Suppose an individual plays a gambling game where it is possible to lose $1.00$, break even, win $3.00$, or win $10.00$ each time she plays. The probability distribution for each outcome is provided by the following table:
Outcome -$1.00 $0.00 $3.00 $5.00 Probability 0.30 0.40 0.20 0.10
The mean outcome for this game is calculated as follows:
$$ \mu = (-1*.3) + (0*.4) + (3*.2) + (10*0.1) = -0.3 + 0.6 + 0.5 = 0.8$$
In the long run, then, the player can expect to win about 80 cents playing this game -- the odds are in her favor.
In the above gambling example, suppose a woman plays the game five times, with the outcomes $0.00$, -$1.00$, $0.00$, $0.00$, -$1.00$. She might assume, since the true mean of the random variable is $0.80$, that she will win the next few games in order to "make up" for the fact that she has been losing.
Unfortunately for her, this logic has no basis in probability theory. The law of large numbers does not apply for a short string of events, and her chances of winning the next game are no better than if she had won the previous game.
If she plays the game 500 times a day for next 30 days = 15000 games. Wouldn't she will definitely win because it is a large number now?