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 08-03-2017, 11:31 AM #1 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,155 Flush Probability – Combinatorial Confusion As one of a series of articles/blogs I’m writing on basic poker mathematics, I ran into a problem for a piece simply titled “On Flushes.” I thought it would be useful to show how to calculate the probability that a Hold ‘em player will have a flush by the river without any thought of folding. From the respected Wikipedia and Wizard of Odds webb sites, I ‘knew’ the probability to be 3.0255%. 10 PM Method 1. The first approach that came to mind was to use the following formula: Pr(Flush) = Pr(double suited)*Pr(flush|double suited) + Pr(single suited)*Pr(flush|single suited) When I did the math and after careful checking, I got 2.9208%. WTF – I’m low. “Oh, I see the problem, I forgot to include the possibility that the flush could be only board cards.” 10:30 PM Method 2. I now added the possibility of only a board flush and got 3.0566%. WTF^2 – I’m high. I then realized there is an easier way. 10:45 PM Method 3. Since you don’t care where the flush occurs, I need only consider that of 7 cards dealt, at least 5 are of one suit. When I did this I got 3.0566% again. WTF^3. Now being an old-hat at these type of calculations, I knew that the most likely reason for a result bigger than expected is double counting some events. But after some thought, I couldn’t find any such possibility. So, either Wikipedia and the Wizard are wrong or I have some double count or error still to be found – time to go to bed and see what tomorrow brings. 1:30 AM Method 4. Woke up with a jolt. Got it! Getting the probability of at least 5 of 7 dealt cards of one suit was done correctly. But, some of these so-called flushes are actually straight flushes and cannot be included for they have a higher ranking value. What I had done was equivalent to including sets in calculating the probability a player would have a pair by the river. (I’m sure many readers would have already figured out my error.) When I subtracted out the straight flush probability I got the correct result. BINGO! Cliff Notes and Lessons Learned 1. Probability calculations can be tricky 2. It is easy to exclude some events, double count some events or (as I did) include wrong events 3. Search the Internet to see if the problem is already solved but posted results are not always correct so use them only as a check if the source is questionable. 4. Try to do the problem at least two ways for another check 5. Post the problem on this forum if you are still unsure
 08-03-2017, 12:53 PM #2 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,681 Re: Flush Probability – Combinatorial Confusion Very nice. Thanks for posting.
08-04-2017, 04:16 AM   #3
David Sklansky

Join Date: Aug 2002
Posts: 14,411
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by statmanhal When I subtracted out the straight flush probability I got the correct result.
Are you sure? For instance if you handled the seven card straight flush wrong your answer might still be the same to four decimal places.

 08-04-2017, 10:49 AM #4 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,155 Re: Flush Probability – Combinatorial Confusion Wizard and Wikipedia: 0.030254941228 Me: ------------------------0.030254941228
 08-04-2017, 12:16 PM #5 David Sklansky Administrator     Join Date: Aug 2002 Posts: 14,411 Re: Flush Probability – Combinatorial Confusion Ok. I was just pointing out that your cliff note errors could extend to the straight flush subtraction. If you feel confident that you did it right, fine. But of course the agreement with Wizard and and Wikipedia could conceivably mean you all overlooked the same thing. Sorry, can't help myself.
 08-04-2017, 01:14 PM #6 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,155 Re: Flush Probability – Combinatorial Confusion Sure that's possible but I would say pretty unlikely. Anyway, I can send you the Excel file which shows all the calcs. PM me if interested.
 08-04-2017, 06:54 PM #7 David Sklansky Administrator     Join Date: Aug 2002 Posts: 14,411 Re: Flush Probability – Combinatorial Confusion The number of ways is, I think (13C7) (seven card flush in hearts) + (13C6)x(39) (six card flush in hearts) + (13C5)x(39C2) (five card) minus 9x (46C2) (uncounterfeited straight flushes in hearts) minus (47C2) (Royal in hearts) Times 4. Over (52C7). What's that?
08-04-2017, 07:18 PM   #8
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,155
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by David Sklansky The number of ways is, I think (13C7) (seven card flush in hearts) + (13C6)x(39) (six card flush in hearts) + (13C5)x(39C2) (five card) minus 9x (46C2) (uncounterfeited straight flushes in hearts) minus (47C2) (Royal in hearts) Times 4. Over (52C7). What's that?
=1011911 / 133784560

Above *4 = 0.030254941228

That’s exactly what I ended up doing.

The slighty trickly part was recognizing that a royal flush has a slightly higher occurrence probability than individual straight flushes.

08-09-2017, 09:33 AM   #9
blackspoker
centurion

Join Date: Feb 2017
Posts: 158
Re: Flush Probability – Combinatorial Confusion

Picture source: https://wizardofodds.com/games/poker/
Quote:
 Originally Posted by statmanhal Me: ------------------------0.030254941228
Since you did exact calculations (and not aprox.), do you mind explaining me why the number from the picture is not the same as yours (exact number)?

 08-10-2017, 04:36 AM #10 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,681 Re: Flush Probability – Combinatorial Confusion My initial guess, and it is only a guess, is that the probs in the Wizard's table are via simulation and therefore not exact. Even at 1 player (where deriving the exact probs is not very difficult). The Wizard does provide exact probs for 7-card stud hands which, unless I am totally missing something due to the lateness of the hour, should also be the exact probs for 1-player hold-em hands. There it is reported that a player will get a flush (excluding royal and straight flushes) exactly 4,047,644 deals out of a possible grand total of 133,784,560 deals. This fraction agrees with what statmanhal reported above.
08-10-2017, 10:54 AM   #11
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,155
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by whosnext My initial guess, and it is only a guess, is that the probs in the Wizard's table are via simulation and therefore not exact. Even at 1 player (where deriving the exact probs is not very difficult). The Wizard does provide exact probs for 7-card stud hands which, unless I am totally missing something due to the lateness of the hour, should also be the exact probs for 1-player hold-em hands. There it is reported that a player will get a flush (excluding royal and straight flushes) exactly 4,047,644 deals out of a possible grand total of 133,784,560 deals. This fraction agrees with what statmanhal reported above.
Right. The other possibility is that the Wizard did the calculation in steps and first converted the combo results into probabilities and then combined. Depending on the rounding, results will differ very slightly from the 7-card stud results which are based on total combos. I think this is more likely.

Blackspoker should ask the Wizard to explain the difference.

 08-10-2017, 12:50 PM #12 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,681 Re: Flush Probability – Combinatorial Confusion At the risk of beating a dead horse, the number of hands that make a flush, excluding royal or straight flushes, in 7-card stud (or 1-player hold-em) is, I think, given by: 4*{C(13,7)+C(13,6)*C(39,1)+C(13,5)*C(39,2)-[C(5,5)*C(47,2)+9*C(5,5)*C(46,2)]} which, if I evaluated this correctly, is 4,047,644, the number also given by the Wizard. Of course there are C(52,7)=133,784,560 total deals possible. So the true exact fraction is pretty well known at this point.
08-11-2017, 05:47 AM   #13
blackspoker
centurion

Join Date: Feb 2017
Posts: 158
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by whosnext At the risk of beating a dead horse, the number of hands that make a flush, excluding royal or straight flushes, in 7-card stud (or 1-player hold-em) is, I think, given by: 4*{C(13,7)+C(13,6)*C(39,1)+C(13,5)*C(39,2)-[C(5,5)*C(47,2)+9*C(5,5)*C(46,2)]} which, if I evaluated this correctly, is 4,047,644, the number also given by the Wizard. Of course there are C(52,7)=133,784,560 total deals possible. So the true exact fraction is pretty well known at this point.
Thank you for taking your time and thank you for answering my question (and thanks to the OP also).

The reason why I posted my previous post was: After I read this thread for the first time, I went checking and googling, because I was curious why has royal flush higher accourance probability than other straight flushes –make sense now that I know… Then I find the table that has slightly different numbers that OP calculated and I was curious why is that… so posted table here and waited for explanation… now I know..

Now, I checked your calculations (also checked almost every possible site that I could find on the internet that has calculated flush probability and result is allways the same;like yours and OP; so correct)…..also makes sense that probability numbers at holdem are the same as 7 card stud…

I have some other question….I do not understand why are some things in the way they are.... I mean yes, I can calculate this, if I found method somewhere(for example here), but it is important for me that I understand exactly why results are in the way they are….so....

This is from wikipedia:
I Quote: »The Ace-high straight flush or royal flush is slightly more frequent (4324) than the lower straight flushes (4140 each) because the remaining two cards can have any value; a King-high straight flush, for example, cannot have the Ace of its suit in the hand (as that would make it ace-high instead).«

All nine (not royal flush) straight flushes has 4140 each….? All together this is 37,260 as required…..

This is what I expected before reading mentioned quote from wikipedia: same number for 8 non-royal straight flushes… expect from A to 5 straight flush…

Why is A to 5 straight flush same frequent as the other 8 non-royal straight flushes (it is »blocked« down)? Simple and logical explanation? You people can probably explain this faster than I would figure it out (I do not want to lose one week to figure it out why some calculations ended in the way they do)..

I hope you understand what I am asking... Since I am allready on this forum, I would like to learn the most I can (and the fastest). Thank you for answers in advance.

Last edited by blackspoker; 08-11-2017 at 06:12 AM.

08-11-2017, 10:30 AM   #14
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,155
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by blackspoker Why is A to 5 straight flush same frequent as the other 8 non-royal straight flushes (it is »blocked« down)? .
Given a royal flush – 5 cards- any two of the 47 remaining cards can be had to complete the 7-card hand. But with a lower straight flush, say 10 high, the jack of the suit cannot be one of the remaining two cards for that would make a jack high straight flush. Thus for any straight flush less than ace-high, there are “only” 46 allowable cards for the remaining 2 to be included. This holds true for an ace to five straight flush, with the 6 of the suit not allowable. The “blocked down” concept is not applicable here.

08-11-2017, 10:52 AM   #15
Didace
Carpal \'Tunnel

Join Date: Nov 2009
Posts: 13,556
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by blackspoker Why is A to 5 straight flush same frequent as the other 8 non-royal straight flushes (it is »blocked« down)?
In this case, don't think of it as an ace. It is a 1.

08-11-2017, 04:07 PM   #16
blackspoker
centurion

Join Date: Feb 2017
Posts: 158
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by statmanhal Given a royal flush – 5 cards- any two of the 47 remaining cards can be had to complete the 7-card hand. But with a lower straight flush, say 10 high, the jack of the suit cannot be one of the remaining two cards for that would make a jack high straight flush. Thus for any straight flush less than ace-high, there are “only” 46 allowable cards for the remaining 2 to be included. This holds true for an ace to five straight flush, with the 6 of the suit not allowable. The “blocked down” concept is not applicable here.
Thank you for reply… Now I understand

Last edited by blackspoker; 08-11-2017 at 04:24 PM.

 08-12-2017, 01:47 AM #17 PTLou Jellybean     Join Date: Nov 2008 Location: Charlotte, NC Posts: 3,566 Re: Flush Probability – Combinatorial Confusion my first ever post in probability forum got deleted. high probability that I should not post ITT ever again. what are odds of flopping exactly 10 High flush in spades in holdem ? Same question, but if two other players in full ring game have two spades each.
08-12-2017, 10:01 AM   #18
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,155
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by PTLou what are odds of flopping exactly 10 High flush in spades in holdem ? Same question, but if two other players in full ring game have two spades each.
There is some ambiguity in your first question so I’ll interpret it as follows:

What is the probability that after the flop a designated player has a ten high spade flush?

Pr = C(1,1)*C(8,4)/C(52,5) =0.000027,

or about 1 in 37,000 hands.

The second question has even more ambiguity (e.g. can the spade holders have the ten of spades) so I’ll pass on this question but in any case the probability will be small.

08-14-2017, 04:07 PM   #19
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,155
Re: Flush Probability – Combinatorial Confusion

Quote:
 Originally Posted by statmanhal What is the probability that after the flop a designated player has a ten high spade flush? Pr = C(1,1)*C(8,4)/C(52,5) =0.000027, or about 1 in 37,000 hands. .
Happened to glance at my reply today, two days after it was posted and damn, if I didn’t make the same mistake as I made earlier. The formula posted counts the number of combos of getting the ten of spades and four other spades lower than ten. But, it includes the ten high spade straight flush, one combo too many (T9876sssss). So the correct answer is

Pr =( C(1,1)*C(8,4)-1)/C(52,5) =0.000027 (Yeah, no difference to 6 decimals)

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