Calculating Expected Value in Poker?

Sinfonia

Trying to fully understand the calculation of expected value.

I know how to calculate it generally, but can't seem to get my head around applying it to poker. Is it done the same way? If anyone can explain it to me, I'd greatly appreciate it...

Cheers,
SG

ArmaniJeanss

Lets say you shortstacked on the button, 1000 chips, blinds 100/200 You decide to openpush with 23s from the button.

You apply % chances to what happens next.

e.g, 50% chance you win the blinds so you end up with 1300 50% of time.

50% chance you get called, with a 30% chance of your hand winning 2100 and a 70% chance of losing (0 chips).

So your e.v.
is (.50 * 1300)
+ (.15 * 2100)
+ (.35 * 0) = 965 chips, so this would be a -EV push.

Sinfonia

Perfect, thank you.

May I also ask the exact same question as above,
but instead of "Expected Value", insert "Variance"?





Okay, just to make sure I have this EV thing correct:

say it's Hold'em, i have AdKd

board is 4d 9c 8h Jd

We'll say I'm aiming for the flush only,
and ignoring implied odds or the possibility bluffing etc.

Pot: 1000
Opponent bets: 350

Roughly 4-1 to hit the flush

E(x)= 0.2(1700)+0.8(0)=340

This is 10 less than my call for 350, so this is -EV and thus an incorrect call?
Is that right?

Say the bet was 250

E(x) 0.2(1500)+0.8(0)=300

So that's 50 more than my call of 250, so this is +EV and thus a correct call?
Is that right?

5starpool

Variance just means that on occasion you run hot, and on occasion you run cold. Over the long term you can get a general idea of how much you win/lose, but at timees you can go on a good or bad streak that is out of character with your long term results.

Sinfonia

5starpool wrote:

Variance just means that on occasion you run hot, and on occasion you run cold. Over the long term you can get a general idea of how much you win/lose, but at timees you can go on a good or bad streak that is out of character with your long term results.

so when people are talking about variance they're not using it as a measurement, just a general word?

5starpool

SumGuy wrote:

so when people are talking about variance they're not using it as a measurement, just a general word?

They just mean that over time it will even out according to the laws of probability, but right now you might be getting a kicking and always outdrawn (or the other way around). There is no variance in general other than specific odds once the turn of the card is all that is left.

Sinfonia

5starpool wrote:

They just mean that over time it will even out according to the laws of probability, but right now you might be getting a kicking and always outdrawn (or the other way around). There is no variance in general other than specific odds once the turn of the card is all that is left.

ok, gotcha now, cheers:)

Sinfonia

SumGuy wrote:

Okay, just to make sure I have this EV thing correct:

say it's Hold'em, i have AdKd

board is 4d 9c 8h Jd

We'll say I'm aiming for the flush only,
and ignoring implied odds or the possibility bluffing etc.

Pot: 1000
Opponent bets: 350

Roughly 4-1 to hit the flush

E(x)= 0.2(1700)+0.8(0)=340 (or should it be 0.2(1700)+0.8(-350))?

This is 10 less than my call for 350, so this is -EV and thus an incorrect call?
Is that right?

Say the bet was 250

E(x) 0.2(1500)+0.8(0)=300

So that's 50 more than my call of 250, so this is +EV and thus a correct call?
Is that right?

Sorry to force this, but I noticed it was fading into the archives!

ArmaniJeanss

SumGuy wrote:

Sorry to force this, but I noticed it was fading into the archives!

Yes, your maths is correct.
Another way of doing this quickly is divide the pot by 4 (1350/4) = 337.50, this is the most you'd want to pay to see this river (given zero implied odds and an exact 4/1 shot). You are asked to pay 350 so it would be a -EV call.

Ditto your 2nd example, except it would be a +EV call.

Mr. Flibble

Say there are two winning players, Al and Bob, who have played a million hands and have made an equal amount of money.

Al tends to play like a rock, rarely getting involved in big pots unless he has the nuts. In an average session he tends to only lose a little or win a little.

On the other hand Bob plays a wild and aggressive game. He gets his money in on any big draw and is always in a lot of big pots. He tends to wither lose big or win big each night.

You would say Bob plays high variance poker and Al does not. If you saw a graph of their winnings Al's would be fairly steadily with only small zigzags. Bob's on the other hand would jump up and down like mad, but overall average of each graph would be the same.

Some types of poker require you to get involved in big hands often where you will only have a small edge if you wish to make money. You will need a larger bankroll for these games so your graph doesn't ever zig below a buyin and you are busto. These are high variance games. Other games won't and you will need a smaller bankroll in relation to the buyin.

NickyOD

It's a long time since we had a friendly, helpful, sarcasm free thread like his on boards.

Sinfonia

Ok, I saw this on another thread, and I got sorta confused

Gholimoli on another thread wrote:

guy has $200 in his stack ,

he makes it $10 to play,
i suspect he has AA,KK and i have 73o but i know if he flop two pair or better i will stack him :
wow lets see

i think the odds of flopping two pair is something like 50:1(2%) (could be wrong about the odds here).

so we assume everytime we hit two pair we will stack him (winning $200) and everytime we dont we lose our $10.
lets do a simple EV for this genius idea:

i win $200 2% of the time so $200x2%=4
i lose $10 98% of the time so -$10x98%=-9.8

so over all my play has the EV of -5.8

So I would have thought that it would have been
$200*2%=4
$0*98%=0

making EV=4

because your outcomes are to have either $200 or $0 (like in my inquisitive examples above)
so which is actually correct?



Also, thank you Armani, 5starpool, and Mr. Flibble for clearing all this up for me.
And Nicky, just for being Nicky