# How to Calculate Expected Value in Poker

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• Share Expected value in poker is commonly referred to as Expectation or EV. The term expected value had originated from the world of probability mathematics. It basically means a given scenario’s average outcome over the long run.

### How to calculate expected value?

In order to do that, every possible outcome has to be considered. Each of these outcomes has to be multiplied by the probability of the outcome taking place. Then all these numbers have to be added together. The whole thing sounds really confusing, so let me use an example to make it clearer.

Suppose you possess a randomized and very ordinary six sided die. And after rolling the die, you want to calculate the expected value by applying the theory given above. The results are given below. (I assume you are aware of the concept of probability)

If you roll a 6, the probability will be 1/6.
If you roll a 5, the probability will be 1/6.
If you roll a 4, the probability will be 1/6.
If you roll a 3, the probability will be 1/6.
If you roll a 2, the probability will be 1/6.
If you roll a 1, the probability will be 1/6.

Now you have to multiply these values by their respective probabilities. The results are as follows.

1 x 1/6 = 1/6
2 x 1/6 = 2/6
3 x 1/6 = 3/6
4 x 1/6 = 4/6
5 x 1/6 = 5/6
6 x 1/6 = 6/6

Now you have to add all these together and the result will be as follows.

1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5

Therefore, as you can see, 3.5 is the expected value of that randomized die. Suppose, the casino guys had weighted the die. This means that the chances of any one value showing up aren’t the same as in the previous case of a neutral die. Let’s assume that the casino guys have weighted it such that at least half the times, the number 6 will show up. For the other attempts, the remaining 5 values can show up randomly with equal probability. That means, the probability of 6 showing up is ½ and the probability of each of the remaining values showing up is 1/10. The result will be as follows.

1 x 1/10 = 1/10
2 x 1/10 = 2/10
3 x 1/10 = 3/10
4 x 1/10 = 4/10
5 x 1/10 = 5/10
6 x ½ = 3

When you add all these up, the resultant sum equals 4.5. I hope now you can understand the reason behind other numbers having only a 10% chance of showing up.

### Expected Value in Poker

So far, I was discussing the dice. Let us now move on to cards, as in poker.

The poker strategies that you normally witness are of two types – psychological (since poker is a mental game) and non psychological (since poker is also based on mathematics). Most of the non psychological poker strategies have their roots based in expected value. Suppose you are in a game where nobody has raised the pot and you are limping along with the help of very medium pairs.

Many of your opponents are also limping like you. There may be a positive expected value in a game like this. And in the world of poker, the biggest dilemma from a mathematical point of view, is to take the decision that yields the highest expected value. At times, the EV turns out to be negative. However, it is far less negative compared to other courses of action.

Let me now describe a very common scenario to explain the relationship between poker and expected value. Suppose you are in a game of Texas Hold’em. Your position is heads up on the river. The cards you are holding are J and A of hearts. The display on the score board is A 10 5 8 3.

Your position is first. \$10 is the big bet. \$100 is the worth of the pot. The question is, should you bet in this situation?

For the sake of the argument, let us suppose any two cards could be held by your opponent and if he doesn’t possess a club, chances are that he will fold, always. And let us also suppose that he will use his club to call a bet. He may also move on to two bets if the cards on him are Q or K. If you check in this situation, your opponent will check without the clubs or bet with the help of any club.

Now comes the mathematics part. We know that he is holding any two cards. Therefore, it is extremely likely that he might be holding each of those individual clubs. Let us also suppose that he is not holding both sets of clubs; because if he did, he would most probably raise the turn with them.

In the first scenario, you ought to bet. And if he calls your bet, the hand on him will be an inferior one. We can tell this for sure, because if he had a better hand, he would’ve placed a bet earlier. Your opponent can call with possibly 6 different kinds of clubs. Therefore, you stand the opportunity of winning some extra \$10 at least six times. The total number of clubs available is eight. So he would probably call six out of eight times.

\$10 x 6/8 = \$ 7.5

You possess a worse hand now and he has raised. You will lose \$10.

Negative \$10 x 2/8 = Negative \$2.5

The expected value of betting in this case is as follows.

\$7.5 + Negative 2.5 = \$5