What is the Expected Value?

1/3 — Three Key Factors That make a Casino Strategy “Impossible” to beat.

Hey, I am WolfOfAvax,

I used to play online poker for a living and I am currently working as a business developer and advisor in crypto projects that I think can bring innovation to the world.

Recently I’ve joined BetSwirl’s team, hoping to help them create a safe, fair and fun environment for players around the globe.

Here is an educational article I wrote explaining some important terms in the betting industry so unaware people can get in touch with Game Theory critical concepts. I hope you’ll enjoy it !

This is the first article of the series:

1. Expected Value (EV)
2. Variance
3. Bankroll management (Risk of Ruin)

Expected Value (EV)

Let’s start and give a quick introduction to what EV (expected value) is.

EV is the predicted value of a variable, which we calculate by summing all possible values of that variable multiplied by the probability of its occurrence.

Let’s examine/demonstrate 3 different examples so we completely understand what this EV is all about.

1. Coin Flip / Coin toss (2 sides fair coin)
2. Die roll (6 sides fair die)
3. Roulette (0–36 fair European roulette)

Even though EV is a statistical term, it is usually used to examine the profitability of a “bet”, meaning if the bet is profitable, a loss-making or break-even bet in the “long term”. It’s also easier to demonstrate examples with $ being the examined variable so let’s do that.

Coin Flip / Coin toss (2 sides fair coin)

A fair coin assumes the probability of each outcome is exactly 50% meaning that if we toss the coin infinite times we know that the sample of the coin landing on the A or B side will be exactly equal.

Now let’s assume player1 places a 10$ bet vs player2 on the outcome of the coin flip. Obviously, each player bets on a different side and the winner takes all of the bets (his 10$ bet + the other 10$ of the player who lost = 20$).

Using the EV definition above we can demonstrate what is the EV of this bet for each of the players.

So player1 chooses side A, if he wins, he pockets 20$ (securing a 10$ profit since half of the prize pool was his 10$ bet), and if he loses he suffers a 10$ loss which was his initial bet.

Given that we know the probability of each outcome we can use the EV formula so the EV calculation for Player1 would be:

EV1 = (amount P1 wins when he wins) * (chance of him winning) + (amount P1 “wins” when he loses) * (chance of him losing)

Replacing the variables with numbers we get:

EV1 = 10$*50% + (-10$)*50% = 0$

So the EV of this bet = 0$ (break-even bet) which makes sense since P1 will win/lose equally many times the exact same amount.

Die roll (6-side fair die)

Let’s give a slightly more complex example so we understand the importance of EV in making profitable decisions using a common fair 6-sided dice.

Again die being fair means that we expect each side of the die to have equally many chances to get rolled, so the probabilities of each side being rolled equal ⅙ since there are 6 different outcomes equally possible.

So let’s assume I propose you a bet that looks like this: you roll the die and if the outcome is 1 or 2 or 3 or 4 or 5 you get 100$ but if you roll 6 you give me 600$, is this a profitable bet for you? For me? Is it a breakeven bet? How would you decide?

There is this common misconception in betting that makes players get confused by the frequency of winning, meaning that in this example you would win ⅚ times while I’d only win ⅙ times or you would win 5 times more often than me. That sounds really good, winning 5 times more than your opponent surely would make this bet profitable for you. Not really. Let’s see why.

Again the ONLY way for you to decide correctly about the profitability of this bet would be to calculate the EV of this bet.

Similarly, with the Coin Flip example, we should add all the possible outcomes in $-value multiplied by the chance of them occurring. So your EV would look like:

EV = (amount you’d win when you roll 1) * (chance of you rolling 1) + (amount you’d win when you roll 2) * (chance of you rolling 2) + (amount you’d win when you roll 3) * (chance of you rolling 3) + (amount you’d win when you roll 4) * (chance of you rolling 4) + (amount you’d win when you roll 5) * (chance of you rolling 5) + (amount you’d “win” when you roll 6) *(chance of you rolling 6)

As explained above each side has ⅙ chance of being rolled and when you win you’d win 100$ and when you’d lose you’d lose 600$ so replacing the variables in the equation of this EV calc we’d get:

EV = 100$ * ⅙ + 100$ * ⅙ + 100$ * ⅙ + 100$ * ⅙ + 100$ * ⅙ + (-600$) * ⅙ = 100$ * ⅚ + (-600$) * ⅙ = 500$ / 6–600$ / 6 = -100$ / 6 = -16.66$

So by calculating this EV you can see that despite you winning 5x times than I would, this bet would cost you 16.66$ per roll on average, meaning that if you decided to take that bet X times the expected outcome for you would be losing X*16.66$. With this example, it’s obvious that the frequency of winning or losing does not guarantee a positive or negative EV.

Roulette (0–36 fair European roulette)

Let’s also examine the casino example so we demonstrate how much $EV does casino/house get out of a simple roulette bet.

Assuming a player bets 10$ in 1 out of 37 (equally) possible roulette outcomes, what is his EV of that bet? Roulette rewards you with 36 times the bet amount if you guess the number correctly so with similar logic we can write the EV of this bet like this:

EV = (amount we bet*35) * (chance of guessing the number correctly) + (amount we “win” when we don’t guess correctly) * (chance of not guessing the number correctly)

So replacing the variables in the equation of this EV calc we’d get:

EV = 10$ * 35 * 1/37 + (-10$) * (36/37) = 9.459$ — 9.729$ = -0.269$

As we see this means that roulette players are actually always placing a -EV bet (translates into a house edge of approx 2.7% since players would lose $0.269 on average for every 10$ bet).

Examining this result a bit more we can see that roulette is being fair when the player loses (meaning the player bets 10$ and loses 10$) but generates an edge vs the player when the player actually wins, that’s because the player’s guess has a chance of 1 / 37 to be correct but when it is correct he is only being rewarded with 36 times his bet (risk/reward ratio). If roulette were rewarding the winning player with 37x times his betting amount then the EV (for both the player and the house) would be 0$ meaning that this bet would be a break-even bet for both sides.

On the contrary, if the house was paying the player > 37x his bet, the bet would generate profit for the player and loss for the house (obviously this type of casino would go bust very very soon).

The exact same case occurs when players bet on black/red colors, similarly when they lose they lose 100% of their bet but when they win they are rewarded with 2x their bet which would be “fair” if black/red had 50% chance of occurring when in reality they have less than that (18 / 37) because of zero (green).