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Poker Statistics
Poker
Statistics - Holdem |
| |
Seven-Card
Stud High/Low |
Texas
Hold'em |
Omaha
Hold'em |
| |
| Improving
a pocket pair on the flop or beyond |
Discussion:
odds of flopping a set - no pair on board
50C3
= 19,600
total possible flops
2C1
= 2
ways of choosing the set card
12C2
= 66
ways of choosing two distinct ranks for the two other cards on the
flop
2 *
66 * 4 * 4 = 2,112
we mutilpy by four to allow for the four possible suits for each
of the two chosen ranks
2,112
/ 19,600 = .1078
the odds of flopping a set with no pair on board is 10.78%
back
to top
Discussion:
odds of flopping a set - with pair on board
2C1
= 2
ways of choosing the set card
12C1
= 12
ways of choosing the rank for the pair on board
4C2
= 6
ways of choosing the two cards of the paired rank
2 *
12 * 6 = 144
144
/ 19,600 = .007346
the odds of flopping a set with a pair on board (full house) is
.73%
back
to top
Discussion:
odds of flopping a full house - trips on board
12C1
= 12
ways of choosing the rank of the trips
4C3
= 4
ways of choosing the three cards of that rank
12
* 4 = 48
48
/ 19,600 = .002448
the odds of flopping trips on board (full house) is .24%
back
to top
Discussion:
odds of flopping quads
1 *
1 * 48 = 48
We can choose the first two cards in only one way: they must be
the quad cards. The other card can be any of the 48 remaining cards.
48
/ 19,600 = .002448
the odds of flopping quads is .24%
Discussion:
odds of turning a set
2 /
47 = .0426
since there are two cards in the deck out of remaining 47 that will
give you a set.
Discussion:
odds of turning quads
1 /
47 = .0213
since there is just one card in the deck out of 47 that will give
you quads.
back
to top
Discussion:
odds of having a set by the river -no pair on board
2C1
= 2
ways if choosing the set card
12C4
= 495
we must choose four distinct ranks
2 *
495 * 4*4*4*4 = 253,440
each of the four different ranks can be chosen in four ways
253,440
/ 2,118,760 = .1196
your chance of having a set by the river with no pair on board is
11.96%
back
to top
Discussion:
odds of having a set by the river -one pair on board
2C1
= 2
ways of choosing the set card
12C3
= 220
ways of choosing three distinct ranks (A,B,C) since one of the ranks
will be paired on board
4C2
= 6
ways of choosing the two cards among the rank that is paired on
board
2
* 220 * 4*4*6 = 42,240
42,240
* 3 = 126,720
since the paired rank may be A, B or C we must multiply by three
126,720
/ 2,118,760 = .0598
your chance of having a full house by the river with one pair on
board is 5.98%
back
to top
Discussion:
odds of having a set by the river -two pair on board
2C1
= 2
ways of choosing the set card
12C2
= 66
ways of choosing two distinct ranks, both of them will be paired
on board
4C2
= 6
ways of choosing the two cards among the rank that is paired on
board
2 *
66 * 6*6 = 4,752
4,752
/ 2,118,760 = .002242
your chance of hitting a set and the board pairing up twice is .22%
back
to top
Discussion:
odds of having a set by the river -trips on board
2C1
= 2
ways of choosing the set card
12C2
= 66
ways of choosing two distinct ranks (A,B); one of the ranks will
be tripped on board
4C3
= 4
ways of choosing the trips among the four card of the tripped rank
2 *
66 * 4*4 = 2,112
2,112
* 2 = 4,224
the tripped rank can be A or B
4,224
/ 2,118,760 = .001993
your chance of hitting a set and the board tripping up is .20%
back
to top
Discussion:
odds of having a set by the river -quads on board
2C1
= 2
ways of choosing the set card
12C1
= 12
ways of choosing the rank for the quads
4C4
= 1
we must choose all four cards of that rank
2 *
12 * 1 = 24
24
/ 2,118,760 = .00001132
your chance of hitting a set and quads hitting the board is .0011%
back
to top
Discussion:
odds of having quads by the river -no pair on board
12C3
= 220
ways of choosing three distinct ranks to go along with your quads
1 *
1 * 220 * 4*4*4 = 14,080
each of the three ranks can be chosen in four ways
14,080
/ 2,118,760 = .006645
your chance of hitting quads and the board not containing any other
pair is .66%
back
to top
Discussion:
odds of having quads by the river -pair on board
12C2
= 66
ways of choosing two distinct ranks (A,B) to go along with your
quads
4C2
= 6
we must choose two cards among the paired rank
1 *
1 * 66 * 4*6 = 1,584
1,584
* 2 = 3,168
we mutiply by two since the paired rank may be rank A or rank B
3,168
/ 2,118,760 = .001495
your chance of hitting quads and there being another pair on board
is .15%
back
to top
Discussion:
odds of having quads by the river - trips on board
12C1
= 12
ways of choosing the rank of the trips to go along with your quads
4C3
= 4
we must choose three cards among that rank
1 *
1 * 12 * 4 = 48
48
/ 2,118,760 = .0000265
your chance of hitting quads on board and the board tripping up
is .0022%
back
to top
Discussion:
odds of flopping a set (or quads)
50C3
= 19,600
total possible flops
48C3
= 17,296
flops that do not contain any card of the pocket pair rank
19,600
- 17,296 = 2,304
flops that contain one or more of the pocket pair rank
2,304
/ 19,600 = .1176
the odds of flopping a set or better is 12%
back
to top
Discussdion: odds of having a set
(or quads) by the river
50C5
= 2,118,760
total possible boards
48C5
= 1,712,304
boards that do not contain any card of the pocket pair rank
2,118,760
- 1,712,304 = 406,456
boards that contain one or more of the pocket pair rank
406,456
/ 2,118,760 = .1918
the odds of having a set or better by the river is 19%
back
to top
Odds
of improving a pocket pair in HE |
| hand |
percentage |
odds
against it |
| 1.
odds of flopping a set - no pair
on board |
10.78
% |
8.28:1 |
| 2.
odds of flopping a set - with pair
on board (full house) |
00.73
% |
135.99:1 |
| 3.
odds of flopping a full house - trips
on board |
00.24
% |
415.67:1 |
| 4.
odds of flopping any set (1 or 2) |
11.51
% |
7.69:1 |
| 5.
odds of flopping quads |
00.24
% |
415.67:1 |
| 6.
odds of flopping any full house or better (2, 3 or 5) |
01.21
% |
81.65:1 |
| 7.
odds of flopping any set or quads
(1,2 or 5) |
11.75
% |
7.51:1 |
| 8.
odds of turning a set |
00.04
% |
|
| 9.
odds of turning quads |
00.02
% |
|
| 10.
odds of having a set by the river
- no pair on board |
11.96
% |
7.36:1 |
| 11.
odds of having a set by the river
- one pair on board |
05.98
% |
15.72:1 |
| 12.
odds of having a set by the river
- two pair on board |
00.22
% |
453.55:1 |
| 13.
odds of having a set by the river
- trips on board |
00.20
% |
499.00:1 |
| 14.
odds of having a set by the river
- quads on board |
00.0011% |
90,908.09:1 |
| 15.
odds of having any set by the river (10,11,12,13,14) |
18.36
% |
4.45:1 |
| 16.
odds of having quads by the river
- no pair on board |
00.66
% |
150.52:1 |
| 17.
odds of having quads by the river
- pair on board |
00.15
% |
665.67:1 |
| 18.
odds of having quads by the river
- trips on board |
00.0022% |
45,453.55:1 |
| 19.
odds of having any quads by the river (16,17,18) |
00.82
% |
120.95:1 |
| 20.
odds of having any set or quads
by the river (15 or 19) |
19.18
% |
4.21:1 |
| 21.
odds of having a set or quads by the river - no trips or quads
on board (10,11,12,16,17) |
18.97
% |
4.27:1 |
| 22.
odds of having any full house or better by the river (11,12,13,14,19)
|
07.22
% |
12.85:1 |
back
to top
|
Hitting
a pair or better on the flop or beyond |
| Discussdion:odds
of flopping one pair (when holding two distinct ranks)
6 *
(44C2 = 946) = 5,676
5,676
/ 19,600 = .2896
you will flop one pair 29% of the time (that's 1 out of 3.5 flops)
Discussdion:
odds of flopping two pair (when holding two distinct ranks)
3 *
3 * 44 = 396
396
/ 19,600 = .0202
you will flop two pair 2% of the time (that's 1 out of 50 flops)
Discussdion: odds of flopping trips
(when holding two distinct ranks)
(3C2
= 3) * 44 = 132
132
* 2 = 264
the trips could be of rank A or rank B
264
/ 19,600 = .0135
you will flop trips 1.4% of the time (that's 1 out of 74 flops)
Discussdion: odds of flopping a
full house (when holding two distinct ranks)
(3C2
= 3) * 3 = 9
9 *
2 = 18
the trips could be of rank A or rank B
18
/ 19,600 = .0009184
you will flop a full house .09% of the time (that's 1 out of 1,088
flops)
Discussdion: odds of the flop
not containing any of your cards' rank (when holding two distinct
ranks)
44C3
= 13,244
the flop must exclude the six cards of the rank that you are holding
(50-6=44)
13,244
/ 19,600 = .6757
you will catch no help as far as pairs are concerned 68% of the
time
Discussdion: odds of catching some
help as far as pairs are concerned (when holding two distinct ranks)
5,676
+ 396 + 264 + 18 = 6,354
adding all the flops that give you some help
6,354
/ 19,600 = .3242
you will catch some help as far as pairs are concerned 32% of the
time
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|
Hitting
the Straight in Holdem |
| Discussdion:
odds of flopping an open-ended straight draw when holding JT
there
are three kinds of flops that will give you an open-ended
straight draw. x cannnot be any of the eight cards that complete
the straight
1-
KQx (x cannot be A or 9)
2- Q9x (x cannot be K or 8)
3- 98x (x cannot be Q or 7)
For
the first kind of flop KQx, assuming that x is not a K or Q, we
have 4*4*34=544 for all the possible flops (50-4-4-4-4=34). If the
x is a
K or Q, then we have the following possibilities KQK KQQ. Each of
these
can be selected in (4C2 = 6) * 4 = 24 ways. Adding it up we get
544 +
24 + 24 = 592 for the total number of flops that will give you an
open-ended draw of the first kind KQx.
Each
of the other kinds of flops can be combined in the same number of
ways. 592 * 3 = 1,776.
1,776
/ 19,600 = .0906
you will flop an open-ended straight draw with JT 9% of the time
As
you can see, some of these flops will contain a pair, which
decreases the value of your draw since a full house is now possible.
To
calculate the flops that do not contain a pair we eliminate the
48 flops that
are paired for each of the above three forms. 544 * 3 = 1,632
1,632
/ 19,600 = .0833
you will flop an open-ended straight draw that does not contain
a pair
8% of the time.
back
to top
odds of flopping the straight
when holding JT
there
are four kinds of flops that will give you a straight.
1-
AKQ
2- KQ9
3- Q98
4- 987
Each
of the above flops can be combined in 4*4*4 = 64 ways. Since we
have foure kinds of flops, we calculate 64 * 4 = 256 to yield the
number
of flops that will give you a pat straight.
256
/ 19,600 = .0131
you will flop a straight 1.3% of the time
back
to top
odds of having a straight by
the river when holding JT
The
straight can be based on any of the four "forms" listed
above (AKQ,
KQ9, Q98, 987). For each of these forms, we must choose two "kickers".
Lets explore the possiblities for form 1. x cannot be any A, K or
Q.
1-
AKQxx no pair
2- AKQxx with pair
3- AKQJT
4- AKQAA
5- AKQKK
6- AKQQQ
7- AKQAK
8- AKQAQ
9- AKQKQ
10- AKQAx
11- AKQKx
12- AKQQx
**************************************
1- AKQxx no pair
4*4*4
* (8C2 = 28) * 4*4 = 28,672 if it's not a J or T
4*4*4 * 3*32 = 6,144 if one is a J
4*4*4 * 3*32 = 6,144 if one is a T
total:
40,960
*********************************************
2- AKQxx with pair
4*4*4
* (8C1 = 8) * (4C2 = 6) = 3,072
4*4*4 * (3C2 = 3) = 192 pair of jacks
4*4*4 * (3C2 = 3) = 192 pair of tens
total:
3,456
**********************************************
3- AKQJT
4*4*4
* 3*3 = 576 if it's J and T
total:
576
**************************************************
4- AKQAA
5- AKQKK
6- AKQQQ
4*4
* (4C3 = 4) = 64
total:
64 * 3 = 192
*********************************************************
7- AKQAK
8- AKQAQ
9- AKQKQ
4 *
(4C2 = 6) * (4C2 = 6) = 144
total:
144 * 3 = 432
********************************************************
10- AKQAx
11- AKQKx
12- AKQQx
4*4
* (4C2 = 6) * 38 = 3,648
total:
3,648 * 3 = 10,944
**************************************************************
40,960
+ 3,456 + 576 + 192 + 432 + 10,944 = 56,560
total number of boards that contain AKQ
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--------------------------------------------------------------------------
Now let's explore all the boards that contain KQ9 that have NOT
already
been counted. This means we must eliminate all combinations that
contain an ace since that will put an AKQ on board
1-
KQ9xx no pair
2- KQ9xx with pair
3- KQJT9
4- KQ9KK
5- KQ9QQ
6- KQ999
7- KQ9KQ
8- KQ9K9
9- KQ9Q9
10- KQ9Kx
11- KQ9Qx
12- KQ99x
********************************************
1- KQ9xx no pair
4*4*4
* (7C2 = 21) * 4*4 = 21,504 if it's not a J or T
4*4*4 * 3*28 = 5,376 if one is a J
4*4*4 * 3*28 = 5,376 if one is a T
total:
32,256
*********************************************
2- KQ9xx with pair
4*4*4
* (7C1 = 7) * (4C2 = 6) = 2,688
4*4*4 * (3C2 = 3) = 192 pair of jacks
4*4*4 * (3C2 = 3) = 192 pair of tens
total:
3,072
**********************************************
3- KQJT9
4*4*4
* 3*3 = 576 if it's J and T
total:
576
**************************************************
4- KQ9KK
5- KQ9QQ
6- KQ999
4*4
* (4C3 = 4) = 64
total:
64 * 3 = 192
*********************************************************
7- KQ9KQ
8- KQ9K9
9- KQ9Q9
4 *
(4C2 = 6) * (4C2 = 6) = 144
total:
144 * 3 = 432
********************************************************
10- KQ9Kx
11- KQ9Qx
12- KQ99x
4*4
* (4C2 = 6) * 34 = 3,264
total:
3,264 * 3 = 9,792
**************************************************************
32,256
+ 3,072 + 576 + 192 + 432 + 9,792 = 46,320
total number of boards that contain KQ9
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-------------------------------------------------------------------------------
(AKQ, KQ9, Q98, 987)
When
calculating boards that contain Q98, we must eliminate
combinations that contain a K since that has already been counted
in the form 2
calculation but an Ace is allowed.
When
calculating boards that contain 987, we must eliminate
combinations that contain a Q since that has already been counted
in the form 3
calculation but an Ace and/or King is allowed.
Therefore,
form 3 (Q98) and form 4 (987) have the same number of
combinations as form 2 (KQ9)
In
each of the forms, there are 576 combinations that will contain
a
straight on board and you will therefore be "playing the board"
and you
will be unable to win the entire pot at a showdown, althought you
won't
lose either unless your opponent has a bigger straight.
For
the purpose of our calculation, we will eliminate these
combinations
AKQ
56,560 - 576 = 55,984
KQ9 46,320 - 576 = 45,744
Q98 46,320 - 576 = 45,744
987 46,320 - 576 = 45,744
55,984
+ 45,744 + 45,744 + 45,744 = 193,216
total possible straights with which you will be playing your hole
cards
to form the straight and you will therefore have the chance of winning
the entire pot.
193,216
/ 2,118,760 = .0912
your chance of having a straight by the river when holding JT is
9.1%
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|
Hitting
a Flush in Holdem |
| Discussion:
odds of flopping a flush draw when holding two suited cards
(11C2
= 55) * 39 = 2,145
2,145
/ 19,600 = .1094
your chance of flopping a flush draw is 11%
back
to top
Discussion: odds
of flopping a flush when holding two suited cards
(11C3
= 165)
165
/ 19,600 = .0084
your chance of flopping a flush is .8% (you will flop a flush once
out
of 119 flops)
back
to top
Discussion:
odds of having a flush by the river when holding two suited cards
The
flush can be formed in one of the following ways
1-
three-flush
2- four flush
3- five flush
**************************
1- three-flush
(11C3
= 165) * (39C2 = 741) = 122,265
****************************
2- four flush
(11C4
= 330) * (39C1 = 39) = 12,870
****************************************
3- five flush
(11C5
= 462)
*****************************************
122,265 + 12,870 + 462 = 135,597
adding it all up
135,597
/ 2,118,760 = .0640
you will have a flush by the river 6.4% of the time
***********************************************
back
to top |
Odds
of being dealt any of Sklansky's group 1-8 starting hands |
Discussion: odds of getting dealt a group-1 hand (AA, KK,
QQ, JJ, AKs)
52C2
= 1,326
total number of ways to catch the two hole cards
4C2
= 6
number of ways to catch any specific pocket pair
(4C1
= 4) * 1 = 4
ways
of catching any specific two ranks (AK in our case) suited. We multipy
by one because once the first card is selected, the other card MUST
be of the complimentary rank and of the same suit.
(4C1
= 4) * 3 = 12
ways of catching any specific two ranks non-suited. We mutiply by
three because once the suit of the first rank is selected, we can
only choose three suits for the other rank in order to form an offsuit
combination.
AA-
6
KK- 6
QQ- 6
JJ- 6
AKs- 4
6 +
6 + 6 + 6 + 4 = 28
adding it all up
28
/ 1,326 = .0211
your chance of catching a group-1 hand is 2.1%
back
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Discussion:
odds of getting dealt a group-2 hand (TT, AQs, AJs, KQs, AK)
TT-
6
AQs- 4
AJs- 4
KQs- 4
AK 12
6 +
4 + 4 + 4 + 12 = 30
adding it all up
30
/ 1,326 = .0226
your chance of catching a group-1 hand is 2.3%
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to top
Discussion:
odds of getting dealt a group-3 hand (99, JTs, QJs, KJs, ATs, AQ)
99-
6
JTs- 4
QJs- 4
KJs- 4
ATs- 4
AQ- 12
6 +
4 + 4 + 4 + 4 + 12 = 34
adding it all up
34
/ 1,326 = .0256
your chance of catching a group-3 hand is 2.6%
back
to top
Discussion:
odds of getting dealt a group-4 hand (T9s, KQ, 88, QTs, 98s, J9s,
AJ, KTs)
T9-
4
KQ- 12
88- 6
QTs- 4
98s- 4
J9s- 4
AJ- 12
KTs- 4
4 +
12 + 6 + 4 + 4 + 4 + 12 + 4 = 50
adding it all up
50
/ 1,326 = .0377
your chance of catching a group-4 hand is 3.8%
back
to top
Discussion:
odds of getting dealt a group-5 hand (77, 87s, Q9s, T8s, KJ, QJ,
JT, 76s, 97s, Axs, 65s)
77-
6
87s- 4
Q9s- 4
T8s- 4
KJ- 12
QJ- 12
JT- 12
76s- 4
97s- 4
Axs- 32
65s- 4
6 +
4 + 4 + 4 + 12 + 12 + 12 + 4 + 4 + 32 + 4 = 98
adding it all up
98
/ 1,326 = .0739
your chance of catching a group-4 hand is 7.4%
back
to top
Discussion:
odds of getting dealt a group-6 hand (66, AT, 55, 86s, KT, QT, 54s,
K9s, J8s, 75s)
66-
6
AT- 12
55- 6
86s- 4
KT- 12
QT - 12
54s - 4
K9s- 4
J8s- 4
75s- 4
6 +
12 + 6 + 4 + 12 + 12 + 4 + 4 + 4 + 4 = 68
adding it all up
68
/ 1,326 = .0513
your chance of catching a group-4 hand is 5.1%
back
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Discussion:
odds of getting dealt a group-7 hand (44, J9, 64s, T9, 53s, 33,
98, 43s, 22, Kxs, T7s, Q8s)
44-
6
J9- 12
64s- 4
T9- 12
53s- 4
33 - 6
98 - 12
43s- 4
22- 6
Kxs- 28
T7s- 4
Q8s- 4
6 +
12 + 4 + 12 + 4 + 6 + 12 + 4 + 6 + 28 + 4 + 4 = 102
adding it all up
102
/ 1,326 = .0769
your chance of catching a group-4 hand is 7.7%
back
to top
Discussion:
odds of getting dealt a group-8 hand (87, A9, Q9, 76, 42s, 32s,
96s, 85s, J8, J7s, 65, 54, 74s, K9, T8)
87-
12
A9- 12
Q9- 12
76- 12
42s- 4
32s - 4
96s - 4
85s- 4
J8- 12
J7s- 4
65- 12
54- 12
74s- 4
K9- 12
T8- 12
12
+ 12 + 12 + 12 + 4 + 4 + 4 + 4 + 12 + 4 + 12 + 12 + 4 + 12 + 12
= 132
adding it all up
132
/ 1,326 = .0995
your chance of catching a group-4 hand is 10%
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Chance
of getting dealt a playable starting hand in HE |
| Hand |
odds
of being dealt |
| group-1
hand (AA, KK, QQ, JJ, AKs) |
2.11% |
| group-2
hand (TT, AQs, AJs, KQs, AK) |
2.26% |
| group-3
hand (99, JTs, QJs, KJs, ATs, AQ) |
2.56% |
| group-4
hand (T9s, KQ, 88, QTs, 98s, J9s, AJ, KTs) |
3.77% |
| group-5
hand (77, 87s, Q9s, T8s, KJ, QJ, JT, 76s, 97s, Axs, 65s) |
7.39% |
| group-6
hand (66, AT, 55, 86s, KT, QT, 54s, K9s, J8s, 75s) |
5.13% |
| group-7
hand (44, J9, 64s, T9, 53s, 33, 98, 43s, 22, Kxs, T7s, Q8s) |
7.69% |
| group-8
hand (87, A9, Q9, 76, 42s, 32s, 96s, 85s, J8, J7s, 65, 54, 74s,
K9, T8) |
9.95% |
| group
1-2 |
4.37% |
| group
1-3 |
6.93% |
| group
1-4 |
10.70% |
| group
1-5 |
18.09% |
| group
1-6 |
23.22% |
| group
1-7 |
30.91% |
| group
1-8 |
40.86% |
| utter
trash |
59.14% |
back
to top |
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back
to top |
SCS
Three-Card Starting Hand Stats |
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