# Lesson 24 – What else did you expect?

On a hot summer evening, a low energy Leo walked into his apartment building without greeting the old man smoking outside. Leo has just burnt his wallet in Atlantic City Roulette game, and his mind has been occupied with how it all happened. He went in with \$500. Ten chips of \$50. His first gamble was to play safe and bet one chip at a time on red. He lost the first two time, won the third time, and lost the next two times. After five consecutive bets, he was left with \$350. In the Roulette game, the payout for red or black is 1 to 1. He started getting worried. Since the payout for any single number is 35 to 1, in a hasty move, he went all in on number 20, just to realize that he was all out.

Could it be that luck was not favoring Leo on the day? Could it be that his betting strategy was wrong? Are the odds stacked against him? If he had enough chips and placed the same bet long enough, what can he expect?

###### Based on his first bet

Imagine Leo bets a dollar at a time on red. He will win or lose \$1 each time. In an American Roulette, there will be 18 red, 18 black and two green (0 and 00) — a total of 38 numbers. Each number is independent, i.e. there is an equal chance of getting any number. The probability of getting a red is 18/38 (18 reds in 38 numbers). In the same way, the probability of getting a black is 18/38, and the probability of getting a green is 2/38.

If the Ivory ball ends in a red, he will win \$1; if it ends in any other color he will lose \$1 – or he gets -\$1. In the long run, if he keeps playing this game with dollar on and off, his expected win for a dollar will be

On average, for every \$1 he bets on red, he will lose 0.05 cents.
###### Based on his Second bet

Now let us imagine Leo bets on a single number where the payout is 35 to 1. He will win \$35 if the ball ends up in his number, or lose the dollar. The probability of getting any number is 1/38 (one number in 38 outcomes). Again, in the long run, if he keeps playing this game; win \$35 or lose \$1, over time, his expected win for a dollar will be

Although the payout is high, one average, for every \$1 he bets on a single number, he will still lose 0.05 cents.

This estimation we just did is called the Expected Value of a random variable. Just like how “mean” is a description of the central tendency for a sample data, the expected value (E[X]) is a descriptive quantity of the central tendency (average behavior) of a random variable X with a probability distribution function (f(x)).

In Leo’s case, X is the random variable describing his payout, x is the actual payout from the house (\$1 or \$35 if he wins, or -\$1 if he loses), and f(x) is the probability distribution or frequency of the outcomes (18/38 for red and 20/38 otherwise, or 1/38 for a single number and 37/38 otherwise).

You will notice that this equation is exactly like the equation for the average of a sample. Imagine there is a very large sample data with repetitions; we are adding and averaging over the groups.

Poor Leo expected this to happen but didn’t realize that the table is tilted and the game is rigged.

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