Percentages

[IanV]

I'm readin HE for Advanced Players by Sklansky and Malmuth. Came across the second par. on page 169 where the percentage of time you'll win a pot is discussed.
My question is: how do I calculate the percentage of time I will win a pot?

[elitegimp]

Play a lot of hands and divide the number you win by the number you play (and multiply by 100 to turn it into a percentage). If you play tightly in a loose game, you should win less than 10% of the time, and I would guess 6-8 would be a decent range.

Checking my meager PT stats, through 7504 hands I've won 561 after the flop... that's 7.4 or 7.5% (not counting times it is folded to me pre-flop, which doesn't happen too often at .5/1 limit /images/graemlins/smile.gif)

edit: re-read your question, and am now re-reading the paragraph mentioned in the book /images/graemlins/smile.gif

edit 2: to calculate the percentage of times you win, you count your outs (cards that will give you the best hand). You also count all the cards you can look see (your hand, cards on the board, cards accidentally mucked face up). Suppose you have x outs, with y cards left in the the deck. You will hit x/y of the time, or (x/y)*100 percent of the time (since x has to be less than or equal to y, this percent is less than or equal to 100). If you are on the flop, and have to cards to come, it is usually easiest to calculate the percentage that you don't hit your outs: (y-x)/y on the turn, and then (y-1-x)/(y-1) on the river -- multiply these together, subtract from one (if you _don't_ win 0.25 of the time, or 25% of the time, you do win 0.75 of the time, or 75%) and multiply by 100 to get it as a percent.

edit 3: one tricky bit is to figure out which cards to come will give you the best hand. In the example in the book, if your opponent has KT then only a 9 or an 8 helps (and the 9 only helps if a K doesn't also come). Since there are 3 9s left in the deck (you have one in your hand!), and there are 4 8s in the deck, you have 7 outs. Assuming you know your opponent has KT, you have seen 7 cards... so your probability of winning is roughly 1-(38/45)*(37/44) = 0.29 or 29%. If your opponent has 72, a T, 9, or 8 gives you the best hand, so you have 10 outs, then you win 1-35/45*34/44 = 0.40 or 40% of the time. However, if your opponent has 76, then only an 8 can help you, giving you 4 outs. You win 1-41/45*40/44 = 0.17 or 17% of the time.

Typically, you don't know exactly what your opponent has. However, if you think your opponent will fold KT here and raise 76 or an A, then you can guess that you have 10 outs (assuming you bet and get called). Also, since you don't know what your opponent has, you only consider 5 cards seen (two in your hand, three on the flop), so your odds for hitting one of ten outs is 1-(37/47)*(36/46)=0.38 or 38%. However, if you check this flop, you may fear a hand like KT and therefore you wouldn't want to count the 3 Ts as outs... so now you only have 7 outs and a 1-(40/47)*(39/47)=0.28 or 28% chance of winning.

Apologies if this makes no sense, it's late and I should go to bed /images/graemlins/smile.gif