I have KK on the button.

is there around 1% chance that sb or BB has AA? 0.96% to be exact.

how did someone figure that out?

52 cards in a deck divide by 1 x 51 divide by 1 = 2652 divide by 2 = 1326 starting hands.

Ah-Ad, Ah-Ac, Ah-As, Ad-Ac, Ad-As and Ac-As. 6 combinations of AA.

1326 divide 6 = 221

221:1 = 0.45%

so theres a 0.45% SB has AA.

but hang on theres only 50 cards in the deck since we have KK.

50/1 x 49/1 =
1225 starting hands

1225 x 6 = 204:1

0.48%

do we just times 0.48 x 2 because we have 2 opponents left in the hand to get 0.96% chance of sb or bb having AA?

if we had 88 on the button we open then got 3 bet by SB is there a 34:1 or 2.8% chance hes got a higher pair?

1225 / 36 (6 combinations of AA, KK, QQ, JJ, TT, 99 each totaling 36 combinations).

34:1 or 2.8% 3% rounded.

so we can believe we have the best hand pre flop with 88 against SB and BB about 94% (94.4%) of the time. is this right?

There is a 1 in 220 chance that a player has AA. So the odds that SB does not have AA is 220 in 221, or 99.548%. The odds of SB and BB both not having AA is 99.548%^2, or 99.0971%. So the likelihood that at least one of SB or BB has AA is 100% - 99.0971%, or .903%. It is easier to calculate the odds of an event not happening, and then subtract that from 100% to find the odds of the event happening.

Note: just read your comment about removing 2 kings from the deck. If you do that, using the method I describe above, you end up with a .9772% chance of SB or BB having AA

If you have a particular combo of KK, then there are 1225 other possible combos for an opponent. Six of those make AA. 6/1225 = 0.49%
If you need two players to both not have aces, then the maths is a little bit more complicated, but it works out as about a 1% chance of one of them having the nuts.

If you get dealt KK in a full ring (9-handed) game there's about a 4% chance that someone else has aces. It happens so infrequently that it's not worth stressing over. Besides, the chance that you run KK into AA is just as likely as someone else running KK into your AA. It's just an inevitable cooler that happens from time to time.

99+ is indeed 36 combos out of 1225. 36/1225 = 2.94%

But that number is not particularly useful in a strategic sense, because the BB could be 3-betting worse hands too, like AK, AQ, KQs, QJs etc. The hand doesn't end pre-flop with someone saying "I have the best hand, so I win".

You should be thinking in terms of equity and post-flop playability. If villain was only playing 99+ (which would be a weird/bad strategy), I'd rather have 65s than 88, because it has more equity/playability post-flop.

This is more complicated than it might first appear for the player hand probabilities are not independent. For example, one player can have one ace, which reduces the chance the other player has AA. I did both an analytical solution and a simulation (10 million trials) and both got 99.02% for neither player having AA.

In this case, assuming independence will be a decent approximation for neither player having AA.

Small probabilities and small samples you can just use a linear approximation. 0,5% * 2.

.

Wouldn't the more pertinent question be, "what are the chances the SB has AA once he 5 bets you?"


To be more serious, chance someone has X or Y isn't that interesting. They haven't acted and sure, they have a random hand. They fold most of <random> to your first raise. With KK, you only care after significant action... at that point, the villain's hand range is much smaller.

There are two good ways to get an accurate approximation. The first is what OP and Kelvis did, simply multiplying one villain's probability by 2 since there are 2 villains. The next is what Spewing showed, treating the hands as independent and calculating the chance of it not happening.

For getting the exact answer, there is only one sane method, which is to use the inclusion-exclusion principle. It makes this problem much simpler.

OP, what you did isn't quite the exact answer because it double-counts the possibility that both villains have AA. To make it exact, we subtract that possibility so that it's only counted once.

From start to finish: 2*6/1225 - 1/C(50,4) = 0.97915762 %

(That's larger than what OP wrote only because he rounded wrong.)

Also, +1 to what DougL said.

Getting 3-bet increases the chance (unless they 3-bet 100%), but the calculation for before they've acted is as follows:

2⋅36/1225 - [C(6,2)⋅6²/3 + 6]/C(50,4) = 267/4606 ≈ 5.8%

Again I started with what you did and then subtracted the chance of both players getting it.

C(6,2)⋅6²/3 is the #ways they can be dealt two different pairs. I divided by 3 because if the cards AAKK are dealt, we only want to count AA|KK cases as opposed to AK|AK. Only one out of 3 distributions is AA|KK. You can either do it like I did and divide by 3 on this step, or you can wait until the end and make the denominator C(50,4)⋅3 or C(50,2)⋅C(48,2)/2. But then you'd have to multiply the 6 same-pair combos by 3 so that its 3 cancels with the denominator's, since all 3 ways to distribute AAAA result in AA|AA. In other words that would look like [C(6,2)⋅6² + 6⋅3] / [C(50,4)⋅3]