say I have 99 and my opponent has 99 (just us 2) and I want to figure out how often overcards (10sJsQsKsAs) will hit the flop compared to how often undercards (anything 8 and below) will hit.

So I know that on the first card the chances of it being an overcard is 1 in 2.6, but what are the chances of an overcard coming down when drawing 3 cards?

please include formula and explanation, thank you

~50% you will have pp below TP,32% you will have weak pair,18% over pair.This is the formula

lol thanks but i really need the formula do you know how to?? lmao i cant figure it out

∩ means AND
U means inclusive OR
Let the chance of X happening be denoted P(X)
Let P(X) given event Y be denoted P(X | Y)
Let the n^th card being an overcard be denoted as A_n
Let the n^th card being an undercard be denoted as B_n

The chance of at least one undercard coming down are

P(B₁ U B₂ U B₃) = 1 - (P(A₁) * P(A₂ | A₁) * P(A₃ | A₁ ∩ A₂))

P(A₁) = 20/48 (20 of 48 remaining cards are overcards, = 1 in 2.4 not 1 in 2.6 because the four 9s are not possible choices)

P(A₂ | A₁) = 19/47 (19 of 47 remaining cards are overcards)

P(A₃ | A₁ ∩ A₂) = 18/46 (etc.)

P(B₁ U B₂ U B₃) = 1 - (20 * 19 *18) / (48 * 47 * 46)

P(B₁ U B₂ U B₃) = 1 - 0.0659 = 0.9341

I guess the dude with tha nick "do the math" knows about maths

what on earth did you just say?
i can feel my brain sizzling

okay so what exactly is .9341 is that odds or a percentage? :S still lost lmao

93.41% or a little over 14:1 odds.

oh wow thanks, i would really love to know how you got that but to me that formula was gibberish... I wish there were a google translate for math lmao thanks though

You have 99 and your opponent has 99

There are 48 unseen cards and 20 of those are over cards to your 99's (4 each of the 5 ranks, A, K, Q, J, T) and the other 28 must be undercards.

There are 48C3 possible flops (where nCr is the number of ways you can select r items from a set of n items assuming the order in which you pick them is irrelevant).

That's 48C3 = 17296 possible flops.

And there are 28C3 = 3276 flops made up entirely of undercards.

So p(flop is all undercards) = 3276/17296 = 0.189

So p(flop contains at least one OVERcard)
= 1- 0.189
= 0.811

call it 81%

odds against of 0.811 : 0.189
roughly 4.3:1

You could also fire up TheOddsOracle from ProPokerTools and ask it how often the flop is 3 undercards:

ProPokerTools Odds Oracle Results (2.25 Professional)
Holdem, Generic syntax
PLAYER_1 99
PLAYER_2 99
600000 trials (randomized)

How often does the board match range [8-2][8-2][8-2]
18.9152% (113491)

Here, I'll try words instead of formulas (but you asked for the formula first.)

I think you see how the chance of an overcard for the first card are 20/48 = 1/2.4 = .4167 = 41.67%. There's 20 overcards out of 48 cards still available.

To express that as odds against, you need to make a ratio of failures to successes. That's easy to do. 28 failures : 20 successes = 28:20 = 7:5 against (divde both sides of the ratio by their common factor 4).

You can always convert odds to the form X:1 by dividing the each number by the second: 7:5 = 1.4:1. (7/5 = 1.4, 5/5 =1) You can also invert that decimal above, and subtract the one that will be on the right of the ratio: 1/.4167 = 2.4, 2.4-1 = 1.4 -> 1.4:1 odds

And look: when you add the two side of the ratio you get the total number of outcomes reduced to a unitary denominator 1.4+1 = 2.4 = 2.4/1 That's the inverse of the 1/2.4 we saw above. It's the inverse because above we were talking about success (chance of), but we calculated the odds as odds against.

So, if the first card is an overcard, there are now 19 overcards left in the deck, out of 47 total cards left. Repeat with the new numbers. If the second card is an overcard, then 18/46 cards left are overcards. To get the odds of three related things happening together, multiply the chances of each together.

So why are we taking about the chance of flopping all overcards if we really want to find the chance of getting at least one undercard? Well, because when you get at least one undercard, you didn't get all overcards, and it is easier to calculate all of one than the exceptions.

Either you get all overcards or you don't. The total probability of two exclusive choices is 1. The probability of the complete set of alternative possibilities is 1. So the chance of getting NOT all overcards is 1 - (the chance of getting all overcards). Getting NOT all overcards means the same as getting at least one undercard.

Yeah, from his posts I didn't think he was ready for permutations and combinations.

thanks for the detailed explanation I get it now

thanks for the explanation and link

You should probably just read some basic probability textbook if you're really interested in this sort of thing.