Hi Guys,

I tried to google this and came up with scratch. I have had two straight flushs with suited connectors (KQ and 67) in less than a month, when I've played for barely 30 hours and I play very few hands (maybe 8 or 9 hands an hour). This seemed rather improbable to me, and so I was wondering. I realize the odds are not the same (there are no higher flushes than the royal flush with kq, whereas there are quite a few of them with 67), but since my casino pays 200 dollars for a straight flush and I play 3-6 limit tables, it is quite significant to my calculations at times.

Thank you for your time and knowledge,

Best,

Nir.

With a suited connector (JT – 54) you'll get a straight flush roughly once every 500 boards.

See http://www.propokertools.com/pql:

In general, if you elect to play a suited connector and always stay until the river you're about to get a straight flush roughly once every 660 Boards (probability 0.15%).

You're dealt a suited connector once every 26 hands (probability 0.04%).

Thus if you always play any suited connector, you should expect to see a straight flush roughly once every 17,000 hands (probability 0.006%).

Let's just assume you played 1,000 hands (roughly the amount you might see at a live table over 30 hours) and always played any suited connector. Then the probability of hitting more than 1 straight flush is (1 - BinomDist(1;1000;0.006%;1)) = 0.17% (1 : 600).

If you fold half of your suited connectors preflop the probability is (1 - BinomDist(1;1000;0.003%;1)) = 0.04% (1 : 2,400).

54s - 98s:
(4*C(46,2) + 1*45) / C(50,5)
=~ 0.0019752119

T9s:
(4*C(46,2) + 1*46) / C(50,5)
=~ 0.0019756839

JTs:
(3*C(46,2) + 1*C(47,2)) / C(50,5)
=~ 0.0019756839
(same as T9s by Pascal's triangle identity )

QJs:
(2*C(46,2) + 1*C(47,2)) / C(50,5)
=~ 0.0014871906

43s:
(3*C(46,2) + 1*45) / C(50,5)
=~ 0.0014867186

KQs:
(1*C(46,2) + 1*C(47,2))/ C(50,5)
=~ 0.00099869735

32s:
(2*C(46,2) + 1*45) / C(50,5)
=~0.00099822538

AKs:
1*C(47,2) / C(50,5)
=~ 0.00051020408

2As:
(1*C(46,2) + 1*45) / C(50,5)
=~ 0.00050973211

Note those include str8 flushes using 1 or 2 hole cards. To remove 1 hole card str8 flushes, remove the 1*45 or 1*46 terms, but it's a small difference.

Thank you, Bruce- that's amazing to know (of course I didn't understand most of the math or I could have done it myself - but that makes me all the more grateful!) --one thing I didn't understand. With a suited connector I get the straight flush every 500 boards but if I stay to the river with the same suited connector I should get it every +600 boards?? what is the difference? Or did you mean, every 600 boards it will be by the turn that I make the str8 flush?

At any rate, thank you so much!! I will play a lot more of my suited connectors now (normally I only play 1 in 3, in later positions or when they are 9t and above... I almost never play 23 34 or 45 )

Basically, 54s-JTs each make 4 str8 flushes, e.g., JTs makes A high, K high, Q high, J high. You can see that QJs and 43s only make 3, KQs and 32s make 2, and A2s and AKs make 1. Then just multiply by the number of ways to choose the other 2 cards, but the trick is to exclude the next higher card of the str8 flush. Otherwise you'd have the next higher str8 flush. So there will be 46 cards remaining instead of 47, except when we make the A high str8 flush, and then there are 47. So we multiply by C(47,2) or C(46,2) ways to choose those cards. That's 47*46/2 or 46*45/2. The other added number is for 1 card str8 flushes, but this doesn't change much. Then divide by the total number of 5 card boards C(50,5) = 50*49*48*47*46/(5*4*3*2*1).


That was sixhigh. He ran a simulation of all the suited connectors together. Of course that will hit less often than 1 in 500 which is only for 54s-JTs which make 4 str8 flushes. The others make 3/4, 1/2, and 1/4 as many, so they bring down the average.


23 and 34 are usually avoided because they make fewer str8s and often the low ends of str8s, called the "idiot ends", and low flushes. Also, play suited connectors for their ability to make str8s and flushes, not just str8 flushes which are rare.

what about say 97s 86s etc?

can someone help:
two people go all in every hand in heads up NL how often will they combined make a straight flush with two hole cards?
i.e. 62s on 345xxx would count etc but 6sJx on 5437ssss doesnt etc

I think you are asking what is the probability of making a straight flush in hold-em with a suited one-gapper where you must use one or both of your hole cards.

It might be easiest to simply walk through every case. Of course, there are C(50,5) = 2,118,760 possible 5-card boards for any 2-card hole card combination, so that is the denominator in the probability. What follows are the numerators.

AQ
A-T: KJT on board -> C(3,3)*C(47,2) = 1081
Q-8: JT98 on board, not K -> C(4,4)*C(45,1) = 45
5-A: 5432 on board, not 6 -> C(4,4)*C(45,1) = 45
Total = 1171

KJ
A-T: AQT on board -> C(3,3)*C(47,2) = 1081
K-9: QT9 on board, not A -> C(3,3)*C(46,2) = 1035
J-7: T987 on board, not Q -> C(4,4)*C(45,1) = 45
Total = 2161

QT
A-T: AKJ on board -> C(3,3)*C(47,2) = 1081
K-9: KJ9 on board, not A -> C(3,3)*C(46,2) = 1035
Q-8: J98 on board, not K -> C(3,3)*C(46,2) = 1035
T-6: 9876 on board, not J -> C(4,4)*C(45,1) = 45
Total = 3196

J9
A-T: AKQT on board -> C(4,4)*C(46,1) = 46
K-9: KQT on board, not A -> C(3,3)*C(46,2) = 1035
Q-8: QT8 on board, not K -> C(3,3)*C(46,2) = 1035
J-7: T87 on board, not Q -> C(3,3)*C(46,2) = 1035
9-5: 8765 on board, not T -> C(4,4)*C(45,1) = 45
Total = 3196

T8
A-T: AKQJ on board -> C(4,4)*C(46,1) = 46
K-9: KQJ9 on board, not A -> C(4,4)*C(45,1) = 45
Q-8: QJ9 on board, not K -> C(3,3)*C(46,2) = 1035
J-7: J97 on board, not Q -> C(3,3)*C(46,2) = 1035
T-6: 976 on board, not J -> C(3,3)*C(46,2) = 1035
8-4: 7654 on board, not 9 -> C(4,4)*C(45,1) = 45
Total = 3241

97
K-9: KQJT on board, not A -> C(4,4)*C(45,1) = 45
Q-8: QJT8 on board, not K -> C(4,4)*C(45,1) = 45
J-7: JT8 on board, not Q -> C(3,3)*C(46,2) = 1035
T-6: T86 on board, not J -> C(3,3)*C(46,2) = 1035
9-5: 865 on board, not T -> C(3,3)*C(46,2) = 1035
7-3: 6543 on board, not 8 -> C(4,4)*C(45,1) = 45
Total = 3240

86
Q-8: QJT9 on board, not K -> C(4,4)*C(45,1) = 45
J-7: JT97 on board, not Q -> C(4,4)*C(45,1) = 45
T-6: T97 on board, not J -> C(3,3)*C(46,2) = 1035
9-5: 975 on board, not T -> C(3,3)*C(46,2) = 1035
8-4: 754 on board, not 9 -> C(3,3)*C(46,2) = 1035
6-2: 5432 on board, not 7 -> C(4,4)*C(45,1) = 45
Total = 3240

75
J-7: JT98 on board, not Q -> C(4,4)*C(45,1) = 45
T-6: T986 on board, not J -> C(4,4)*C(45,1) = 45
9-5: 986 on board, not T -> C(3,3)*C(46,2) = 1035
8-4: 864 on board, not 9 -> C(3,3)*C(46,2) = 1035
7-3: 643 on board, not 8 -> C(3,3)*C(46,2) = 1035
5-A: 432A on board, not 6 -> C(4,4)*C(45,1) = 45
Total = 3240

64
T-6: T987 on board, not J -> C(4,4)*C(45,1) = 45
9-5: 9875 on board, not T -> C(4,4)*C(45,1) = 45
8-4: 875 on board, not 9 -> C(3,3)*C(46,2) = 1035
7-3: 753 on board, not 8 -> C(3,3)*C(46,2) = 1035
6-2: 532 on board, not 7 -> C(3,3)*C(46,2) = 1035
Total = 3195

53
9-5: 9876 on board, not T -> C(4,4)*C(45,1) = 45
8-4: 8764 on board, not 9 -> C(4,4)*C(45,1) = 45
7-3: 764 on board, not 8 -> C(3,3)*C(46,2) = 1035
6-2: 642 on board, not 7 -> C(3,3)*C(46,2) = 1035
5-A: 42A on board, not 6 -> C(3,3)*C(46,2) = 1035
Total = 3195

42
8-4: 8765 on board, not 9 -> C(4,4)*C(45,1) = 45
7-3: 7653 on board, not 8 -> C(4,4)*C(45,1) = 45
6-2: 653 on board, not 7 -> C(3,3)*C(46,2) = 1035
5-A: 53A on board, not 6 -> C(3,3)*C(46,2) = 1035
Total = 2160

A3
A-T: KQJT on board -> C(4,4)*C(46,1) = 46
7-3: 7654 on board, not 8 -> C(4,4)*C(45,1) = 45
6-2: 6542 on board, not 7 -> C(4,4)*C(45,1) = 45
5-A: 542 on board, not 6 -> C(3,3)*C(46,2) = 1035
Total = 1171

I think these are correct as I tried to be both methodical and careful. But if anybody sees anything that looks like it could be wrong, don't hesitate to say so.

thank you

I think you are asking what is the probability of one or both players in HUNL making a straight flush using both of their hole cards.

Principle of Inclusion-Exclusion (PIE) tells us in this case:

Prob(One or Both) = 2*Prob(One) - Prob(Both)


Prob(One)

Prob(One) is the probability of a single player making a straight flush in hold-em (assuming every hand goes to "showdown") using both of his hole cards. Throughout, of course, a royal flush is considered a straight flush.

I wrote three straightforward programs to tally the number of these cases in which this happens. For simplicity I gave the player two spade hole cards. Clearly player's hole cards must be suited and sufficiently close together and three cards of the same suit must appear on the board that make him a SF (without making a higher SF with fewer than two hole cards).

Case 1: 5 cards of suit appear on board
2,170 cases * 4 suits
= 8,680 deals

Case 2: 4 cards of suit appear on board
710 cases * 4 suits * 39 other cards for the other board card
= 110,760 deals

Case 3: 3 cards of suit appear on board
100 cases * 4 suits * C(39,2) other cards for the two other board cards
= 296,400 deals

TOTAL = 415,840 deals

To derive the probability of this happening we need to consider how many total one-player NLHE deals are possible. So we have the probability:

= 415,840 / [C(52,7)*C(7,5)]
= 415,840 / 2,809,475,760
= .0001480134 (approx 1 in every 6,756 deals)


Prob(Both)

Here we seek the probability of both players in a two-person hold-em game each make a straight flush using both of their hole cards on the same deal. Clearly both players have suited cards in the same suit and each "interact" with the board so that they each make a straight flush using both of their hole cards (without making a higher straight flush using fewer than two hole cards).

I wrote three parallel tally programs to derive these figures. Note that in these programs the first player always had the highest hole card since, of course, everything would be the same if the two players' hands were switched (meaning that below we need only consider half of all possible deals for the denominator).

Case 1: 5 cards of suit appear on board
403 cases * 4 suits
= 1,612 deals

Case 2: 4 cards of suit appear on board
83 cases * 4 suits * 39 other cards for the other board card
= 12,948 deals

Case 3: 3 cards of suit appear on board
8 cases * 4 suits * C(39,2) other cards for the other two board cards
= 23,712 deals

TOTAL = 38,272 deals

To derive the probability of this happening we need to consider how many total two-player NLHE deals are possible (where, as explained above, we should consider two deals with the same board and exact same two player hands switched as counting as only one deal). So we have the probability:

= 38,272 / [C(52,9)*C(9,5)*3!!]
= 38,272 / 1,390,690,501,200
= .0000000275201 (approx 1 in every 36,337,022 deals)


Prob(One or Both)

Putting it all together:

Prob(One or Both) = 2*Prob(One) - Prob(Both)

= 2*.0001480134 - .0000000275201

= .0002959992 (approx 1 in every 3,378 deals)


Simulations

I also ran some simulations to help estimate/confirm these figures. It should be pointed out that the simulation programs used the exact same underlying logic as the various tallying programs, so they do not provide as "independent" information as we would like.

First simulation was the 1-person case to estimate Prob(One). In one billion deals of one-person hold-em, the player made a straight flush using both hole cards in 147,947 deals. This represents a probability estimate of .000147947 which is close to the probability derived via the tallying programs of .0001480134.

Second simulation (separate from the first simulation) was the 2-person case to estimate Prob(One or Both). In one billion deals of HUNL, one or both players made a straight flush using both of their hole cards in 296,951 deals. This represents a probability estimate of .000296951 which is close to the probability derived via the tallying programs of .0002959992.

By the way, in this second simulation there were a grand total of 31 deals in which both players made straight flushes using both hole cards which is consistent with the probability derived via the tallying programs.

If anything is unclear or looks wrong, please let us know. Of course, other approaches or results are more than welcome.

Hello Guys,

I see high math skills around here, @whosnext and @BruceZ's approaches looks impressive to me.
Can anyone help me to find out the exact probability of each of the following hands hits straight flush until the river using two hole cards mandatorily?

AKss
AQss
AJss
ATss
A5ss
A4ss
A3ss
A2ss
KQss
KJss
KTss
K9ss
QJss
QTss
Q9ss
Q8ss
JTss
J9ss
J8ss
J7ss
T9ss
T8ss
T7ss
T6ss
98ss
97ss
96ss
95ss
87ss
86ss
85ss
84ss
76ss
75ss
74ss
73ss
65ss
64ss
63ss
62ss
54ss
53ss
52ss
43ss
42ss
32ss

Appreciate if someone can help, thanks in advance! Cheers

did you find your answer for this?

So can I conclude if I play this range:

23s+
24s+
25s+
8Qs+

I have 1 in every 7762 deals a straight flush and those deals also include K3o etc which don't qualify for the straight flush with a suited connector?