Percentage on group of cards

[SlowHabit]

Hi all,

I am a math newbie when it comes assigning groups of hands to different percentages. For example, the top 10% of all hold 'em is what and what? Like what Sklansky group does it beong to? Where can I find more info on this?

Thanks.

[AaronBrown]

I'm not sure what you mean.

There are 52*51/2 = 1,326 possible starting hands in hold'em. 10% of that is 133. The best starting hands, in terms of probability of winning at a 10-player table if everyone stays in until showdown, are pairs 9 and above; AK, AQ and KQ; and if the cards are suited only, A with 8 or above, K with 9 or above, Q with T or above, J with 9 or above and T9. You can get each of the six pairs 6 ways (36), each of the three combinations 16 ways (48) and each of the twelve suited combinations 4 ways (48). Add them up and get 132, or 10% of starting hands.

If you look instead at how much people actually win with different starting cards, you add 88 and AJo to the hands above, and remove A8s, K9s, QTs, J9s and T9s.

[bachfan]

The "top 10% of hands" is, as Aaron suggests, not a precise concept on its own. If you are looking for a reasonable ordering of hands for a full ring game, I have computed a reasonable approximation.

details on the generation here

Begin ordering (hand in parens = soooted)
AA
KK
QQ
JJ
(AK)
TT
AK
(AQ)
99
(AJ)
AQ
88
(AT)
(KQ)
AJ
77
(KJ)
(QJ)
(KT)
KQ
(A9)
AT
66
(A8)
(QT)
(JT)
KJ
(A7)
(A5)
(K9)
(A4)
(A6)
55
(Q9)
(A3)
(J9)
KT
QJ
A9
(T9)
(K8)
(A2)
(K7)
44
A8
QT
(Q8)
JT
(J8)
(K6)
(98)
(T8)
(K5)
A7
(K4)
K9
A5
33
(K3)
A4
Q9
(87)
(Q7)
(T7)
(Q6)
(K2)
(J7)
A6
(97)
(Q5)
A3
J9
T9
22
K8
A2
(Q4)
(76)
K7
(86)
(96)
(J6)
(J5)
K6
(Q3)
(Q2)
(T6)
(65)
K5
(75)
Q8
(54)
J8
(J4)
T8
98
(85)
(95)
K4
(J3)
(64)
(T4)
(T5)
87
Q7
K3
(J2)
(74)
97
J7
(53)
Q6
(T3)
K2
(94)
T7
(84)
(43)
(63)
Q5
86
(T2)
(93)
76
Q4
(92)
96
(73)
J6
Q3
(52)
65
J5
T6
(82)
(42)
(83)
Q2
75
54
J4
(62)
85
(32)
95
(72)
J3
T5
T4
64
J2
53
74
84
T3
43
94
T2
93
63
92
73
52
42
83
82
62
32
72