Why an Ace always comes...

[Masquerade]

The scenario. Late in a tourney. It's folded to the button who allins. The small blind folds and you have 88. Correctly putting him on overcards, probably a big Ace, you call and he flips up. AJo. You know you're a 55-45 fav but an Ace duly shows up and you're out. But you got your money in as the favourite ... or did you?

The odds of at least one Ace showing up in the next five cards, given that you've seen four cards, is easy to calculate: 28.6%. (The same for a Jack of course). But the odds of at least an Ace or a Jack are almost exactly 50%. [The probabilities arent linearly additive because there is some overlap.] Of course not all these cases win as the 8s can improve as well and you need to use an odds calculator but it gives the rough picture.

Now we are going to make an extreme assumption. We are going to model a weak online NL player as one who plays ANY ace. Obviously this isnt totally realistic but let's see show that affects the analysis. Let's also assume a 10 player table.

As none of the other 8 players entered the pot we can assume that the three remaining Aces must all be in the 32 remaining cards. This raises the probability of at least one Ace showing up by river to an incredible 41%. When we correctly account for the Jack as well, the probability of AJ catching goes from 50% to 60%, and overall becomes approximately a 55-45 favourite over 88!

These figures of course reflect the extreme assumption that as noone else played, no-one had an Ace. But that's not SO far off the mark, particularly in the later stages of a tournament when any ace is likely to be played into an unopened pot - particularly at lower limits. Do we need to rethink tournament strategy accordingly? And is this why an Ace always seems to come, busting our pair?

[LuvDemNutz]

What if he has KQ?

[adanthar]

Late in a tourney, more than a couple of people who play any ace are wildly unlikely to be present, even at a low buyin (though maybe not in a $1 or something.)

[locutus2002]

I have often thought that this is the case.

If only 1/2 the aces are playable (AQ,AK in EP, A9 in MP, and AX in LP) then only about ~1/2 an ace has gone out of the deck in 16 cards if 8 people fold. This raises the odds of getting an ace to about ~1/3; the odds for the jack are unchanged.

[locutus2002]

I think it would be interesting to see a distribution of the remaining cards in a deck after 8 folds which would be quite easy to do if you had access to poker data from any site.

[kdotsky]

This has already been done - Greenstein talks about it in his book. On his website he has more detailed results along with the code he used to simulate it. Basically, he programmed in the typical starting requirements for each position and randomly dealt out cards to a 9-person table except for the blinds, which received AK and a pocket pair every hand. If someone was dealt a suitable hand, they raised and the trial is over. However, every time everyone folded to the blinds they went all in against each other. The results show that AK beats most underpairs in this situation. AKs can be as high as 55.4% against 22. Only JJ and TT were always favorites against all combinations of AK.

A related concept - he gives everyone random hands, and checked the percentage of times the BB has AA when everyone folds to him. You're 65% more likely to flip over AA in the BB when everyone has folded.

The point of all this was just to show that the odds and math as we think of them are tainted because they are dependent on other factors. The use of this knowledge isn't useful during gameplay though, as Barry states.

Taken from http://www.barrygreenstein.com/extras.htm

Averages for AK against a pair

(Example of this calculation for AK offsuit versus a small pair: There are 6 suit combinations for the pair. One combination will have no suits matching the AK offsuit. One will have both suits matching the AK offsuit. Four combinations will have one suit matching one of the AK’s suits.)

Not very relevant, two players in a heads up game:
Pair Percent win AK suited Percent win AK unsuited

Deuce 49.8933% 47.3515%
3 49.2061% 46.6252%
4 48.5869% 45.9695%
5 48.0344% 45.3835%
6 47.6942% 45.0113%
7 47.6897% 45.0161%
8 47.5241% 44.8372%
9 47.4270% 44.7290%
10 45.9370% 43.1225%
Jack 45.9664% 43.1485%
Queen 46.0486% 43.2417%
King 34.1064% 30.1202%
Ace 12.1405% 6.8281%


The following table is more relevant since it takes into account other players folding:

Everyone folds to the blinds:
Pair Percent win AK suited Percent win AK unsuited

Deuce 55.4144% 52.7147%
3 54.5226% 51.7303%
4 54.2049% 51.4650%
5 53.0757% 50.9556%
6 52.6798% 50.6427%
7 53.1181% 49.9561%
8 52.1371% 49.9713%
9 51.7050% 48.9705%
10 49.8178% 46.6155%
Jack 49.7355% 46.9116%
Queen 50.3328% 46.8328%
King 38.2566% 35.2465%
Ace 12.3512% 7.5166%

[burningyen]

[ QUOTE ]
What if he has KQ?

[/ QUOTE ]
Folded hands are more likely to contain a K or a Q than an A.

[HoldingFolding]

In the mid/late stages of a tournament, I think the flip side of this thinking might help. A scenario you might consider is shortish stack going all in from ep, a medium stack coming over the top in mp and you holding a mid/low pair in lp. If you are also shortishly stacked calling may be +EV if you think there's a probability that shorty's going for it with a weak ace and mp has a better one. I obviously don't have a viable sample size, but I've tripled up like this a few times.