Probability of getting poker flush

Matty1973

Critical Thinker
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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.
Click to expand...

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.
Click to expand...

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.
Click to expand...

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.
 
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