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 07-28-2011, 11:35 PM #1 helpamexican newbie   Join Date: Aug 2008 Posts: 42 Math odds of a flop Ok well I have done the math my self and have came to the same answer my buddy is convinced I'm wrong. What are the odds of three of a kind flopping on the board tens or higher including aces.
 07-29-2011, 01:10 AM #2 Sherman Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,764 Re: Math odds of a flop So TTT, JJJ, QQQ, KKK, or AAA on the flop? I'm just going to assume we are dealing 3 cards out and no one has any cards in their hand. Obviously in practice people are more likely to hold T+ kinds of cards when a flop is seen so these numbers are going to be over-estimates of reality. For TTT it is 4/52 * 3/51 * 2/50 = .000181. It is the same for JJJ, QQQ, KKK, and AAA. So we can just multiply that number times 5. 5 * .000181 = .000905. So about 9 times in 10,000 if you deal three cards out of a shuffled 52 card deck will they be TTT, JJJ, QQQ, KKK, or AAA. But in practice, the odds will probably be even lower as people are more likely to see flops when they hold cards that are T+ and thus the probability a flop in Hold'em or even Omaha is like that is almost certainly even lower.
 07-29-2011, 10:52 AM #3 Keasbey Nights stranger   Join Date: Dec 2009 Location: South Florida Posts: 5 Re: Math odds of a flop Good answer by Sherman. Just to make this a bit more realistic, if we take the average six-hand hold'em game plus the burn card, there will be 52 - 13 = 39 cards left in the deck for the flop. Thus, the probability of a flop like this is 5 * (4/39) * (3/38) * (2/37) = 0.002188. So assuming that nobody in your six-hand hold'em game has a ten or higher and assuming that a ten or higher was not burned, you'll see trips flopped 0.2188% of the time. Just for fun, if we assumed that one ten, one jack, one queen, one king, and one ace were dealt to the hole cards of your various opponents / burned, the probability of seeing trips on the flop is now 5 * (3/39) * (2/38) * (1/37) = 0.0005471, or 0.05471%. This statistic is a decently close ballpark; to get any closer, you'd either have to get into some crazy conditional probability, or you'd have to Monte Carlo it. Best of luck with your playing! Last edited by Keasbey Nights; 07-29-2011 at 11:09 AM.
 07-29-2011, 12:11 PM #4 Sherman Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,764 Re: Math odds of a flop Well, I suggest the way to find the actual answer in practice is to use a database of real flop data. Spadebidder has one, but he is a busy guy with other priorities than answering questions like these. But depending on how much effort it would take, he might be able to count up the number of times the flop falls into one of these categories divided by the total number of flops he has in his database.
 07-29-2011, 12:24 PM #5 tringlomane veteran   Join Date: Jun 2011 Posts: 2,483 Re: Math odds of a flop The burn card does not matter. The hole cards definitely do though. Ignoring hole cards, Sherman's math is correct. To get the most accurate result, you should consult a poker hand database though.
 07-29-2011, 01:04 PM #6 Keasbey Nights stranger   Join Date: Dec 2009 Location: South Florida Posts: 5 Re: Math odds of a flop Normally, the burn doesn't matter because we don't know what it is, but if we assume that it's not a ten through ace, then my math is valid for the purpose it's trying to serve. Realistically, nobody can calculate the right answer based on simple probability; I was just trying to ballpark a more accurate answer than simply disregarding hole cards.
07-29-2011, 01:05 PM   #7
Actually Shows Proof

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Re: Math odds of a flop

Quote:
 Originally Posted by Keasbey Nights Good answer by Sherman. Just to make this a bit more realistic, if we take the average six-hand hold'em game plus the burn card, there will be 52 - 13 = 39 cards left in the deck for the flop. Thus, the probability of a flop like this is 5 * (4/39) * (3/38) * (2/37) = 0.002188. So assuming that nobody in your six-hand hold'em game has a ten or higher and assuming that a ten or higher was not burned, you'll see trips flopped 0.2188% of the time.
This is wrong. The distribution of the burn cards and hole cards will be close to random, making them almost irrelevant for flop distributions. You have to divide by the total unknown cards, not the total undealt cards.

Here's some flop data:

There actually are some small removal effects since folds are not random and flops get seen more frequently when players hold high cards, but the effect is small. I'll give an empirical answer from a large database in a little while.

Last edited by spadebidder; 07-29-2011 at 01:15 PM.

07-29-2011, 01:33 PM   #8
Keasbey Nights
stranger

Join Date: Dec 2009
Location: South Florida
Posts: 5
Re: Math odds of a flop

I'm aware that the distribution will be random, but I voluntarily elected to consider them known on a whim just to approximate those removal effects. I know it's technically incorrect math, and I alluded to that in my original post when I said:

Quote:
 ...assuming that nobody in your six-hand hold'em game has a ten or higher and assuming that a ten or higher was not burned, you'll see trips flopped 0.2188% of the time.
It's just a bit of a further expansion upon the question. Of course, if we treat the burn and hole cards as random and unknown, then of course Sherman's answer is indisputably correct.

Thanks for sharing your flop data, by the way.

 07-29-2011, 01:38 PM #9 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,903 Re: Math odds of a flop Ok. It turns out that the total number of all ranks triplets flopped is dead on expectation and isn't affected by card removal effects. However, there is a skew in the ranks because of the standard rank bias whereby users tend to see flops more often when they hold high cards, meaning low cards flop a little more often. So if we look at just TTT through AAA flopping, if it were purely random then Sherman's calculation of .09045% is correct. Looking at 9-player cash games, 96 million flops seen, the actual number is .08871% or about 98% as often as purely random would predict. The other side of the coin is that flops with the bottom 5 ranks of triplets (222 to 666) happen about 102% as often as purely random would predict. The middle ranks of 777 to 999 happen just about exactly the expected amount. In heads-up holdem the skew is even more, down to about 97% of the expected amount for T to A. In 6-max it's in between. Here's the raw rank distribution charts: Last edited by spadebidder; 07-29-2011 at 01:49 PM.
07-29-2011, 01:43 PM   #10
Sherman
Carpal \'Tunnel

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Location: Psychology Department
Posts: 7,764
Re: Math odds of a flop

Quote:
 Originally Posted by spadebidder Ok. It turns out that the total number of all ranks triplets flopped is dead on expectation and isn't affected by card removal effects. However, there is a skew in the ranks because of the standard rank bias whereby users tend to see flops more often when they hold high cards, meaning low cards flop a little more often. So if we look at just TTT through AAA flopping, if it were purely random then Sherman's calculation of .09045% is correct. Looking at 9-player cash games, 96 million flops seen, the actual number is .08871% or about 98% as often as purely random would predict. The other side of the coin is that flops with the bottom 5 ranks of triplets (222 to 666) happen about 102% as often as purely random would predict. The middle ranks of 777 to 999 happen just about exactly the expected amount. In heads-up holdem the skew is even more, down to about 97% of the expected amount for T to A. In 6-max it's in between.
Seems about what we would expect. Very cool. Thanks!

 07-29-2011, 01:51 PM #11 Keasbey Nights stranger   Join Date: Dec 2009 Location: South Florida Posts: 5 Re: Math odds of a flop Thanks for posting that analysis, spadebidder. I'll keep that in mind for the future.
 07-29-2011, 02:39 PM #12 helpamexican newbie   Join Date: Aug 2008 Posts: 42 Re: Math odds of a flop Heres how I did the Math since all cards are unknown in deck first card of deck is 52/20(10s,j,q,k,a)=2.6 second card of deck 51/3 (whatever card flops remaing cards= 17 50/2 (remaing cards to complete the trips flopped) 25 then I took 2.6 x 17 x 25= 1105 SO i came up with 1105/1 anyways the bet we made is I say its closer to 1000-1 then it is 10000-1
07-29-2011, 03:46 PM   #13
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Re: Math odds of a flop

Quote:
 Originally Posted by helpamexican Heres how I did the Math since all cards are unknown in deck first card of deck is 52/20(10s,j,q,k,a)=2.6 second card of deck 51/3 (whatever card flops remaing cards= 17 50/2 (remaing cards to complete the trips flopped) 25 then I took 2.6 x 17 x 25= 1105 SO i came up with 1105/1 anyways the bet we made is I say its closer to 1000-1 then it is 10000-1
Your math is correct at 1/1105 (which means 1104 to 1).

That is the same answer Sherman gave in decimal form, in post #2.

Last edited by spadebidder; 07-29-2011 at 03:57 PM.

07-30-2011, 02:26 AM   #14
R Gibert

Join Date: Jan 2006
Posts: 924
Re: Math odds of a flop

Quote:
 Originally Posted by helpamexican Ok well I have done the math my self and have came to the same answer my buddy is convinced I'm wrong. What are the odds of three of a kind flopping on the board tens or higher including aces.
I'll show you how to do this calculation in your head. My answer will differ from some of the other answers for 3 reasons:
• I'm ignoring card removal effects.
• I'm assuming that you have knowledge of your own cards.
P(AAA) ≈ (4/200)% = 0.02%
And since there are 5 possibilites for X i.e. A, K, Q, J and T, then for all X >= a ten,
P(XXX) ≈ (5*4/200)% = 0.1%
The constant 200 is an approximation of C(50, 3)/100 = 19600/100. I'm dividing C(50, 3) by 100 so as to produce an answer in percent, so 200 is just an approximation of 196.

All I did in the above to calcs is count the number flops—excluding order of the cards flopped—and dividing by 200.

For comparison, you can calculate exact values by dividing by 196. This gives
P(XXX) = (5*4/196)% ≈ 0.1020408163%

P(XXX) ≈ ((4*4 + 1)/200)% = 0.085%
P(XXX) ≈ (4*4/200)% = 0.08%
P(XXX) ≈ ((3*4 + 1 + 1)/200)% = 0.07%

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