Poker Probability – Part 4: Expected Value in Poker

So far, we’ve learned all about probability in poker. We know how to predict outcomes with a fairly high degree of accuracy. But how does this help us to make a profit at the tables? The answer lies in expected value.

Expected value is like a window into the future. Using it, we can predict our long-term profits and losses with stunning accuracy. We can use EV in poker to exploit our opponents, make winning bets, and prevent losses. So let’s find out how to put EV to work for us at the tables.

What Is Expected Value?

Formally, the expected value for a distribution is the sum of the value of each outcome multiplied by its probability. For example, say we have the following probability distribution:

D = {($5, 1/6), ($3, 1/6), ($1, 1/6), ($-4, 1/3)}

This distribution represents a game where we bet on the outcome of rolling a 6-sided die. The event and reward chart looks like this:

• Roll a 1: win $5
• Roll a 2: win $3
• Roll a 3: win $1
• Roll a 4: lose $4
• Roll a 5: lose $4
• Roll a 6: lose $4

We can calculate the EV of participating in this die toss easily. For each possible outcome, we multiply the reward or loss by the chance of the outcome occurring. Then, we sum the calculated value for each outcome to obtain our expected value.

It’s easier than it sounds. Check it out:

EV(roll die) = $5*(1/6) + $3*(1/6) + $1*(1/6) + (-$4)*(1/3)
EV(roll die) = $5/6 + $1/2 + $1/6 – $4/3
EV(roll die) = $1/6 = $0.1666…

So our expected value for rolling the die is approximately seventeen cents. Cool, but what does that mean?

What Does Expected Value Represent?

Expected value tells us how much money we can expect to win or lose by making a particular bet. For example say we have a 35% chance of winning a poker hand, and our opponent bets $10 into a $12 pot on the river. How much is a call worth in this spot? How much money will we win or lose by calling?

Our opponent’s bet made the pot $22. So our prospective winnings total $22. A call would cost us $10, which means we’d lose $10 when the hand doesn’t go our way. To find the exact amount a call is worth here, we just multiply the value of each outcome by the probability of its happening:

EV(call) = P(win)*($win) – P(lose)*($lose)
EV(call) = (.35)*($22) – (.65)*(10)
EV(call) = $7.7 – 6.5
EV(call) = $1.2

So a call in this situation is worth exactly $1.20. That means every time we make this call, we expect to win a dollar and change. The expected value is positive, and so it’s a good bet.

What if we changed the payouts a little? Say the pot is now $5, and our opponent bets $6 on the river. Our chance of winning remains the same as before. What’s a call worth now?

Our opponent’s bet made the pot $11, so that’s our payoff. A call costs $6, which is our potential loss. Plugging these values into the expected value formula, we get:

EV(call) = P(win)*($win) – P(lose)*($lose)
EV(call) = (.35)*($11) – (.65)*($6)
EV(call) = $3.85 – $3.9
EV(call) = -$0.05

So it turns out that a call here is worth negative five cents. That is, we lose five cents every time we make the call. Thus we should decline to call, opting to fold instead. After all, the EV of a fold is always 0.

Hopefully this makes clear what expected value represents in poker: the worth of a particular bet. Knowing the expected outcome of our actions helps us to take profitable lines all the time at the tables. But there’s a twist: sometimes, the EV of a bet won’t quite match up with the outcomes you observe in the short term.

Expected Value in the Long Term

It’s entirely possible to win money by making a losing bet. This sounds like a contradiction – after all, you either win or you lose, right? Wrong! Here’s why.

Expected value, like probability, is only meaningful over the long term. EV isn’t a crystal ball, and can’t tell us what the outcome of our next bet will be with certainty. What it can tell us is the amount of money we’ll win or lose taking a particular action over a large sample of trials.

So it’s entirely possible for a bet with a negative expected value to net us positive wins in the short come. It’s also possible for a bet with positive expected value to lose us money in the short term. In the long term, the amount we win or lose from each bet will converge on its respective EV. But since probability is a measure of chance, there are swings involved; and we may do better or worse “than we should” in the short term.


1. Consider this $1/2/ NL Holdem hand. The pot is $50 and you’re on the turn. You have nothing but four-to-a-flush that you’re hoping to make on the river. If you hit, you win. If you miss, your opponent wins. The probability of hitting your flush is 9/46. Your opponent bets $48.

a) What is the expected value of calling?
b) Which is more profitable: a call or a fold?

2. Consider the following $1/2 NL Holdem hand. It’s the turn and the pot is $36. You have AA, and know you will win unless your opponent makes his flush on the river. The probability of the river card completing your opponent’s flush is 9/46.

a) How much should you bet to make it correct for your opponent to fold?
b) Say your opponent calls and wins. Has he made a profit or a suffered a loss in the long run?

Answers to Exercises

1a.    $win = $48 + $50 = $98
$loss = $48
P(win) = 9/46
P(loss) = 1 – 9/46

EV(call) = (9/46)*($98) – (37/46)*($48)
EV(call) = $19.17 – $38.61
EV(call) = -$19.44

1b. If you call, your expected value is negative $19.44. You lose about twenty dollars. The EV of a fold is always $0. Thus you should fold – your flush draw isn’t worth it this time.

2a. Let the variable x represent the amount you bet. Then we get the equation:

EV(opp call) = P(win)*($36+$x) – P(lose)*($x)
EV(opp call) = (9/46)*($36+$x) – (37/46)*($x)

Now we solve for X to obtain the point at which our opponent’s EV would be zero:

0 = $7.04 + (9/46)$x – (37/46)$x
0 = $7.04 – (14/23)$x
$7.04 = 0.61$x

So a bet of $11.54 is our opponent’s break-even point. That means as long as we bet a minimum of $11.55, calling our bet is unprofitable for our opponent.

2b. If our opponent calls any bet above $11.55, his play has a negative expectation. That means he will loses money in the long run every time he makes the call.

If he nevertheless wins the pot by calling a bet greater than $11.55, that’s fine – he still loses in the long run. Variance has provided him with a bit of short-term good luck; that good luck will invariably come back to haunt him in the future, as his expectation evens back out.

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