3

Also: is the probability so small that this has never happened?

  • 1
    Everything that can happen in poker has happened. Nothing is that unlikely. – WW. Oct 9 '14 at 9:36
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    While it's true that there are more possible combinations than have been experienced on pokerstars, not all cards are truly meaningful in a hand, so the extra combinations provided by those combos can be ignored. It doesn't often matter what exact blanks the other 7 people have when you and your villain have competing straight flushes. That can collapse the parameter space by a lot. So, while every possible exact hand has certainly not been played, it's not at all unlikely that straight flush over straight flush occurs on an occasional basis. – Chris Farmer Oct 9 '14 at 16:43
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    @ChrisFarmer I agree, straight flush over straight flush has occurred, but there are meaningful situations that are so unlikely that they probably have not occurred. I suspect that all meaningful heads up combination have occurred, but I doubt that royal flush VS straight flush VS quad aces VS full house VS full house VS full house VS nut flush has ever occurred in a fairly dealt game, though it it a possible scenario. – Paul Oct 9 '14 at 17:04
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    Well, the nut flush would be the royal flush, so I think that's impossible. But your point is noted. :) – Chris Farmer Oct 9 '14 at 17:10
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    At the Sahara Poker Room sometime in the 80's in a seven card stud game, the pot was split between two players both holding royal flush's. to vaguely answer the question about this ever happening from op. – Jon Oct 15 '14 at 19:41
5

The only way to lose with a King high straight flush is to a royal flush of the same suit. That means that the KQJT must all be community cards. The last community card is either the 9 or you have it as a pocket card.


Case 1: 9 is on the board. The odds of this are the same as getting dealt a royal flush in 5 card stud:

20/52 * 4/51 * 3/50 * 2/49 * 1/48 = 1/649740


Case 2: 9 is not on the board. The odds of this are calculated in a similar way, except we pick from 16 cards initially, since we exclude the four 9s and we require that the last card is not the 9 or the Ace of the relevant suit, so we need to multiple by 46/48:

16/52 * 3/51 * 2/50 * 1/49 * 46/48 = 23/1624350 ~ 1/70624

Then we want the 9, which we have a 2/47 chance of getting, but we also don't want the Ace, so we have:

2/47 - 2/47*1/47 = 91/2162

and odds of both us getting the 9 without the Ace and the KQJT appearing on the board are:

91/2162 * 23/1624350 = 1/1677900


The chance of one of these cases happening is:

 1/649740 + 1/1677900 = 163/76344450 ~ 1/468370


What you want to know though, is what is the chance of one of these things happening and an opponent having the Ace of that suit. That depends entirely on the number of players dealt into the hand. If you were heads up, the odds would be:

 163/76344450 * 2/45 = 163/1717750125 ~ 1/10538344

A one in 10 million chance is small, but not nearly small enough to believe that it hasn't happened in poker before. Against 8 other players the chances would be much higher, since any of them could have the Ace:

 163/76344450 * 16/45 = 1304/1717750125 ~ 1/1317239
| improve this answer | |
  • How can you have a 2/47 chance of getting a single card - 9 or flush? – paparazzo Nov 16 '16 at 20:44
  • @Paparazzi There are 47 cards left in the deck (since there are 5 on the board) and you get dealt 2 of them. – Paul Nov 16 '16 at 20:47
  • You can't lose to quads with a straight flush. A straight flush beats quads. – Paul Nov 16 '16 at 20:49
  • OK that is a good answer – paparazzo Nov 16 '16 at 20:50
  • And it isn't 1/47 + 1/46. You can't get dealt the same card twice, so you would have to multiply that 1/46 by the chance of not getting dealt it on the first card. In other words: 1/47 + 46/47*1/46 = 2/47. – Paul Nov 16 '16 at 20:50
1

Two ways:

  • K high flush is on the board
    And villain as A
  • K - T on board and hero is holding 9
    And villain as A

Come at it from combinations
Start with board

K high flush on the board
number of boards = 4 / combin(52,5) = 0.0000015 = 1/649740
villain has A = 2/47 = 0.043 = 1/23.5
combined = 0.000000065 = 1 / 15268890

4 straight on the board
number of boards = 4 * (52-4-2) / combin(52,5) = 0.0000739 = 1 / 13536
villain 9 = 2/47 = 0.0426
hero has ace = 2/45 = 0.0444
combined = 0.000000134 = 1 / 7468479

combined combined 0.000000199 = 1 / 5015329

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0

Actually, wouldn't the odds be calculated much differently based on what game you are playing? If you are playing Texas hold'em as above and the King-Ten suited sit on the board. If you are holding the 9 in your hand, the odds of some one holding the ace are 1 in 45 given the 5 cards on the table and the 2 in your hand. Or am I way oversimplifying the math?

| improve this answer | |
  • I calculate the odds assuming Texas hold'em. The odds are 2/45 of someone holding the Ace, since they have 2 pocket cards, not one, which is why I multiplied by 2/45 on this line: 163/76344450 * 2/45 = 163/1717750125 ~ 1/10538344 – Paul Oct 31 '14 at 17:00
  • The 2/45 also only applies in the cases where you know that neither of the hero's cards are the ace, and the remaining board card is not the ace, which is a necessary condition for losing the hand. – Paul Oct 31 '14 at 17:09
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    The question was tagged with "nlhe" so it's safe to assume texas holdem. – Chris Farmer Nov 3 '14 at 16:09
0

If you just deal out five-card poker hands to yourself and one opponent and simply show them down, your odds of losing with a K-high straight flush are 1,533,938 to 1. If you're playing 9-handed Texas hold'em and playing the K-high straight flush on the board, your odds of losing are about 2 to 1.

For your game, it's somewhere between those two extremes. :-)

If you want to know how often something like that happens in a typical game, it's quite likely, because most poker games today are hold'em, so I'd expect a small cardroom to see one or two such hands a year; a big casino many more.

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-1

The probablity of winning will change depdning on the number of players...

It's anywhere from 4.4% to 40%

Regardless of the number of players, there is only 1 possible hand combination that makes that loosing with a king high straight flush possible....

The Board needs to be

KQJ10x

That's King, Queen, Jack, 10 all same suit and X is any other card

And you have 9 of that suit, and villian has A of that suit

For heads up, villian has 2 hole cards, so that's 2 chances for either one being the A and there 45 remaining unknown cards...

so villian has 2 out 45 chance of having the A...

that's about 4.4% of the time.... (2 divide by 45)

There are 52 cards in the deck, you have 2, and there are 5 on the board, so that's 7....

that leaves 45 cards....

and villian has 2 so = 2/45 or 4.4%

You do the same process with each added player....

2 way = 2 = 4.4% 3 way = 4 = 8.8% 4 way = 8 = 13.3% 5 way = 10 = 17.7% 6 way = 12 = 22.2% 7 way = 14 = 26.7% 8 way = 16 = 31% 9 way = 18 = 35.540% 10 way = 20 = 40%

So the chances of Winning with a King high flush is therefore any where from 4.4% - 40% depening on the number of players.

and best explanation is from wiki

The Law of Large Numbers

According to the ergodic hypothesis, given an infinite universe, every event with non-zero probability - however small - will eventually occur. Or put another way: given enough chances, even the most unlikely event is certain to happen.

When talking about the improbable, it's easy to ignore the cases where the event does not happen. People are naturally self-centered and think about their own experience first: from any one individual's point of view, the odds of winning the lottery are minuscule and the odds of finding someone with the same birthday are exactly as you'd expect.

But when considered in a more comprehensive and inclusive way, the true odds are revealed. For example, the probability of one particular mutation during evolution may be tiny, but there are billions of mutations happening continuously and being sorted by natural selection. Because of all these chances, that one minute possibility isn't really unlikely at all. It's a certainty.

We tend to pay attention to the improbable things that do happen - and never to the improbable things that don't happen and don't defy the odds. This particular cognitive bias is an important aspect of the Black Swan theory of improbable events. We may be staggered by an event with a 1 in a million odds, but completely ignore that at least 999,999 other 1-in-a-million events just happened to have not occurred. This is often boosted by a form of post hoc fallacy that explains the event that happened but discounts the events that don't - analogous to rolling a die but only ever telling someone or acknowledging the roll when it's a 6, indeed the die may be invisible and no one will know it's being rolled until it shows a 6.

In short, one-in-a-million events happen all the time.

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-3

A bicycle straight flush should beat a K-9 due to the ace.

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  • The ace counts as low, not high. – Herb Jan 12 '17 at 0:25
  • That's the way it is in Pai Gow in Nevada. But in poker (everywhere), if the ace being being used as a 1 for a straight, then it only ranks as a 1. – Lee Daniel Crocker Oct 10 '17 at 17:33

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