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.

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.

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!

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.

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.

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.

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:
http://www.spadebidder.com/category/flop-analysis/

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.

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:

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.

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:

Seems about what we would expect. Very cool. Thanks!

Thanks for posting that analysis, spadebidder. I'll keep that in mind for the future.

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.

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.
• I'm approximating the answer.

If neither of your cards are broadway cards, then

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%

If your hand has one broadway card, then

P(XXX) ≈ ((4*4 + 1)/200)% = 0.085%

If your hand has 2 paired broadway cards, then

P(XXX) ≈ (4*4/200)% = 0.08%

If your hand has 2 unpaired broadway cards, then

P(XXX) ≈ ((3*4 + 1 + 1)/200)% = 0.07%

This is a really old topic but looking for the odds. Something fun to add into this but add in the odds of seeing KKK twice on the flop within 100 hands of each other lol. Just happened tonight on ACR.

Ignoring what cards are held by the players, the chance KKK will fall on the flop is P3k = 4/22,100 = 0.000181 or about once every 5525 times.

There are (at least) two ways of defining the problem. I’ll address both:

1. Given a randomly selected 100 flops, what is the probability that KKK will occur at least twice.

Pr =1 – Pr(KKK falls either 0 or 1 time in 100 flops)

= 1 - ( 1- P3k)^100 – 100*P3k *(1-P3k)^99 = 0.00016

2. Given KKK falls on the flop, what is the probability a KKK flop occurs at least one more time in the next 100 trials

Pr =1- (1- P3k)^100 = 0.01794

haha awesome man thanks. Came back to add some more fun, 3rd day in row KKK on ACR. was about 150 hands in.