EV is also referred to as expectation and is the most important aspect of each move you make at a poker table.
To explain I'll take some examples of random gambling propositions first before moving onto some common poker situations. First, so we all know how odds are typically written I will give you an example. If your odds of winning any proposition are 4:1, this means you will win 1 in 5 times. The 4 represents the times you will lose and the 1 represents the times you will win.
The easiest proposition is the flip of a coin which is 50/50. If I offered you 1:1 odds on the flip of a coin for any amount of money this is considered a fair bet. Over 100 coin flips you can expect to win 50 times and lose 50 times. Your expectation or EV is 0. Now if i offered you 2:1 odds on the flip of a coin for your $1 (I'll be paying you $2) you have +EV. I think this is obvious. Over 100 coin flips you will win $2 -50 times and lose $1 -50 times. Therefore you win $100 and lose $50 giving you an expectation or EV of +$50 or $.50 per coin flip.
Dice also offer a great gambling proposition. You will roll a 7 more often than any other number. If you roll a 1,2,3,4,5, or 6 you will need to roll a 6,5,4,3,2, and 1 respectively to make a 7. Another way of reasoning this is that there are 36 combinations you can make (6*6) and there are 6 ways to roll a 7. So we have 6 ways to make a 7 and 30 ways to make something else.Therefore, your odds against are 30:6 or 5:1, so if I offered you 5:1 on the roll of a 7 your EV=0. (You can take the time to figure out all the odds if you'd like, here's the amount of ways to roll different numbers: 6 or 8 -> 5 ways, 5 or 9 -> 4 ways, 4 or 10 -> 3 ways, 3 or 11 -> 2 ways. 2 and 12 -> 1 way.
Now if I offer you $10 to your $1 to roll a 7 here's how the math works out. Your odds are 5:1 and I'm offering 10:1. Over 6 rolls you lose $1 five times and make $10 one time. You made $5 in 6 rolls, now divide 6/5=1.2. So your EV is +$1.20.
Here's how it works in poker, the odds of hitting a flush on the river are 4.111:1, for the sake of easier math we're going to say it's an even 4:1. If you're opponent is all in on the turn for his last 500 in a pot of 1k, you're being offered 3:1 odds as it will cost you 500 to have a chance to win 1500 (500 all-in plus 1000 pot). Now we know this is -EV but how much can we expect to lose assuming all flush cards are outs and we have no other outs?
Our odds against are 4:1 so we will lose 4 times and win 1 time in every 5 tries. We stand to win 1500 once while losing 500 four times. So in five tries we've lost 500 giving us an EV of -$100 per try.
Now let's say our villian moved in for his last 500 into a pot of 2k. It costs us 500 four times like we stated previously, but now we win 2500 one time. So after 5 tries we are up $500 and our EV is +$100.
Now, here is an example of when EV debates become much more complex and hypothetical. Suppose you're in the cutoff with T9 off suit and a stack of 2k. Blinds are 200/400, everybody has you coverd but not by much. We're going to assume everybody folds 50% of the time and you win $600. The other 50% of the time you are a 2:1 underdog (remember we're being very hypothetical) so you stand to lose your stack twice and double up once in 3 called all-ins. Since this is only the case 50% of the time you have a 0.33 x 0.5 = 16.5% chance of doubling up. The chance of going broke is 0.66 x 0.5 = 33%. Of course our numbers are repeating so we'll say it's 33.5% we lose.
Distribution for this move looks like this: We stand to lose 2k 33.5% of the time. We stand to win 2k 16.5% of the time. We win $600 50% of the time.
Therefore in 100 tries we've lost: 2000 x 33.5 = 67k and we've won 600 x 50 = 30k + 2000 x 16.5 = 33k for a total of 63k
So the move in this very hypothetical situation is -EV, we've lost 4k over 100 hand so our EV is -40.
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