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Old 09-06-2005, 03:28 AM
PokerAce PokerAce is offline
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Join Date: Jan 2005
Posts: 385
Default Re: Redefining Aggression - PokerAce Hud

Here's what I'm trying to accomplish. Basically, I'm turning aggression into a percentage. 0% would be the most passive play possible. 100% would be the most aggressive play possible.

Would this not be pretty accurate to define aggression?

I think we need to define what is aggressive and what is passive.

Betting = Aggressive
Raising = Aggressive
Check Raising = Aggressive

Calling = Passive
Check Calling = Passive
Check Folding = Passive

Check without Call/Fold = Neutral


After discussing things with my friend some more, we came up with this new calculation:

(times_bet + times_raised) / (times_bet + times_raised + times_called + times_folded)


Let's do an example. We'll ignore preflop play for now. This player is aggressive, so he check raises the flop, bets the turn, but check folds the river.

That's 1 bet, 1 raise, and 1 fold. Doing the formula:

(1 + 1) / (1 + 1 + 0 + 1) = 67%

If the player bets the river, then it's:

(2 + 1) / (2 + 1 + 0 + 0) = 100%

If the player calls the river, then it's:

(1 + 1) / (1 + 1 + 1 + 0) = 67%


Another example. A player check calls the flop, check calls the turn, and makes his flush so he check raises the river:

That's 2 calls and 1 raise. That would be:

(0 + 1) / (0 + 1 + 2 + 0) = 33%

If he bets the river, then it's:

(1 + 0) / (1 + 0 + 2 + 0) = 33%

If he misses and check folds the river, then it's:

(0 + 0) / (0 + 0 + 2 + 1) = 0% (no aggression at all)


Let's say we have a maniac who bets his flush the whole way. Bet the flop, bet the turn, bet the river.

That's 3 bets, nothing else:

(3 + 0) / (3 + 0 + 0 + 0) = 100% aggression.


So does this make sense or am I completely out of my mind?
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