Simulation Check- No Better Ace Kicker or Pair

I wrote a brute force program to consider all possibilities.

My program says that there are 1,099,164 out of 1,381,800 deals in which neither other player has a better ace or a pair.

1,099,164 / 1,381,800 = 79.5458%

To be honest I did this very hurriedly with a minimum of checking so there is a not insignificant chance that this result is bogus. But I'll post it anyway in case it is helpful.

Okay. I got 79.6% with much fewer simulations so I would say we're both right (or wrong!).

I don't know if I could do this analytically but I don't see why it is not doable.

Anybody?

I haven't done a "brute force" combinatorical derivation in awhile so here goes. Not everybody will divide the cases the way I did but this is what seemed natural to me. The cases will reflect the ranks of the other two players' four cards. The tallies will reflect "prohibiting" pairs and Aces with higher kickers. Hopefully most things will be self-explanatory.

Case 1: 4 ranks not A,8
= C(11,4)*C(3,1)*C(2,1)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 506,880

Case 2: 4 ranks with both A,8
= C(11,2)*C(2,1)*C(4,1)*C(4,1)*C(3,1)*C(3,1) + C(6,2)*C(2,1)*C(2,1)*C(4,1)*C(4,1)*C(3,1)*C(3,1) + C(6,1)*C(5,1)*C(2,1)*C(4,1)*C(4,1)*C(3,1)*C(3,1)
= 33,120

Case 3: 4 ranks A no 8
= C(6,3)*C(3,1)*C(2,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1) + C(6,2)*C(5,1)*C(2,1)*C(2,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1) + C(6,1)*C(5,2)*C(2,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 103,680

Case 4: 4 ranks 8, no A
= C(11,3)*C(3,1)*C(2,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 190,080

-----

Case 5: 3 ranks not A,8
= C(11,3)*C(3,1)*C(2,1)*C(2,1)*C(4,2)*C(4,1)*C(4,1)
= 190,080

Case 6: 3 ranks AA8x
= C(6,1)*C(2,1)*C(2,1)*C(3,2)*C(3,1)*C(4,1)
= 864

Case 7: 3 ranks A88x
= C(6,1)*C(2,1)*C(2,1)*C(3,2)*C(3,1)*C(4,1) + C(5,1)*C(2,1)*C(2,1)*C(3,2)*C(3,1)*C(4,1)
= 1,584

Case 8: 3 ranks AAxy
= C(6,2)*C(2,1)*C(2,1)*C(3,2)*C(4,1)*C(4,1)
= 2,880

Case 9: 3 ranks 88xy
= C(11,2)*C(2,1)*C(2,1)*C(3,2)*C(4,1)*C(4,1)
= 10,560

Case 10: 3 ranks A8xx
= C(6,1)*C(2,1)*C(2,1)*C(3,1)*C(3,1)*C(4,2)
= 1,296

Case 11: 3 ranks Axxy
= C(6,1)*C(5,1)*C(2,1)*C(2,1)*C(3,1)*C(4,1)*C(4,2) + C(6,2)*C(2,1)*C(2,1)*C(2,1)*C(3,1)*C(4,1)*C(4,2)
= 17,280

Case 12: 3 ranks 8xxy
= C(11,2)*C(2,1)*C(2,1)*C(2,1)*C(3,1)*C(4,1)*C(4,2)
= 31,680

-----

Case 13: 2 ranks not A,8
= C(11,2)*C(2,1)*C(2,1)*C(4,2)*C(4,2)
= 7,920

Case 14: 2 ranks AA88
= C(2,1)*C(2,1)*C(3,2)*C(3,2)
= 36

Case 15: 2 ranks AAxx
= C(6,1)*C(2,1)*C(2,1)*C(3,2)*C(4,2)
= 432

Case 16: 2 ranks 88xx
= C(11,1)*C(2,1)*C(2,1)*C(3,2)*C(4,2)
= 792

If I did everything correctly, I get the grand total of the above sub-cases to equal 1,099,164 which was the total found via the brute force program.

Cool. Your complex analysis confirms my initial thought to use simulation was correct.

In case anyone is interested, the genesis of my question is research I’m doing on what I call a Preflop Wonder Hand, which is an AX hand where no opponent has an ace with a higher kicker or a pair. Such a hand will usually have a showdown equity advantage over opponent hands.

For example, A8o is a monster hand against KQs, being a 54/46 favorite. If a second opponent has 89s, then the showdown equity of A8o is 33% and KQs does a bit better with a 37% equity. Still, the Wonder hand will have +EV if the hand was checked down.

Having an estimate of a particular Ax hand being a Wonder hand, you can then do an EV analysis. For example, for the A8 hand against 2 opponents, there is about a 20% probability that one or both have a pair or an Ace with a higher kicker. If you make estimates of opponent better holdings (e.g., one opponent’s range is 77+), you can then use the occurrence probabilities and showdown equites to get a first-cut EV estimate and hopefully develop strategy for achieving good profit with these Wonder hands.

I get 79.59%, or an extra 600 combos compared to whosnext's count.

P(opposite) = P(A9+) + P(pair) - P(both)
=
2*5*3*4/(50C2) - 3[(5C2)(4^2) + 5*6]*(2/3)/(50C4)
+ 2[11(4C2) + 2*3]/(50C2) - [(11C2)(4C2)^2 + 11*3 + 11*2*3(4C2) + 3^2]/[3(50C4)]
- 5*3*4*72/[3(50C4)]

I used inclusion-exclusion for P(A9+) and again for P(pair), so three times total.

@whosnext I haven't compared notes much, but I notice you're missing at least one case: one rank aka both villains same pair, of which there are 66 combos with your denominator. Since your code agreed with your calcs, I wonder if we have a different interpretation of the problem.

I'm trippin, I forgot you were counting the opposite thing I was in my calculation (until the end when I subtracted everything from 1).

So then yes, most likely the error is in my calculation. I'll return to this another time.

One more result that may interest some.

The relationship between the occurrence probability and the ace kicker is linear according to my simulation results.

For example, for 1 and 2 opponents they are as follows:

One Opponent: Prob = 0.0098 * K + 0.8138, K = kicker

Two Opponents: Prob. = 0.0174 * K + 0.6581

Found my error:

The 72 should be 67. I lazily reused the 72 from the previous line without considering whether I should. It changes because once one of the villains gets an Ax hand, there are of course fewer pair combos the other villain can make: 10*6 + 2*3 + 1.

Now my result exactly matches whosnext's.

Since my brute force program takes just a few minutes to run, I ran all the 2-opponent cases.

You will see that the kicker effect is not quite linear but pretty close. Anytime that multiple events deal with shared cards, there will always be a card removal effect (which will make the overall effect non-linear). In this case it is fairly small.

Thanks for catching that. As said above my program is so freaking simple (essentially just one line of checking each opponent's ranks) that there aren't many places for it to go wrong.

Of course, what took me an hour of combinatoric derivations took far far less time using the very powerful Principle of Inclusion-Exclusion.