# Texas Hold'em

July 12, 2004 8:47 PM Subscribe

What are the odds of a straight flush beats four of a kind on the river in Texas Hold'em?

I think he means, specifically, what are the odds that one person will have three of a kind and another will have four cards of a straight flush after the turn, and then the river comes up making both the four of a kind and a straight flush?

posted by kenko at 9:28 PM on July 12, 2004

posted by kenko at 9:28 PM on July 12, 2004

Well, at least 8 cards have already been dealt (leaving at most 44 cards). There's only one card that will make four of a kind and a straight flush, so it's at best 1 in 44, right?

posted by willnot at 9:35 PM on July 12, 2004

posted by willnot at 9:35 PM on July 12, 2004

impossible to say without knowing how many other players there are and what the previously dealt cards were.

posted by falconred at 9:39 PM on July 12, 2004

posted by falconred at 9:39 PM on July 12, 2004

Well, at least 8 cards have already been dealt (leaving at most 44 cards). There's only one card that will make four of a kind and a straight flush, so it's at best 1 in 44, right?

Well, at least 8 cards have already been dealt (leaving at most 44 cards). There's only one card that will make four of a kind and a straight flush, so it's at best 1 in 44, right?

Yes, that's pretty much all you can do.

For more information, and to save this post from being a simple--and, let's be honest, somewhat boring--math problem, go to www.cardplayer.com and check out the Hold 'Em calculator. Cardplayer is probably the best poker magazine in the world and they offer all their subscription content for free. This way you can plug any of millions of different Hold 'Em combinations into the calculator and check the odds for your disappointing defeats (or stirring, miracle victories).

Here's to languid, refreshing rivers.

posted by The God Complex at 9:52 PM on July 12, 2004

*Well, at least 8 cards have already been dealt (leaving at most 44 cards). There's only one card that will make four of a kind and a straight flush, so it's at best 1 in 44, right?*

that holds for the outcome of the last card, but the odds are much much much lower when you consider the chances against ever arriving at the initial configuration of 3 of a kind vs 4 card straight flush where the 3 of a kind happens to be in either the high or low card that would complete the straight flush...

posted by juv3nal at 10:00 PM on July 12, 2004

That's true, jun3nal. If it was an open-ended-straight-flush draw then the odds would be 1 in 22 instead of 1 in 44.

The question of what the odds of ever arriving at that situation are, however, beyond my ability to compute--or, for that matter, wholly beyond any desire I have to do so. I'm sure the odds are astronomically low, but I'm not sure if there's any legitimate use for such calculations (calculating the odds of someone making the straight flush could be very important, however, in order to examine how someone played/bet their hand).

Of possible interest to some is the "bad beat jackpot" offered at most major casinos (Foxwoods, etc.). Generally speaking, if you lose a four of a kind (usually one where you have two of the four in your hand) to a higher four of a kind or a straight/royal flush you are awarded the bad-beat jackpot, which grows every day. In some cases the amount of the bad-beat jackpot far exceeds any money lost for the hand (upwards of thousands of dollars); also, there's often a stipulation that anyone involved in the hand receives some of the money, even if they folded after the flop.

posted by The God Complex at 10:11 PM on July 12, 2004

The question of what the odds of ever arriving at that situation are, however, beyond my ability to compute--or, for that matter, wholly beyond any desire I have to do so. I'm sure the odds are astronomically low, but I'm not sure if there's any legitimate use for such calculations (calculating the odds of someone making the straight flush could be very important, however, in order to examine how someone played/bet their hand).

Of possible interest to some is the "bad beat jackpot" offered at most major casinos (Foxwoods, etc.). Generally speaking, if you lose a four of a kind (usually one where you have two of the four in your hand) to a higher four of a kind or a straight/royal flush you are awarded the bad-beat jackpot, which grows every day. In some cases the amount of the bad-beat jackpot far exceeds any money lost for the hand (upwards of thousands of dollars); also, there's often a stipulation that anyone involved in the hand receives some of the money, even if they folded after the flop.

posted by The God Complex at 10:11 PM on July 12, 2004

In a hand of five-card draw last weekend I lost a natural king-high flush to four aces for $40, after he took two cards.

posted by nicwolff at 11:14 PM on July 12, 2004

*That*sucked.posted by nicwolff at 11:14 PM on July 12, 2004

The gist of it is that the straight flush is twice as likely since it can come at either end.

posted by Space Coyote at 2:31 AM on July 13, 2004

posted by Space Coyote at 2:31 AM on July 13, 2004

Not necessarilyâ€”you could have 4H 5H 7H 8H, and be waiting for the six of hearts.

posted by kenko at 7:00 AM on July 13, 2004

posted by kenko at 7:00 AM on July 13, 2004

Actually, the odds don't change for an inside vs. open-ended straight flush draw. Remember we are dealing with the river - just one card. There is only ever one card that can make both the 4 of a kind and the straight flush, regardless of where in the straight it is. But all that is incidental.

Keep in mind that the one player already has a set of three and as such has the best hand at this point. However, the player that has 4 to the straight flush also by definition has 4 to the flush, which if made will beat the trips. The operative question therefore isn't "what are the odds that 4 of a kind, made on the river, will lose to straight flush, also made on the river?" to which the answer is, as others snarkily pointed out, "always," which is to say that the question is a tautology - if one event happens the other must also. The important question is "what are the odds that trips will lose to a straight flush draw, where the card required for the 4 of a kind is in the same suit as the flush draw and will complete the straight flush for the opponent." This detail is important because without the 4th card completing the straight flush but only comleting the flush, the odds improve slightly for the hand with trips, because thats one more card in the deck that will make him the winner: 4 of a kind beats a flush. In this case, however, that card is a loser for the hand with trips because the straight flush will beat the 4 of a kind.

The answer is that the trips will lose, believe it or not, 79.5% of the time, even though it is the best hand going into the river.

posted by ChasFile at 7:55 AM on July 13, 2004

Keep in mind that the one player already has a set of three and as such has the best hand at this point. However, the player that has 4 to the straight flush also by definition has 4 to the flush, which if made will beat the trips. The operative question therefore isn't "what are the odds that 4 of a kind, made on the river, will lose to straight flush, also made on the river?" to which the answer is, as others snarkily pointed out, "always," which is to say that the question is a tautology - if one event happens the other must also. The important question is "what are the odds that trips will lose to a straight flush draw, where the card required for the 4 of a kind is in the same suit as the flush draw and will complete the straight flush for the opponent." This detail is important because without the 4th card completing the straight flush but only comleting the flush, the odds improve slightly for the hand with trips, because thats one more card in the deck that will make him the winner: 4 of a kind beats a flush. In this case, however, that card is a loser for the hand with trips because the straight flush will beat the 4 of a kind.

The answer is that the trips will lose, believe it or not, 79.5% of the time, even though it is the best hand going into the river.

posted by ChasFile at 7:55 AM on July 13, 2004

The scenerio is six players, to start. My hand is 8h and 10h. Flop comes 7s, 7d, 9h. Betting. Stay in to see one more. Down to three players. Turn is the 6h. More betting. Still three players. River is 7h. Crazy betting and the call. I beat a Ah and 7c four of a kind, and a full house of 9's and the 7's. I couldn't believe my luck and was trying to figure what were the odds of that were.

posted by brent at 8:10 AM on July 13, 2004

posted by brent at 8:10 AM on July 13, 2004

In that case, only one card (7h) could complete both the straight flush and four-of-a-kind. At that point, you only knew 6 cards, leaving 46. The odds were 1 of 46.

posted by mischief at 8:21 AM on July 13, 2004

posted by mischief at 8:21 AM on July 13, 2004

First of all, why are you playing an 8 10 with 5 opponents? Second, Why are you calling bets with 3 to a flush and 2 opponents?

The answer is you got insanely lucky. I won't do this w/o a calculator that can do combinatorics, because doing it out by hand would be stupid boring. Also, would you like the odds you win at each stage in the hand without knowing the other hands, or the odds that 3 players make any full boat, any 4 of a kind, and any straight flush all in the same hand, or the odds that this specific hand ever happens, or what? You need to be more specific. In the latter case, at least, it is something on the order of:

1/(52! - 41!)

posted by ChasFile at 8:28 AM on July 13, 2004

The answer is you got insanely lucky. I won't do this w/o a calculator that can do combinatorics, because doing it out by hand would be stupid boring. Also, would you like the odds you win at each stage in the hand without knowing the other hands, or the odds that 3 players make any full boat, any 4 of a kind, and any straight flush all in the same hand, or the odds that this specific hand ever happens, or what? You need to be more specific. In the latter case, at least, it is something on the order of:

1/(52! - 41!)

posted by ChasFile at 8:28 AM on July 13, 2004

The question is better asked by Kenko. Or wasn't the Mel Gibson "Maverick" the same type of 4 of a kind and straight flush, and what would the odds be of that event happening?

As for why play that hand I guess to see one more card after the flop. After the betting I thought both players were trying to buy the pot, but I just figured I had the better hand and chances, thinking that neither had higher pairs than what's showing.

posted by brent at 9:55 AM on July 13, 2004

As for why play that hand I guess to see one more card after the flop. After the betting I thought both players were trying to buy the pot, but I just figured I had the better hand and chances, thinking that neither had higher pairs than what's showing.

posted by brent at 9:55 AM on July 13, 2004

In that case, willnot had the right answer to kenko's version of the question.

Its important to note that this is the god-perspective, and assumes knowledge of both hands; specifically that the card(s) you need to complete the straight flush are not in your opponent's hand. Granted that is impossible given the preconditions you and kenko apply (that the river make both hands), but it is nonetheless an unrealistic and generally unhelpful number. A better one is the odds based on your perspective, with only your 6-card knowledge of what's been played: 1 in 46.

Again, the answer very much depends on the question you ask, but based upon the one kenko asked, its 1 in 44. I get the feeling, however, that you are looking for satisfyingly impossible odds, like "one in 14 septillion," or something. In that case, you need to ask "What is the probability that one hand will be a 4 of a kind and the other will be a straight flush, and that niether player will complete their hands until the last card is played, in any given 2-player hand of texas hold 'em?" That number would be very small indeed, and like I said without a TI-83 nearby I wouldn't want to attempt it.

posted by ChasFile at 10:29 AM on July 13, 2004

Its important to note that this is the god-perspective, and assumes knowledge of both hands; specifically that the card(s) you need to complete the straight flush are not in your opponent's hand. Granted that is impossible given the preconditions you and kenko apply (that the river make both hands), but it is nonetheless an unrealistic and generally unhelpful number. A better one is the odds based on your perspective, with only your 6-card knowledge of what's been played: 1 in 46.

Again, the answer very much depends on the question you ask, but based upon the one kenko asked, its 1 in 44. I get the feeling, however, that you are looking for satisfyingly impossible odds, like "one in 14 septillion," or something. In that case, you need to ask "What is the probability that one hand will be a 4 of a kind and the other will be a straight flush, and that niether player will complete their hands until the last card is played, in any given 2-player hand of texas hold 'em?" That number would be very small indeed, and like I said without a TI-83 nearby I wouldn't want to attempt it.

posted by ChasFile at 10:29 AM on July 13, 2004

Actually, the odds don't change for an inside vs. open-ended straight flush draw. Remember we are dealing with the river - just one card. There is only ever one card that can make both the 4 of a kind and the straight flush, regardless of where in the straight it is. But all that is incidental.

Actually, the odds don't change for an inside vs. open-ended straight flush draw. Remember we are dealing with the river - just one card. There is only ever one card that can make both the 4 of a kind and the straight flush, regardless of where in the straight it is. But all that is incidental.

Umm, yes they do. If you have 6789 of hearts, then the 10 or 5 makes your straight on the river, giving you twice the odds to make your straight flush.

*First of all, why are you playing an 8 10 with 5 opponents? Second, Why are you calling bets with 3 to a flush and 2 opponents?*

The first one is impossible to answer, but the second is not. With a small enough bet, the implied pot odds if you hit an open-ended straight draw (which he had) are fairly high. You have a 32% chance (roughly) of hitting your straight if you call the next two bets, which would beat most hands at this point, unless you put someone on a 9-7. With that many people in the have the implied odds to call and take a shot at your straight.

posted by The God Complex at 10:39 AM on July 13, 2004

This thread is closed to new comments.

posted by trharlan at 9:05 PM on July 12, 2004