525=) 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 77768 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)
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 (525) 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.
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 (525=) 47 unknown Cards.
P(Two pair  one pair AND change 3) =
(Combin(3;2)*3*(47212) +
Combin(4;2)*9*(47222))/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. 77568). You have 47 remaining unknown Cards. This 47 unknown Cards contains a pair (77), 3 three of a kind (555, 666, 888) and 9 Four of a kind (1111,2222,3333,4444,9999,…,13131313).
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 77568 and you throw away 568. Then the possibilities to get two pair with either 55, 66 or 88 combined with 77 is:
55x
5x5
x55
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*(47212)
added by
66x
6x6
x66
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*(47212)
added by
88x
8x8
x88
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*(47212)
=3*3*(47212)
=3*Combin(3;2)*(47212)
=378
In the same manner
You got this hand 77568 and you throw away 568. Then the possibilities to get two pair with either 11,22,33,44, 99,1010,1111,1212 or 1313 combined with 77 is:
11xx
1x1x
1xx1
x11x
x1x1
xx11
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*(47222)
added by
22xx
….
etc..
=9*6*(474)
=9*Combin(4;2)*(4722)
=2214
Q.E.D.
P(Three of a kind  one pair AND change 3) =
[Combin(2;1)*Combin((472);2)
Combin(2;1)*3*Combin(3;2)
Combin(2;1)*9*Combin(4;2)]/16215=
2*(990954)/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. 77568). You have 47 remaining unknown Cards. This 47 unknown Cards contains a pair (77), 3 three of a kind (555, 666, 888) and 9 Four of a kind (1111,2222,3333,4444,9999,…,13131313).
You have two 7’s that will give you the Third 7 (Combin(2;1))and 472 other cards to fill the Combin(45;2) remaining hand.
You need to subtract the possible house you can get with either 55, 66 or 88. F.ex. 77755. 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 11, 22, 33,44,99,…or 1313. F.ex. 77711You 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 77568 and you throw away 568.
Add the bullet points:
House with the pair (77)
· You have three 555, 666, 888 (3*Combin(3;3))
· and nine 1111,2222, etc (9*Combin(4;3))
House with an extra card in the pair (777)
You can draw the extra 7 in Combin(2;1) ways.
· You have three 555, 666, 888. You can draw 2 out of 3 of these (Combin(2;1)*3*Combin(3;2)).
· And nine 1111,2222, … , 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)*(472))/16215=
45/16215
(approx. 1/360)
Why? You got this hand 77568 and you throw away 568. 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 (525=) 47 Cards.
P(Two pair  one pair AND change 2) =
(Combin(3;1)*(4732) +
9*Combin(4;2) +
2*Combin(3;2))/Combin(47;2)=
186/1081 (approx. 1/6)
Why? You got this hand 77568 and you throw away 68
Combin(3;1)*(4732) : (7755*) 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)*(4732))/Combin(47;2) =
84/1081 (approx. 1/13)
Why? You got this hand 77568 and you throw away 68
Combin(2;1)*(4732) : (7757*) 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 77568 and you throw away 68
Combin(3;2) : 77555 – You can draw two 5 out of tree remaining.
Combin(2;1)*Combin(3;1) : 77575 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
A: <…missing for the moment…>
Suppose that people just throw away cards that don’t destroy any (one pair, two pair etc.)
A: <…missing for the moment…>
I’ve played 32 card deck poker in
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…>
[1] 
http://www.britannica.com/bcom/eb/article/6/0,5716,62116+3,00.html 
[2] 

[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 checkraise (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/cs100sp99/ProgramDocs/P3_sol.htm P3B Output: P is the probability of having exactly one pair in an ncard 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 nsaa” http://www.schoolnet.ca/vppv/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/exercises4.4.html 
[14] 
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