Expected value or expected growth?

Expected value—E(V)—is the probability-weighted value of all possible outcomes of a random variable. In gambling or investing, expected value is defined by the monetary value of each outcome. Expected growth takes the same variables (outcome, probability, and payoff), and includes the effect of compounding.

Of note:

• Expected value may never appear as an actual observation. As with any average, the result describes a collection of observations, rarely an included instance in the sample.
  • E(V) is the probability-weighted mean of possible outcomes; while
  • E(G) indicates the long-term trend, not a final outcome.
• Expected value will always be greater than or (rarely) equal to expected growth. (Where probabilities are 0% and 100%, E(V) and E(G) will be equal;) and
• Most importantly, over-betting on a positive expected value can create negative expected growth. Under-betting slows the growth rate. Over-betting leads to ruin.

What is expected value?

Expected value is the sum of the weighted products of the possible outcomes times the probability of those outcomes. In gambling or investing, those products are multiplied by the payoff of each outcome to find an expected value in dollars.

Imagine a fair six-sided die, but instead of a number on each side, there’s a letter (A through F). Next, imagine that the game is to bet on what letter will come up. If you roll your letter, you win 6x your bet.

You bet on F. What’s the expected value of this bet?

• What are all the possible outcomes? A, B, C, D, E, and F. (Six possibilities.)
• What is the probability of each outcome? All are equal: 1/6 or 0.16 or 16.6%.
• What is the payoff of each outcome? What is each outcome worth?
• A through E are worth -1x. You’ll lose your bet.
• F is worth 6x. You’ll win six times your bet if F appears.
• So, what’s the expected value of the bet?

$$E(V) = -1(\frac{1}{6}) -1(\frac{1}{6}) -1(\frac{1}{6}) -1(\frac{1}{6}) -1(\frac{1}{6}) + 6(\frac{1}{6})$$

$$E(V) = -1(\frac{5}{6}) + 6(\frac{1}{6})$$

$$E(V) = -\frac{5}{6} + \frac{6}{6} = 0.1\overline{6}$$

So how much do you want to bet on each roll?

It’s tempting to think that if the expected value is 0.16 that you should bet 16.6% of your bankroll. If you start with $100, why not bet $16.66 on the first round and 16.6% of the remaining bankroll thereafter? Because that would be way too much.

What is expected growth?

The shortcoming of expected value is that it doesn’t consider the effect of compounding over a large number of trials, especially the effects of negative compounding. That’s where expected growth comes in.

The formula for expected growth is:

$$E(G) = (1 + b * f)^{np}(1 - f)^{nq} - 1, where$$

• 1 represents your bankroll;
• b is the odds* of the bet;
• f is the fraction of our bankroll at stake;
• n is the number of trials;
• p is the probability of a winning outcome; and
• q is the possibility of a losing outcome (q = 1 – p).

So for our letter-sided die above, the expected value of rolling an F is 0.16. What is the expected growth if we bet a fraction of our roll equal to the expected value of 16.6%?

• b is 6. We’ll win six times our bet if “F” comes up;
• f is 16.6% of our bankroll;
• n is 1 for betting just once;
• p is 0.16. There’s a 1/6 chance that we’ll win; and
• q is 0.83. There’s a 5/6 chance that we’ll lose.

$$E(G) = (1 + 6 * 0.1\overline{6})^{1(0.1\overline{6})}(1 - 0.1\overline{6})^{1(0.8\overline{3})} - 1$$

$$E(G) = 0.9643$$

So the expected growth would be a factor of 0.9643 each time that we bet. The trend will be negative.

Each time we play, our bankroll is expected to become 96.43% of what it was. If we start with $100, then after one roll, we’re expected to end up with $96.43.

That was on one roll. But if we keep betting like this, it’s even worse.

How does expected growth include compounding?

The expected growth above was based on betting just once. If we continue to play this game, then the factor we came up with is multiplied with each roll.

If we have the 1-roll growth factor, we can still find out what the most likely outcome is over multiple rolls. Over any number of rolls, we put the expected growth factor to the power of n, the number of rolls. So after 10 bets on this game:

$$E(G) = 0.9643^{10} = 0.6952$$

... our bankroll is expected to shrink by a factor of 0.6952.

If we play this game ten times, our bankroll is expected to be 69.52% of what it was. Starting with $100, we’re expected to end up with $69.52. And if we keep playing long enough, $0.

$$E(G) = 0.9643^\infty = 0$$

Will there be the occasional upward trend in our bankroll? Of course. Expected growth doesn’t predict each outcome. There will be “winning streaks”. But the long-run trend will be negative, eventually ruining us.

This is why Las Vegas will always be successful. In the games that they allow, the house is on the side of positive expected growth over an infinite number of trials while individual players are betting on negative expected growth and hoping for winning streaks. In effect, rational people are betting against emotional ones. That bet always has the same outcome.

With a growth factor of less than one, the long-run expected growth is negative. So the long-run expected value of betting anything on this game is $0.

Fine, now what?

Let's un-math this to make it more practical.

Our lizard brains are often suckered into thinking, "There is a chance of a good outcome here. I want to play that game." This happens often in:

• ... dating, with attractive but unscrupulous people;
• ... investing, with exciting but profitless businesses (TSLA, anyone?);
• ... backcountry skiing, with steep and pristine but perhaps-unstable slopes;
• ... anything, where the upside provides some—but not enough—compensation for the risk.

The important non-math takeaway is this: The possibility of a positive outcome is not enough to take an action. Only the long-term effect of repeating the behavior can tell us the wisdom of a choice. The trend (up or down) of a repeated action is what makes an action, even just once, a wise one or not.

So how do we find opportunities that, if repeated, would lead to an upward trend? Believe it or not, there's a formula for that too.

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* "Odds" in this context are gambling odds. If a game offers 6:1 odds, then a win will pay out six times the bet. In that context, "odds" does not mean the chance of success unless you change the odds into an inverted fraction like \(\frac{1}{6}\)​. However, it's unlikely that the house would offer a payoff that is symmetrical to the risk. Over time, they'd go broke.

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For more detail and examples, check out the following two posts from the Sports Book Review forum:

• Expected value versus expected growth
• Maximizing expected growth