Probability of getting poker flush

Matty1973

Critical Thinker
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Mar 27, 2005
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Me and my friend were playing poker and we were both delt a pair of aces - this meant we were likely tie. My friend (who is not very skeptical) said he thought one of us would get a flush (5 cards of the same suit) and thus not tie. As it happened he was right - 4 clubs came on the table and he won with his Ace of clubs.

Now he is convinced he is psychic.

I believe that the chance of one of us getting a flush is not as remarkable as he believes. However without knowing the exact chane I'm having difficulty convincing him it is probably around the one in twenty rather than the one in a million he believes.

Any maths gurus care to work it out?

Another way of describing it would be the chance of drawing 5 cards from a deck of 48 (aces already gone) and 4 of those 5 being the same suit.

I've been trying to work to out as:

Just look at one suit and then multiply by 4 at the end to get the total.

So for Clubs choose 5 from 48 and then 4 of those must be from the 12 clubs left.

Can't get further than that.
 
Matty1973 said:
I've been trying to work to out as:

Just look at one suit and then multiply by 4 at the end to get the total.
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So far you're doing great.

Matty1973 said:
So for Clubs choose 5 from 48 and then 4 of those must be from the 12 clubs left.
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The key is proper counting. You have to count two things: (1) the total number of possible combinations; (2) the number of positive combinations.

(1) The total number of possible combinations to pick 5 cards out of 48 is "48 over 5"; see Binomial coefficientWP. I'll write it as Bin(48,5). The formula is given by:
Bin(48, 5) = 48! / ( 5! . (48-5)! )
where ! is the factorial, e.g. 5! = 5 . 4 . 3 . 2 . 1 = 120

Writing it out you get Bin(48, 5) = (48 . 47 . 46 . 45 . 44) / (5 . 4 . 3 . 2 . 1)

(2) The number of positive combinations - having 4 clubs - is given by rephrasing: in how many ways can I choose 4 clubs and one other card? So that boils down to the formula:
A1 = Bin(12, 4) . Bin(36, 1)
That disregards, however, the possibility of choosing 5 clubs:
A2 = Bin(12, 5)
So you have to add up those two.

A1 + A2 = (12 . 11 . 10 . 9) / (4 . 3 . 2 . 1) . 36 + (12 . 11 . 10 . 9 . 8) / (5 . 4 . 3 . 2 . 1)
= (36 . 5 + 8) . (12 . 11 . 10 . 9) / (5 . 4 . 3 . 2 . 1)
= 188 . (12 . 11 . 10 . 9) / (5 . 4 . 3 . 2 . 1)

and, as you already noted, we have to multiply this number by 4 to account for the 4 hearts, 4 spades and 4 diamonds cases too.

The chance now is dividing the number from (2) by the number from (1). They now so happen to have the same denominator in their formulas, so we get:

(4 . 188 . 12 . 11 . 10 . 9) / (48 . 47 . 46 . 45 . 44)
= 8933760 / 205476480 = 0.0435

the last rounded to 4 decimals. So you have about 4% chance on this happening.
 
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You know, I would have had trouble working out the math (I love numbers, but for some reason combinations, permutations, and probability always gave me trouble), but I'm surprised it worked out the way it did. I would have thought the chances would have been less.

Looks like your guess of 1 in 20 was nearly dead-on!
 
The suit of the first board card is immaterial. So I would solve by finding the likelihood of the next three cards being the same suit as the first board card. That is 11/47 * 10/46 * 9/45 = .01
There being four combinations of those three cards in the four remaining, the solution would be .01 * 4 = .04 or about 24 to 1.

ETA: Arrgghh! Already beaten to it.
 
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ddt -Thanks for that - it's the making sure they were part of the positive combinations that I still can't quite get my head round - I'll read it through a couple of more times to be sure.

Bobk - Your way might wel be an easier way to show to my friend that it is not as unlikely as he thinks. Although it doesn't include the possibility that the first board card is no good an then next 4 are but that is probably worth ignoring for the sake of simplicity.

Nobby - I'm pretty chuffed my 20/1 was so accurate - although I was rather hoping it would be less - that still might be big enough to convince my friend he is psychic even though he guesses what will happen on virtually every hand. He never remembers the ones where he is wrong! He keeps going on about the time a player had a pair of nines and he predicted another nine would come on the board. Even a very basic approximation of 5*(2/48) comes out at 10/48 or roughly 20% but he still thinks that was an amazing prediction!
 
Matty1973 said:
I believe that the chance of one of us getting a flush is not as remarkable as he believes. However without knowing the exact chane I'm having difficulty convincing him it is probably around the one in twenty rather than the one in a million he believes.
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Compliments on that call, yes!
 
BobK said:
The suit of the first board card is immaterial. So I would solve by finding the likelihood of the next three cards being the same suit as the first board card. That is 11/47 * 10/46 * 9/45 = .01
There being four combinations of those three cards in the four remaining, the solution would be .01 * 4 = .04 or about 24 to 1.
Click to expand...

I'd advise heavily against this method, for any probability problem involving multiple draws. By looking at each draw in succession, you're introducing order into the problem, whereas you're only really interested in the end result without any order in the 5 cards drawn.

By introducing that order, you have to account for the case that the first card is the odd-one-out, that the second card is the odd-one-out, etc., etc. That makes your formula much more complicated than just looking at the end result - the 5 drawn cards - and how many positive combinations are possible.
 
Yeah. I ignored 5 cards, but the difference is slight. You might want to point out there are still 28 ways to end up with a tie. When the board holds a straight flush between queen high and six high.
 
Matty1973 said:
ddt -Thanks for that - it's the making sure they were part of the positive combinations that I still can't quite get my head round - I'll read it through a couple of more times to be sure.
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I'll expand on that a bit.

The key to the counting I did is that the order in which the five cards are drawn is utterly unimportant. The only thing that matters is how many combinations are possible.

The formula Bin(n, k) gives the number of combinations of k things you can draw from an total of n things.

In the case of the positive combinations, there are two subcases.

a) 4 clubs and 1 other card. The 4 clubs are drawn from 12 clubs in total, so that gives Bin(12, 4) different combinations of 4 clubs. The 1 other card is drawn from 36 non-club cards in total, so that gives Bin(36, 1). Bin(36, 1) equals 36.

Every combination of 4 club cards can be combined with every 1 non-club card. So you have to multiply those two numbers:
Bin(12, 4) . Bin(36, 1)

b) 5 club cards. The 5 club cards are drawn from 12 clubs in total, so that gives Bin(12, 5) different combinations.

If you're a regular poker player, you should try the method on more such situations. You'll easily beat out on players like your friend who don't calculate their chances and have a lousy intuition at them to boot. :)
 
BobK said:
Yeah. I ignored 5 cards, but the difference is slight. You might want to point out there are still 28 ways to end up with a tie. When the board holds a straight flush between queen high and six high.
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Good addition.

I'll readily admit I'm not a poker player, so I just went with what the OP told :).
 
Matty1973 said:
Me and my friend were playing poker and we were both delt a pair of aces - this meant we were likely tie. My friend (who is not very skeptical) said he thought one of us would get a flush (5 cards of the same suit) and thus not tie. As it happened he was right - 4 clubs came on the table and he won with his Ace of clubs.
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Excuse me, but what kind of deck of cards are you playing with that allows you to be dealt a PAIR of aces and then wind up with a flush?
 
Timothy said:
Excuse me, but what kind of deck of cards are you playing with that allows you to be dealt a PAIR of aces and then wind up with a flush?
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Sounds like they were playing Texas Hold Em.
 
I assume we are using a standard 52 card deck. Each player is dealt two cards, and four cards are laid on the table. The players then make a five-card hand out the six cards (the two the player holds and the four on the table).

Question: given that there are two players who each hold two aces, what are the chances at least one player will be able to make a flush hand?

Answer: 11/47 * 10/46 * 9/45 = 1.0175%

Hold on. Nevermind. I see now there are five cards on the table.
 
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I skimmed the maths but unless I missed something important the chance of someone getting a flush is four times what you guys have settled on, since four cards of any one suit being dealt would have led to someone getting a flush in this case.

So there was about a 16% chance that the math-illiterate friend would be "proven right". That is about the same as the chance of predicting that a dice will roll a 6 and being "proven right", which is to say that it's not an extraordinary event at all.
 
While somewhat off topic, a couple of years ago I got a royal flush on the flop playing no limit holdem. I was dealt Q J of Hearts, and A K 10 of Hearts came on the flop. Another heart actually came on the river. I slow played it (naturally) and two others stayed in the hand (one of whom also finished up with a heart flush). It finished up a nice little pot.

Just wondering if one of the more mathematically inclined members here could give me the odds of that one, assuming a normal deck and no cheating (I was not the dealer, and I did not know her outside of tournament poker). I imagine that the odds would be rather high.

Norm
 
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Kevin_Lowe said:
I skimmed the maths but unless I missed something important the chance of someone getting a flush is four times what you guys have settled on, since four cards of any one suit being dealt would have led to someone getting a flush in this case.
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You did not only skim the math but also the explanation of what was calculated - that is namely exactly as you describe, the chance of four of one (arbitrary) suit turning up on the table.
 
Matty1973 said:
Nobby - I'm pretty chuffed my 20/1 was so accurate - although I was rather hoping it would be less - that still might be big enough to convince my friend he is psychic even though he guesses what will happen on virtually every hand. He never remembers the ones where he is wrong! He keeps going on about the time a player had a pair of nines and he predicted another nine would come on the board. Even a very basic approximation of 5*(2/48) comes out at 10/48 or roughly 20% but he still thinks that was an amazing prediction!
Click to expand...

You're going about this the wrong way. Affirm to your friend that he is indeed psychic, do not explain the actual numbers to him and ask if he's willing to increase the blinds/limits you usually play for...
 
Matty1973 said:
Now he is convinced he is psychic.

I believe that the chance of one of us getting a flush is not as remarkable as he believes. However without knowing the exact chane I'm having difficulty convincing him it is probably around the one in twenty rather than the one in a million he believes.
Click to expand...


The part I don't understand is your trying to convince him that he is not psychic. I am not a poker player, but from what I understand about the game, having an opponent who believes flushes are a one in a million occurrence is an advantage.
 
robinson said:
http://www.cardplayer.com/poker_odds/texas_holdem

http://www.learn-texas-holdem.com/poker-odds-calculator.htm

Everything you want to know.
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That doesn't help with the question at hand. That link shows how to calculate to odds of winning the hand given a particular deal. It does not show the odds of a particular hand occuring.


On the topic of a royal flush in texas hold-em: I've only been at the table once when one was dealt.
The interesting thing was that the person who got the royal had one other person stay in the hand until the end -- her mother. Her mother stuck around because she had the 9 to go with the t-j-q-k on the table to give her a king high straight flush!
 
Molinaro said:
That doesn't help with the question at hand. That link shows how to calculate to odds of winning the hand given a particular deal. It does not show the odds of a particular hand occurring.
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In this case, the chances of winning are exactly the odds of getting a flush, since in the case of both players holding aces, that is the only hand that will allow one person to win. If you would bother to actually click on the app, you would see all the odds, from the deal until the river. Which is why I said it is what is needed to answer the OPs question.

Matty1973 said:
Me and my friend were playing poker and we were both dealt a pair of aces - this meant we were likely tie.
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Yes. Using either app I linked to, you can see the exact odds at that point. Both players are Win: 2.17%, Tie: 95.65%.

To change that to odds, just call it 96 to one to tie. Or 1 in 50 to win.

Matty1973 said:
I believe that the chance of one of us getting a flush is not as remarkable as he believes. However without knowing the exact chance I'm having difficulty convincing him it is probably around the one in twenty rather than the one in a million he believes.
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Poker players can calculate the odds of every hand as it is played. The web pages that let anyone do it shows exactly what the chances (odds) are, at every point. After the flop, if there were two spades, the player holding the ace of spades is now
Win:
4.55%,
Tie
95.45%

So after the flop, the chances of winning, for the player holding the AoS, is 9 to 200.

If a spade comes on the turn, the odds are now
Win: 20.45%
Tie: 79.55%

Or 1 in 5 to win.

These odds would be the same for any suite, I just used spades.

If there were more than two players, it changes the odds, based on what cards they threw into the muck.
 
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Robinson, your link does not solve my issue at all. Using your first link, with my Q J of Hearts, and a small heart and random second card for my opponent the odds were 75/20/5. I have no problem with this. This did not show me the odds of getting a royal flush in the next three cards. When A K 10 of hearts came out on the flop, the odds of me winning became 100%, which I sort of already knew.

The question for me anyway was what are the odds of that event actually occuring, not the odds of winning the hand.

Norm
 
fromdownunder said:
Robinson, your link does not solve my issue at all. Using your first link, with my Q J of Hearts, and a small heart and random second card for my opponent the odds were 75/20/5. I have no problem with this. This did not show me the odds of getting a royal flush in the next three cards.
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I was only answering the OP, which due to the nature of the hands, the odds of winning are exactly the same as the odds of a flush. In your case, which I had not read, it is probably not the same.

I will go back and read your problem, and see if there is any help to be had from calculators.
 
fromdownunder said:
While somewhat off topic, a couple of years ago I got a royal flush on the flop playing no limit holdem. I was dealt Q J of Hearts, and A K 10 of Hearts came on the flop. Another heart actually came on the river. I slow played it (naturally) and two others stayed in the hand (one of whom also finished up with a heart flush). It finished up a nice little pot.

Just wondering if one of the more mathematically inclined members here could give me the odds of that one, assuming a normal deck and no cheating (I was not the dealer, and I did not know her outside of tournament poker). I imagine that the odds would be rather high.

Norm
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OK, looking at that, there are two different issues. The odds of getting the royal flush, and the odds of winning with it.

The second part is easy. After the flop your odds are 100% to win. The odds of getting a royal flush are 1 in 649,740.source
 
I'm impressed that somebody actually got a royal flush. In my entire life, I have never witnessed anyone, either in person, or on television, catch a royal flush.



















And I am really old.
 
robinson said:
I'm impressed that somebody actually got a royal flush. In my entire life, I have never witnessed anyone, either in person, or on television, catch a royal flush.
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Robinson, thanks for the response to my query, the odds, and the link. I have seen a straight flush on television (but not on the flop), and "live" during a friendly with my kids, but never a royal (only in 7 card poker though). At 650,000:1!. but as with lotto winners, somebody gets to do it, and probably quite often given the number of poker hands that would be played daily throughout the world. But I was (obviously) extremely lucky, and I dined out on the story for weeks afterwards. I wish I could do the same in lotto.

Norm
 
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Matty1973 said:
Me and my friend were playing poker and we were both delt a pair of aces - this meant we were likely tie. My friend (who is not very skeptical) said he thought one of us would get a flush (5 cards of the same suit) and thus not tie. As it happened he was right - 4 clubs came on the table and he won with his Ace of clubs.
Click to expand...

Matty1973 said:
He keeps going on about the time a player had a pair of nines and he predicted another nine would come on the board. Even a very basic approximation of 5*(2/48) comes out at 10/48 or roughly 20% but he still thinks that was an amazing prediction!
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Matty, you do realize when you're playing hold 'em you're not supposed to show your cards to each other, don't you?








;)
 
WildCat said:
Matty, you do realize when you're playing hold 'em you're not supposed to show your cards to each other, don't you?

;)
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My assuption about this was that either one of them was all in and they knew what cards they were talking about, or that it was a post-hoc discussion about what might have been.

Norm
 
robinson said:
In this case, the chances of winning are exactly the odds of getting a flush, since in the case of both players holding aces, that is the only hand that will allow one person to win.
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But one player could get a flush and it could be a tie, if the board had a straight flush that was not king-high, so the chances are not the same, but pretty close.
 
Rasmus said:
You're going about this the wrong way. Affirm to your friend that he is indeed psychic, do not explain the actual numbers to him and ask if he's willing to increase the blinds/limits you usually play for...
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:D
 
robinson said:
I'm impressed that somebody actually got a royal flush. In my entire life, I have never witnessed anyone, either in person, or on television, catch a royal flush.

...

And I am really old.
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I'm quite young (28), but I play poker pretty much for a recreational player, both in live games and in the internet. I've had Royal Flush 4 times, I've played poker (NL Texas Holdem) only 4 years. Being a recreational player I don't know how many hands I've played, but I play roughly 10-15 hours a week, mainly internet heads up games, which helps to explain it.

It's a wonderful game :)
 
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Wildcat - fromdownunder is correct, I was all in with my pockets Aces and he called and won.

I have recently started to write down his predictions in a record (often he says no - if the feeling is not strong - funny he never had 'weak' predictions before I started recording them!). I ask him if what he says is 'on the record' or not and so far all that are have been wrong. Of course he is blaming me for negatively effecting his powers - but it is all in good fun. I have introduced the idea of postive confirmation bias to him and he hasn't rejected the idea totally out of hand so there is hope yet!
 
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If he feels the odds of getting a flush in Hold 'Em are 1 in a million, then I recommend playing Hold 'Em against this friend as often as possible for money.
 
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