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Expected Value (EV)
waynec
Member Posts: 1,023
I found this very interesting article (Written by FPaulsson) online and thought I would pass it on to you guys.
If you have a die, ordinary randomized six-sided die, and apply the above reasoning to find out what the expected value of rolling the die is, you end up with this:
Rolling a "1" has a probability of 1/6.
Rolling a "2" has a probability of 1/6.
Rolling a "3" has a probability of 1/6.
Rolling a "4" has a probability of 1/6.
Rolling a "5" has a probability of 1/6.
Rolling a "6" has a probability of 1/6.
Multiplying the values with their respective probability gives:
1 * 1/6 = 1/6
2 * 1/6 = 2/6
3 * 1/6 = 3/6
4 * 1/6 = 4/6
5 * 1/6 = 5/6
6 * 1/6 = 6/6
Adding them together gives:
1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5
Thus, your expected value of a randomized die is 3.5. What if the die was weighted, so that the number "6" had a 50% chance of coming up? Well, if all the other numbers still had a uniform distribution ("equal chance of coming up in regards to each other"), you get this:
1 * 1/10 = 1/10
2 * 1/10 = 2/10
3 * 1/10 = 3/10
4 * 1/10 = 4/10
5 * 1/10 = 5/10
6 * 1/2 = 3
The sum of which is 4.5. Do you see why all the other numbers now only have a 10% chance of coming up?
Expected Value is the basis for most non-psychogical poker strategies. Like limping with medium pairs if the pot is not raised and there are other players who limp as well - that's a play that may have positive Expected Value. The poker dilemma, mathematically speaking, is to always make the decision that has the highest expected value (for the sake of stringency it might be worth pointing out that the highest expected value may in some cases be negative, but less negative than any other course of action).
To explain how expected value relates to poker, let's work with a (relatively) common scenario. You're playing Texas Hold'em, and you've somehow found yourself heads-up on the river, holding these cards:
A♥ J♣
And the board shows:
A♣ 10♣ 5♦ 8♣ 3♣
You're in first position, the pot is $100, and the big bet is $10. Do you bet?
Let's say, for the sake of argument, that your opponent can hold any two cards and will always fold if he doesn't have a club. Let's also stipulate that he'll call a bet with any club, and make it two bets if he has the K♣ or Q♣. Let's also say that in case you check, he will bet with any club and check with no clubs.
Let's do the math. Since he can hold any two cards, each of the individual clubs is as likely to be in his hand (and let's pretend that he can't have two of them - because we know him well enough to know that he would have raised the turn if he did).
Note: We do not bother adding in the times when he has no club at all, in these scenarios. Your opponent will fold if you bet, and check if you check in these cases, and you will then always win the pot uncontested. For the mathematically curious, this actually has implications on the expected value for the situation as a whole, but not for the specific purpose that we're discussing it here: Determining the correct strategy.
$10 * 6/8 = $7.5
If he raises, we know that you have a worse hand, and you will have lost $10.
-$10 * 2/8 = -$2.5
So your expected value of betting here is $7.5 + (-)2.5 = $5. Not bad.
6 times out of 8, you will win when you call his bet, and 2 times you will lose.
$10 * 6/8 = $7.5
-$10 * 2/8 = -$2.5
Here, again, is your expected value $5. Okay, so checking is as good as betting in this theoretical situation. What if we check with the intention of raising if he bets?
If he has the K♣ you will fold and lose $20:
-$20 * 1/8 = -$2.5
If he has the queen, you will get a showdown, but still lose $20.
-$20 * 1/8 = -$2.5
If he has any other club, you will win $20:
$20 * 6/8 = $15.
Sum: $15 - $2.5 - $2.5 = $10.8.
Concept: Expected Value
"The term "Expected Value" (also referred to as "EV" or "Expectation") is used a lot in poker strategy discussions, and if you've wondered what it means but never dared to ask, this is the article for you! The term originates in math (specifically probability mathematics) and is used to describe the long-term average outcome of a given scenario. In order to calculate expected value, you take every possible outcome, multiply each by the probability of that outcome happening, and then adding those numbers altogether. Sounds tricky? Let's look at an example.If you have a die, ordinary randomized six-sided die, and apply the above reasoning to find out what the expected value of rolling the die is, you end up with this:
Rolling a "1" has a probability of 1/6.
Rolling a "2" has a probability of 1/6.
Rolling a "3" has a probability of 1/6.
Rolling a "4" has a probability of 1/6.
Rolling a "5" has a probability of 1/6.
Rolling a "6" has a probability of 1/6.
Multiplying the values with their respective probability gives:
1 * 1/6 = 1/6
2 * 1/6 = 2/6
3 * 1/6 = 3/6
4 * 1/6 = 4/6
5 * 1/6 = 5/6
6 * 1/6 = 6/6
Adding them together gives:
1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5
Thus, your expected value of a randomized die is 3.5. What if the die was weighted, so that the number "6" had a 50% chance of coming up? Well, if all the other numbers still had a uniform distribution ("equal chance of coming up in regards to each other"), you get this:
1 * 1/10 = 1/10
2 * 1/10 = 2/10
3 * 1/10 = 3/10
4 * 1/10 = 4/10
5 * 1/10 = 5/10
6 * 1/2 = 3
The sum of which is 4.5. Do you see why all the other numbers now only have a 10% chance of coming up?
How does Expected Value relate to Poker?
Now, enough with the dice. We're poker players, let's focus on cards.Expected Value is the basis for most non-psychogical poker strategies. Like limping with medium pairs if the pot is not raised and there are other players who limp as well - that's a play that may have positive Expected Value. The poker dilemma, mathematically speaking, is to always make the decision that has the highest expected value (for the sake of stringency it might be worth pointing out that the highest expected value may in some cases be negative, but less negative than any other course of action).
To explain how expected value relates to poker, let's work with a (relatively) common scenario. You're playing Texas Hold'em, and you've somehow found yourself heads-up on the river, holding these cards:
A♥ J♣
And the board shows:
A♣ 10♣ 5♦ 8♣ 3♣
You're in first position, the pot is $100, and the big bet is $10. Do you bet?
Let's say, for the sake of argument, that your opponent can hold any two cards and will always fold if he doesn't have a club. Let's also stipulate that he'll call a bet with any club, and make it two bets if he has the K♣ or Q♣. Let's also say that in case you check, he will bet with any club and check with no clubs.
Let's do the math. Since he can hold any two cards, each of the individual clubs is as likely to be in his hand (and let's pretend that he can't have two of them - because we know him well enough to know that he would have raised the turn if he did).
Note: We do not bother adding in the times when he has no club at all, in these scenarios. Your opponent will fold if you bet, and check if you check in these cases, and you will then always win the pot uncontested. For the mathematically curious, this actually has implications on the expected value for the situation as a whole, but not for the specific purpose that we're discussing it here: Determining the correct strategy.
Scenario 1: You bet!
If he calls, we know that it will be with an inferior hand because he would have bet a better hand. There are 6 possible clubs that he will call with. So six times, you will win an extra $10. As there are 8 clubs available, the chance of him calling is 6/8 (six out of eight):$10 * 6/8 = $7.5
If he raises, we know that you have a worse hand, and you will have lost $10.
-$10 * 2/8 = -$2.5
So your expected value of betting here is $7.5 + (-)2.5 = $5. Not bad.
Scenario 2: You check, with the intention of calling if he bets.
(As above, you can safely ignore all the times when he has no clubs) 6 times out of 8, you will win when you call his bet, and 2 times you will lose.
$10 * 6/8 = $7.5
-$10 * 2/8 = -$2.5
Here, again, is your expected value $5. Okay, so checking is as good as betting in this theoretical situation. What if we check with the intention of raising if he bets?
Scenario 3: You check, with the intention of raising if he bets
In order to figure this out properly, we now need to stipulate that he will always re-raise with the nuts, so if he does have the K♣, he will re-raise you and you will fold. To avoid adding too much confusion, we will pretend that he will still call your raise with any other club. If he has the K♣ you will fold and lose $20:
-$20 * 1/8 = -$2.5
If he has the queen, you will get a showdown, but still lose $20.
-$20 * 1/8 = -$2.5
If he has any other club, you will win $20:
$20 * 6/8 = $15.
Sum: $15 - $2.5 - $2.5 = $10.8.
Conclusion on Expected Value
In this theoretical situation, your expected value is $6 higher if you check and raise, instead of betting out. To maximize your winnings, therefore, you should always check in this situation, and raise if he bets, because that will give you an average profit that's half a big bet higher than just betting out (or checking and calling). With the relatively small edges that are in effect for poker players, getting those extra 0.5BB in where you can is often the difference between a long term winner and a long term loser.Is this really applicable?
Yes, yes it is. In fact, virtually all that is widely considered "correct" poker is based on calculations like these. Check-raising, bluffing, calling with a decent but not strong hand, they're all plays based on the expected value. Of course, no one (okay, almost no one) actually calculates the exact values in their heads right there at the tables, but the strategies we play by are dictated by these numbers. Understanding how it works is not necessary to learn how to play, but it's necessary in order to review and analyze your own decisions, which is a very powerful way to strengthen your own game: Look at a specific hand, ask yourself "how could I have won more?" and do the calculations. Good luck! "