Playing any Ax or Kx.

[Gregg777]

I was on a table tonight (Party Poker 25NL), with a real aggressive player who was making a killing with Ax and Kx, but primarily Ax.

His strategy was simple, limp in with anything under A10, and pot bet to the end whenever he had an ace (except with obvious straight and flush boards).

Yes, he lost some hands, but he won more than he lost. I was wondering about the percentages of this strategy.

The only time he slowed down was to a strong preflop raiser or a reraise. How sound is this? I am thinking of giving it a shot at the lower limits just to test it out. Any thoughts?

[LokiV]

http://www.pokertips.org/strategy/probabilities.php

Probability no one else has an ace assuming you do at a 10 person table: 25%

Case closed. He was getting lucky. And you picked up his pattern easily, so I hope you waited for the right hand and busted the hell out of him. End of his game.

[Relentless]

Nice link, wasn't aware of some of them like the Ace stats.

[schwza]

yeah, that's a horrible play. what the heck were people paying him off with? be patient till you get AK/AQ and do work.

[ScottTheFish]

[ QUOTE ]
http://www.pokertips.org/strategy/probabilities.php

Probability no one else has an ace assuming you do at a 10 person table: 25%


[/ QUOTE ]

Is this with an Ace on the flop? I'm not a stats expert, but doesn't an ace on the board make it less likely someone else has one in the hole? I doubt with an ace in your hand, and on thflop, the chances of another ace is as high as 25%, though I haven't calculated it.

Anyway obviously playing Ax and Kx is wrong, just wondering about the math

[Zag]

[ QUOTE ]
Is this with an Ace on the flop? I'm not a stats expert, but doesn't an ace on the board make it less likely someone else has one in the hole? I doubt with an ace in your hand, and on thflop, the chances of another ace is as high as 25%, though I haven't calculated it.

[/ QUOTE ]

25% is incorrect. Here is the calculation:

You can see 5 cards, two in your hand and 3 on the board. Assuming there is one ace in your hand and one ace on the board, there are 2 aces remaining of 47 unseen cards, which means 45 non-aces.

In the other players' hands, there are 18 cards and let's just assume that anyone with an ace has stayed to see the flop. Therefore, the chance that NONE of the 18 cards in your opponents hands is an ace is:

45/47 * 44/46 * 43/45 * ... * 29/31 * 28/30 = (29 * 28) / (47 * 46) = 38%

(Consider the cards one at a time. For the first card, there is a 45/47 chance of it being a non-ace. Assuming it is not an ace, then the second card has a 44/46 chance of being a non-ace. etc.))

This is also written as

(45!)(29!) / (27!)(47!) = (using calculator to check my math) 38% (Hooray!)

so the chance that someone DOES have an ace is 62%. (Of course, all things being equal, you are as likely to have the higher kicker as they are. But you know your kicker, so you can calculate that more accurately using that info.)

Note that there is another factor at play here, too -- your win/loss ratio. Against decent players, if you do have a lousy kicker, then the chance you are the only one with top pair is 38%, you will probably win with a bet, and you will have won the size of the pot. However, if someone has a better top pair, or can beat top pair, then you will lose a lot more than the size of the pot, because you will bet and then perhaps call more bets later on. So you would need more of an overlay against decent players, since you stand to gain less than you stand to lose.

Note that this gets reversed if you are a good player against truly awful players. If they will call you down with any pair, but you can get away only losing one pot-sized bet when you are behind, then you can make money with your 38%. I somewhat doubt that even the best player can consistantly get away for only one bet (without folding too many winners), but perhaps he can come close. If you could make your amount won / amount lost ratio better than 2-to-1, it makes sense to play your 38%.