Odds of suited connected cards making a straight flush?

Does anyone know the odds of suited connected starting cards making a straight flush by the river in Texas Hold'em? I play in a game that offers high hand bonuses, so it would be helpful for me to understand the likelihood of such cards making a straight flush. I believe the odds would be different for different suited connectors. I am specifically wondering about the odds of JTs, QJs, KQs, and AKs making a straight flush if played to the river.

You will end up using both cards no matter what since there's no gap in them. For AKs, it should be (3 choose 3)(47 choose 2)/(50 choose 5). KQs is twice that minus As appearing with JT9s, QJs is three times minus KT9s with As minus T98s with Ks, JTs is four times that minus KQ9s with As minus Q98s with Ks minus 987s with Qs. Minuses are due to double counting having king high and ace high simultaneously. Can ignore (3 choose 3) as you always need three specifix cards. I'm ignoring the times the board is straight flushed on it's own.

AKs: (47 choose 2)/(50 choose 5)
KQs: ((47 choose 2) + (46 choose 2))/(50 choose 5)
QJs: ((47 choose 2) + 2*(46 choose 2))/(50 choose 5)
JTs: ((47 choose 2) + 3*(46 choose 2))/(50 choose 5)

Anyone please feel free to correct me as I'm not a combinatorics master, but 95% sure this is right.

Do hands where the board reads a straight flush count? Say you start with QJ of clubs but the board is 4, 5, 6, 7, 8 of hearts.

lol that was fast. no, i didn't count those hands. only hands where you used your hole cards. Since there's no gap, is a little easier since using one card means you always use both and also since they're so high. 45s will sometimes not use the 4.

Some high hand bonuses require both hole cards play, some just one. Almost none would allow no hole cards to play.

No, you must use both hole cards.

Quote: lightningbolts

AKs: (47 choose 2)/(50 choose 5)
KQs: ((47 choose 2) + (46 choose 2))/(50 choose 5)
QJs: ((47 choose 2) + 2*(46 choose 2))/(50 choose 5)
JTs: ((47 choose 2) + 3*(46 choose 2))/(50 choose 5)

Close - well, AK is correct.

In every case, (50)C(5) (what I use for "50 choose 5") is correct for the denominator.
The numerator depends on the two hole cards.
Note that these calculations assume that you must use both hole cards.

Given any two suited cards as hole cards, the probability of making a straight flush is:
N x (47)C(2) / (50)C(5) = N / 1960
where N is the number of ways to make a straight flush with the two hole cards.
N equals:
4, if the cards are adjacent and the higher hole card is between 5 and Jack
3, with 3-4, J-Q, or the cards are 2 apart (e.g. 7-9) and the high card is between 5 and Queen
2, with 2-3, Q-K, 2-4, J-K, or the cards are 3 apart (e.g. 6-9) and the high card is between 5 and King
1, if either card is an Ace, or the cards are 4 apart (e.g. 5-9)

In your four specific cases:
AKs is 1 / 1960
KQs is 2 / 1960 = 1 / 980
QJ is 3 / 1960 = about 1 / 653
JT is 4 / 1960 = 1 / 490

In video poker we would say there is zero chance of getting a straight flush starting from AKs, as we distinguish between a royal flush (generally paying 4000 coins) and a straight flush (paying significantly less).

i'm pretty sure this terminology holds in Texas Hold'em too. On Youtube you can see various times somebody has hit a royal flush in this game --- and they NEVER call it a straight flush.

Quote: BobDancer

In video poker we would say there is zero chance of getting a straight flush starting from AKs, as we distinguish between a royal flush (generally paying 4000 coins) and a straight flush (paying significantly less).

i'm pretty sure this terminology holds in Texas Hold'em too. On Youtube you can see various times somebody has hit a royal flush in this game --- and they NEVER call it a straight flush.

Yes, because in VP there is a higher payout for a Royal Flush than for a straight flush. The same is true in many poker-based table games, such as UTH, where a Royal Flush has a separate payout category. But in Texas Hold-em, a Royal Flush functions simply as an Ace-high straight flush, so game mathematicians have no particular reason to give it a separate hand category. Hence, it is simply an issue of naming convention. IMO, saying that an AKs has a non-zero chance of making a straight flush in Texas Hold-em is perfectly comprehensible and not inherently incorrect.

The highest hand I've ever had in live Texas Holdem was a king high straight flush. I held KJ of diamonds and flopped it with the QT9 of diamonds. I won a whopping $35 in the pot. I've won much more in a single pot making less than a pair. A royal IS nothing more than an Ace high straight flush with a special name.

Yes, for KQs I was excluding an ace high straight flush, for QJs I was excluding K high and A high, for JTs I was excluding Q high K high and A high. It was already counted in the first 47 c 2.

Basically there would not be four suited cards on the board that make you two. So you can't choose between two different straight flushes. For KQs, I wasn't counting a board with 9TJA as it was both a king high and ace high (really just an ace high).

The first straight flush for KQs contains TJA so you choose 47 out of 2 for the other cards and always get an ace high. But if you choose 47 out of 2 for other cards with 9TJ, you include the A in the 47. Then this double counts the TJA as you have boards with 9TJA.

Quote: ThatDonGuy

Close - well, AK is correct.

In every case, (50)C(5) (what I use for "50 choose 5") is correct for the denominator.
The numerator depends on the two hole cards.
Note that these calculations assume that you must use both hole cards.

Given any two suited cards as hole cards, the probability of making a straight flush is:
N x (47)C(2) / (50)C(5) = N / 1960
where N is the number of ways to make a straight flush with the two hole cards.
N equals:
4, if the cards are adjacent and the higher hole card is between 5 and Jack
3, with 3-4, J-Q, or the cards are 2 apart (e.g. 7-9) and the high card is between 5 and Queen
2, with 2-3, Q-K, 2-4, J-K, or the cards are 3 apart (e.g. 6-9) and the high card is between 5 and King
1, if either card is an Ace, or the cards are 4 apart (e.g. 5-9)

In your four specific cases:
AKs is 1 / 1960
KQs is 2 / 1960 = 1 / 980
QJ is 3 / 1960 = about 1 / 653
JT is 4 / 1960 = 1 / 490

OOPS - that is double-counting some of the hands.
For example, if you have 56 suited, the "answer" is counting:
234xx
347xx
478xx
789xx
but, for example, both 234xx and 347xx are counting 2347 suited and K off-suit, and 234xx, 347xx, and 478xx are all counting 23478 suited.

Yeah, 789xx doesn't care what the two other cards are when you use both hole cards. You always have a 9 high. But 478xx can't include the 9 and still be 8 high. 347xx can't include the 8 (this prevents both 8 and 9 high). 234xx can't include the 7 (this prevents 7, 8, and 9). Thanks for checking the work, I always have trouble with these too.