Can someone post the percentage of times you'll face an overcard on the flop when holding the different pocket pairs?

KK 23%
QQ 43%
JJ 59%
TT 71%
99 81%
88 88%
77 93%
66 97%
55 99%
44 99.7%
33 99.9%

This is from page 99 of Mike Petriv's Holdem Odds Book.

These aren't quite right. Assuming the dead cards are random, this is the right answer.

Findingneema is correct given the wording.

Mike Petriv's numbers assume you do not get trips or quads. In that case, the exact probabilities are:

KK 23.43%
QQ 42.88%
JJ 58.72%
TT 71.32%
99 81.06%
88 88.30%
77 93.41%
66 96.76%
55 98.73%
44 99.68%
33 99.98%

you are correct in that he assumes four higher cards are available but are not those numbers the same as what you typed? i mean rounded up, or am i missing something? there is simply not enough difference in any of them to matter....

I'm pretty sure Aaron just added the extra precision in the numbers to show the difference between mine and his.

I am sure findingneema numbers are correct.

I don't get it.
You flop set quite often so 99.98% can't be correct unless you mean :
(flops where overcard flops) / (all flops without a set) but I don't see any use for such number.

Obv for 33 the number is academic, you're generally only continuing if you flop a set. For the hands like 99-KK, the number is pretty meaningful. And since 99% of the time you flop a set, you don't care if there's an overcard or two, Aaron's numbers are slightly more useful than mine. But it really doesn't matter, given that there's no practical difference between them.

I mean he is counting the wrong thing. I think what matters is % of flops when w don't have overpair AND we don't have a set. In other words the % of flops on which we doesn't have a very strong hand.
This number is always higher than % of flopping a set.
For KK it's 20.67%
For QQ it's 37.84%
For JJ it's 51.82%
...
for 33 it's 88.22%

Awesome Post!

AA 0%

I simulated it and I am quite sure this is correct.

Can someone please post the method of calculation for this percentages ?
Let's say odds you'll see an overcard when holding TT, including flops with sets & quads.

Thank you.

A much more useful calculation is the chance for overcards AND you don't flop a set. You don't care about overcards when you hit the set, in fact you welcome them because you might get paid. The chance for at least 1 overcard AND you don't have a set are approximately:

KK .21
QQ .38
JJ .52
TT .63
99 .72
88 .78
77 .82
66 .85
55 .87
44-22 .88

Edit: I see punter beat me to it.

FTR has a great chart detailing the odds of an overcard coming, on each street. You can find it here. It also has a rough explanation of how they got the values.

That was what I was looking for. Thanks.

Thanks findingneema at first I thought I was going crazy as I could not get the numbers, and I didn't want to say they were wrong.

We can also get the answer by building it up

There are 50 choose 3 combos to choose 3 cards from 50 or (50*49*48)/(3*2*1). IE the flop has 19,600 possible flop combinations. This is the denominator or total outcomes. Now we want to find success divided by total outcomes

Then we have 4 aces in the deck so 46 not ace cards (it's 46 as you hold 2 cards in your hand presumably not aces). The odds that one ace shows up on the flop is

A _ _ This is the flop we want. So we have one ace on the flop and 2 not aces [(46*45)/(2*1)]. So there is 46 not aces * 45 not aces. So you get a number 1035 and this would be the number if there was only one ace in the deck but there are 4. So multipled by 4 aces in the deck is 4,140 combinations or 21% chance.

Now there is also the possibility of A A _ coming out So there are 46 combos for that third card. There are also 6 different ways to have AA come up on flop. As there is 4 aces choose 2 ways to pick it.

So 46 *6 which is 276

Then there is 4 choose 3 ways to pick 3 aces or 4 total ways to get an AAA flop.

All together it's (4140+276+4) /19600 or 22.55%

Having said that, the odds are lower if you think your oppnent has an Ace.

That would lower the combinations to 3105+138+1 or 16.5%. So the true odds are 16.5%. Since the odds you're losing when you have KK to an ace is only 16.5%.

Sort of sad that you are the only one who wrote this.

I’m not sure how to interpret DS’s comment.

Anyway, here is the equation to calculate the percentage of having an overcard to your pair on the flop.

Assume the pair is of rank R. Then there are 4*(R-2) cards less than R in the deck and 2 cards equal to R.

The probability of at least one flop card greater than R is then 1 – the probabability that all flop cards are less than or equal to R. This is

Pr= 1-C(4*(R-2)+2, 3)/C(50,3)

Example: You have a pair of tens. There are 4*(10-2) + 2 = 34 cards less than or equal to 10.

Pr = 1 – C(34,3)/C(50,3) = 1 – 0.305 = 0.695

The thread prior to that post was a bit silly, just people listing stats without calculations to back them up. One person lists stats, then another person says, "Those are wrong, here are the correct ones," without providing a reason the others were wrong or why theirs are right.

...AHH, AM I MISSING SOMETHING (JUST FINISHED 24HR GRIND) BUT WHERE IS THE FACTOR FOR NUMBER OF PLAYERS, CARDS ALREADY OUT OF DECK?

Number of players makes no difference because you don't know their cards. Whether the unknown cards are lumped into one big pile or piles of two doesn't change any probabilities. See the sticky

I would appreciate some feedback on my "calculations" here - I'm trying to calculate the probability of getting "a flop I like" with 99. For this exercise I will define "a flop I like" as flopping either a set or an overpair. I know that we could flop an OESD or a monotone flop or a BDFD as well but I'm going to ignore those for now.

There are 50*49*48=117600 total possible flops.
There are 30*29*28=24360 total flops where all 3 cards are 9 or less (4 each of 2 through 8, but only 2 remaining 9s because we have 2 of them).

We just need to add on the number of flops where at least one card is a 9 and at least one of the other 2 is an overcard.

If the first card is a 9 and the second card is an overcard, the 3rd can be anything, so 2*20*48 possibilities. There are 6 total combinations of 9, overcard and anything, so we get 2*20*48*6 = 11,520

So there are 11,520 + 24,360 = 35,880 out of 117600 or 30.5%

Repeating the process for JJ, it's (38*37*36+2*12*48*6)/117600 = (50616+6912) /117600 = 57528/117600 = 48.9%

Repeating for KK, it's (46*45*44+2*4*48*6)/117600 = (91080+2304)/117600 = 79.4%.

Comments?

As a general comment, you should use combinations instead of permutations. You say that there are 50*49*48 flops, but order doesn't matter and, for instance, AcKcQc and KcQcAc are the same flop.

In the quoted passage, you double count some combo. The right number of flops containing at least a 9 and at least an overcard is 1520 (=9120 permutations) and so your total is wrong.

However, you should proceed differently to make sure to not double count combos. Let U be an undercard and X any not-9 card. The correct way to proceed is to count the combos for each of these kinds of flops:

UUU
9XX
99X

As you can see, in this way there isn't any flop that belongs to more than a group and you avoid double counting. I think you can proceed from there (also the above formulation is easily generalizable for any rank).

With this understood and agreed, shouldn't I still get the correct answer if I count permutations because I'm looking for a percentage?

I mean, in the case I'm calculating there, isn't the number of permutations exactly the number of combinations times 6?

Also, in my quoted calculation, can you figure out where I'm going wrong? I mean there are 2 different 9s that could be the first card, there are 20 different overcards that could appear second, and then the third card could be any of the 48 cards that didn't appear in the first or second (including the case 9). I can't figure out what was inaccurate there.

Lastly, while I agree that your UUU/9xx/99x approach is significantly cleaner than my 9s and lower plus at least one 9 and at least one overcard, was my approach incorrect? It seems to me that if I count 9s and lower first, then there are no duplicated permutations when I count at least one 9 plus at least one overcard. Am I missing something?

Thanks for your help
DTXCF