Poker Probability Question

What is the probabilty of getting Aces Full of Jacks, or better, using both hole cards, in Texas Hold Em? I've been all over searching for the answer.....please help.

for one player, or a whole table of hands?

For one player to get at least AAAJJ. I dont think it matters how full the table is....

What range of hands is the player playing?

Quote: hchollenbeck

What is the probabilty of getting Aces Full of Jacks, or better, using both hole cards, in Texas Hold Em? I've been all over searching for the answer.....please help.

Are you trying to figure out bad beat odds?

Quote: mickeycrimm

Are you trying to figure out bad beat odds?

This is my assumption as well

Check out https://wizardofodds.com/games/texas-hold-em/bad-beat-jackpots/

Here is the complete problem that Im facing: I need to figure out the probability for any random player to get Aces full of jacks or better, using both his hole cards, with a random number of opponents (we can say a seven handed table on average), to qualify for a high hand jackpot that I'm trying to incorporate into a poker promotion. (I work for a casino.) The player only needs to get this hand, he doesn't need it beaten etc. Basically it should be a number around 1/700 I'm thinking because the odds of getting ANY full house or better is 1/560.

The odds will be greater than 1/700 for sure.

First the probability of getting any full house is approximately:
C(13,1)*C(4,3)*C(12,1)*C(4,2)*C(11,2)*C(4,1)*C(4,1) / C(52,7) = (13)(4)(12)(6)(55)(4)(4)/ (133,784,560)
= (3,294,720)/(133,784,560)
= 2.46% or about 1 in 40

The probability of getting quads or a straight flush are comparatively negligible so the probability of getting any full house or better can be said to be about 1 in 40.

Now the probability of getting Aces full of Jacks is approximately:
C(1,1)*C(4,3)*C(3,1)*C(4,2)*C(11,2)*C(4,1)*C(4,1) / C(52,7) = (1)(4)(3)(6)(55)(4)(4)/ (133,784,560)
= ( 63,360 )/(133,784,560)
= 0.0474% or about 1 in 2,000

Again the probability of getting quads or a straight flush are comparatively negligible so the probability of getting Aces full of Jacks or better can be said to be about 1 in 2,000.

Now if the requirement is to use both of your hole cards then this will exclude (2/7)+(5/7)(2/6)= approx. 1/2 of the results. Therefore, the probability reduces to (0.0474%)(1 - 1/2) = 0.0226% or about 1 in 4,000.

This also doesn't consider folding. Although if both hole cards are needed, not many of the hands that qualify would fold either.

Folding is also something I hadn't quite considered into the equation, which should variably lower the odds.

Thanks Paisiello, you're the man.