Does expected value change as the number of bets increases?

Question

I am reading about Expected Value in the context of roulette-game. I understand that it is the weighted sum of all the probabilities.

Here is the scenario:

The roulette game consists of a small ball and a wheel with 38 
numbered pockets around the edge. As the wheel is spun, the ball 
bounces around randomly until it settles down in one of the pockets. 
Suppose random variable X represents the (monetary) outcome of a $1 bet on a single 
number ("straight up" bet). If the bet wins (which happens with 
probability 1/38 in American roulette), the payoff is $35; otherwise 
the player loses the bet. 

Expected value for the above bet could be calculated as :

-1 * 37/38 + 35 * 1/38

which comes equal to -0.0526. Now what does it really mean? Can expected value change as the total number of bets increases? If yes, what does the above number actually mean? If it will not change with the increasing number of bets, are there some other factors that could influence the expected value?

Answer 1

Maybe the best way to understand it is the law of large numbers. If you make a million bets, your losses will almost certainly be be $1,000,000 \cdot 0.0526$ with a very small relative margin of error.

In the above statement, increasing the number of bets has the effect of increasing the level certainty and decreasing the relative error.

Answer 2

The expected value of $-0.0526$ corresponds to a bet of $1$. On average you lose $5.26\%$ of the money you bet. If you bet a total of $1000$, whether in one bet or in a series of bets of $1$, the expected loss is $52.60$. This is a consequence of the linearity of expectation.

The probability of particular results will depend on the sequence of bets you make. If you make one bet of $1000$ you will either be $+35000$ or $-1000$. If you make a series of bets you will have many possible results in that range. The probabilities will work out so the average loss is $52.60$

Your bankroll does not matter toward expected value. We assume you can make the required series of bets.

Answer 3

To make this a little easier with the arithmetic, suppose we had a smaller roulette wheel, with numbers $0$ through $9$. If you bet a dollar on any number, you win eight dollars if your number comes up; otherwise, you lose your your one-dollar bet.

The expected gain from this game is

$$ 0.9 \times (-1) + 0.1 \times 8 = -0.9+0.8 = -0.1 $$

That $-0.1$ represents the house percentage, or vigorish ("vig" for short). In the long run, you will lose a dime for every dollar you bet, but it may take a large number of bets for this percentage to become apparent; in the short run, you may lose a different percentage, or even come out ahead.

If you were to play a thousand games, for instance, you would lose, on average, about $100$ dollars. However, there would be some dispersion in the actual amount you would lose; this dispersion or spread is expressed by something called the standard deviation. The standard deviation in this case is about $85$ dollars. Your actual losses would be distributed in a bell curve, such that about $68$ percent of the time, the loss would be within one standard deviation of the average—that is, between $15$ and $185$ dollars. The dispersion is actually quite substantial.