Flush math

Ok, here is what happened. I'm playing in an MTT. Get KK, double up. Get KK again, 3-way all-in now, triple up. Get KK again. All-in, my opp shows A9s. He hits his suit and I'm out. A minute after that, I'm playing my last hand in a ring game and what do I get? KK! Guy goes all-in for $5. I call, he shows QQ. One is of a different suit than my kings. He actually got his flush! So this prompted me to do some math on flushes.

KK vs A9s (both kings not of the suit of the A9)=> 2:1
KK vs QQ (all 4 cards different suit) => 13:3

But how much of this comes from flushes hitting?

Now, I did the math with only 2 known cards, in my examples there 2 more offsuit cards out, so the odds there slightly improve.

So first, 2 suited cards, will hit their flush in 3 instances: flop+turn+river giving 3, 4 or 5 of their suit.

The denominator in all 3 cases is 50x49x48x47x46
D= 254251200

The divisor D1 in the first case of 5 suits is 11x10x9x8x7

The divisor D2 in the second case of 4 suits and 1 offsuit is (11x10x9x8x39)x5
=> the "39" comes from the 'middle' (average) of an offsuit card
=> the "x5" comes from there being 5 different configurations (ie positions in which the offsuit card can be)

The divisor D3 in the third case of 3 suits and 2 offsuits is (11x10x9X39x39)x10
=> the "x10" here equals 10 configurations of which the offsuits can be spread out amongst the suits

So then we get, chance of a flush
= (D1/D)+(D2/D)+(D3/D)
= 0.000218 + 0.006074 + 0.059224
= 0.0655
or 6.55% to get a flush on flop+turn+river if you're holding suited cards.

In the case of one suited card giving a flush, we only need the first two terms so:
=0.006292
or 0.63% to get a flush on flop+turn+river with one suited card.

In the case of two more known cards, the denominator becomes
D=205476480
and the x39 becomes x37, or in the second scenario:
chance=0.00027+0.00713=0.713%
Therefore in the KK vs QQ with one different suited Q, he had 0.713% to hit his flush.


So.. this seem about right? Could someone review my math?

Probability that at least four cards of a specific suit appear on the flop in the KK v QQ scenario:

12*11*10*9*35/48*47*46*45*44 + 12*11*10*9*8/48*47*46*45*44 = 0.0025

Right, you caught one mistake I made, shoulda counted down from 12 instead of 11 in the "one suit" instance. But apart from that, your calculations are wrong.

That is to say, incomplete. You have two terms, and what you calculate is the chance to get 5 suits (second term) or 4 suits and 1 offsuit on the river (first term). The four other occurrences of the offsuit card haven't been accounted for. (which would be: on the turn, first, second, or third in the flop)

To account for them, you need 4 more terms, each substituting the "35" factor for 36, 37, 38, 39. Or you can do what I did, take the average (37) and multiply by 5.

No.

There are 5 cards. It doesn't matter what order they come in.

A final board of AdKd2d8d7h is exactly the same mathematically as a board of 2dKd7hAd8d.


The "35" term is just 48 unknown cards - 13 cards of the suit...
(which now that you mention it should be 36... since only 12 suit cards are left in the deck)

but that doesn't change the math. just substitute 36 for 35.

Let's take this hypothetical situation:

The deck has 6 cards:
Qc Kc Ac
Qs Ks As

Now, a flop can give: 3c, 2c 1s, 1c 2s, 3s.

Odds for 3c: (3/6)(3/6)(3/6)=(1/2)^3=1/8
Odds for 3s: 1/8 (same)

Now, what are the odds for 2c 1s (2 clubs and 1 spade)? According to your theory that the position doesn't matter, this would be:

(3/6)(3/6) [these are the clubs] x (3/6) [this is the spade]
=1/8

Now, 1c 2s has the same odds, so again 1/8.

So we add all the possibilities and we get 4/8=1/2. Where's the other 50%?

So yeah, position matters. In the 2c 1s example, there are three configurations, ie three places the spade can be at. So we have to multiply this by 3. Same for 1c 2s.

Thus we get: 1/8 for 3s and 2c; 3/8 for 2c 1s and 1c 2s

This adds up to 1.


Or to make a long story short.. you really do need to look at all the possible configurations of fitting the unsuited inside the suited cards. With 4 suited and 1 unsuited, there are 5. At the example of 3 suited and 2 unsuited, there are 10 such configurations.

Woops, quite an embarassing mistake I made there! Another minor mistake was that I said you had to use 36, 37 etc.. in fact they're 5 times the same number. Because the number of cards decreases with the number out suits.

But yeah, the position thing is true though What you said about mathematical equivalence, that only counts for when you switch the suited cards amongst themselves; Switching an unsuited with a suited is a mathematically different situation.

You're partly right. Sorry.

2c,1s:
3*2*3/6*5*4 + 3*3*2/6*5*4 + 3*3*2/6*5*4 = 9/20 = 0.45

3s= 3*2*1/6*5*4 = 1/20 = 0.05

Ok my reply is supposed to go after yours there, lol :P

Originally Posted by jackvance

Ok my reply is supposed to go after yours there, lol :P

Yeah, that was kind of weird. I guess you snuck it in while I was editing my post, and it stuck it in before mine somehow.

Demi: I think there's a factor 5 missing in your first term.

Anyway I think that the best way to go about this problem is to use inclusion-exclusion style arguments. Let PX be the probabilities that X particular cards are all of the required flush suit. In the first scenario 2 cards of the suit and 2 offsuit cards are accounted for, so P3 = 11*10*9/48*47*46, ... In the second 1 card of the suit and 3 offsuit are known, so P4 = 12*11*10*9/48*47*46*45, ...

A first approximation of the probability of having at least 4 of the suit on the board is clearly given by 5*P4. The inaccuracy comes from the fact that this counts the instances of all 5 cards having the same suit 5 times, so to get the exact answer we subtract 4*P5. The exact probability is therefore 5*P4 - 4*P5.

A first approximation of the probability of having at least 3 of the same suit is given by 10*P3. This counts each occurance of 4 of the same suit 4 times, so we subtract 3*5*P4 to compensate. Then, hoewver, we count each occurance of 5 of the same suit -5 times, so to get the exact answer we fianlly need to add 6*P5. The exact probability of getting at least 3 of the specified suit is therefore 10*P3 - 15*P4 + 6*P5.

Originally Posted by krimson

Demi: I think there's a factor 5 missing in your first term.

There is. but if I fixed it, everyone correcting me would make no sense.