Was wondering what odds are in a Full Ring NL game when you get dealt AK

that any opponent holds AA or KK?

Thank you.

From simulation, I get that the prob is around 3.89%. (This assumed 8 opponents.)

The gurus over in the Probability Forum would be able to provide an exact answer (and derivation) if you seek that.

If you have AK then there are 3 aces left, making 3 possible combos of aces. There are 1225 holdem hands possible once you've seen your 2 cards, so the chance that one opponent has AA would be 3/1225. The chances of AA or KK would then be 6/1225. That
s 0.4898%

The chance of at least one opponent having AA or KK can be found approximately by
1 - (1 - (6/1225))^N
I say approximately because if one opponent has AA or KK then none of the others can. It's close enough though.

For 9 opponents that is 4.3%, for 8 opponents it's 3.85%

It's actually also a good estimate to say
N*6/1225
because the probability is low and it's only possible for 2 opponents to match. That gives 3.9% and 4.4% respectively for 8 or 9 opponents.

The two estimation methods Rusty showed are always good for these problems (because the chance of multiple occurrences is slim, and card removal effect is slim). Furthermore, we can say a little more: the lower estimate is the lower bound and the higher estimate is the upper bound.

For 8 opponents the exact answer is somewhere between 3.85% exclusive and 3.9% exclusive.
For 9 opponents it's somewhere between 4.3% and 4.4%.

That tells you that the estimates are very close.

Alright, challenge accepted. It shouldn't be too difficult to derive the exact probability. I will try to do it for 8 opponents. Break it down into cases of how many aces or kings are dealt to opponents. C(x,y) is the combination notation.

Prob of at least one of your 8 opponents having either AA or KK when you hold AK is:

= Prob of an opponent having AA + Prob of an opponent having KK - Prob of one opponent having AA and another opponent having KK

Prob of an opponent having AA:

= [C(47,14)*C(3,2)*C(2,2)*C(8,1) + C(47,13)*C(3,3)*C(3,2)*C(8,1)]/[C(50,16)*C(16,2)]

= 11,575,694,957,760 / 590,842,763,469,000

= 1.95918367%

Prob of an opponent having KK is the same.

Prob of one opponent having AA and another opponent having KK:

= [C(44,12)*C(3,2)*C(3,2)*C(8,2)*C(2,2)*C(2,2) + C(44,11)*C(3,3)*C(3,2)*C(8,2)*C(3,2)*C(2,2) + C(44,11)*C(3,2)*C(3,3)*C(8,2)*C(2,2)*C(3,2) +
C(44,10)*C(3,3)*C(3,3)*C(8,2)*C(3,2)*C(3,2)] / [C(50,16)*C(16,2)*C(14,2)]

= 9,805,475,649,060 / 53,766,691,475,679,000

= 0.01823708%

The desired prob can then be determined to be:

= 3.90013026%

P.S. I have been known to make mistakes in these types of calculations so caution is advised.

[8*2*3 / C(50,2)] - [C(8,2)*3^2 / C(50,4) / 3!!] = 3.88189318%

looks like Taylors formula guessing this could be graphed via Area under the Curve.

Thanks much guys!

I doubt this has any relation to Taylor's theorem. It's hypergeometric distribution with a twist. This problem is simplified by applying the principle of inclusion-exclusion and by ignoring terms that cancel out (corresponding to unspecific cards). I started with the upper bound estimate and subtracted what it over-counted (namely, the case of AA and KK being dealt).

Nice!

@whosnext: last night I didn't feel like searching through your long calculation to find the bug. But now I'm bored, so I will.
This part is right.
8*3 / C(50,2) = 1.95918367 %

I see what went wrong: you had to multiply the whole numerator by 2, so as to count player identity (ie the distinction between which villain gets the AA and which one gets KK). The denominator counts that, but player ID doesn't actually matter, so it needs to be either canceled out or not counted anywhere.

If you don't want the denominator to count player ID, it has to be: C(50,16) * C(16,4) * 3!!
or equivalently: C(50,16) * C(8,2) * 15 * 13

Thank you for doing that.

Man we have some phucking mathematicians up in here. Where you guys been hiding ?

LOL

I need a coach that wants to practice on a volunteer basis lol