Suppose you decide to bet $10 on the numbers 1-12. You will win $20 (and keep your $10 bet) if the ball lands falls in slots 1-12; otherwise, you lose your $10. Is this a good game for you? How will you do, on the average, if you play this bet ($10 on 1-12) many times?
First, let's find the probability distribution for the amount of money you win on a single bet. There are two possibilities -- either you win $20 or you win -$10 (win a negative number means you lose +$10). You win if the ball falls in slots 1-12. Since there are 38 slots, each of which is equally likely, the probability of each slot is 1/38, and so the probability of falling in 1-12 (and you win) is 12/38. The probability you lose is 1 - 12/38 = 26/38. In summary, the probability distribution for the amount you win is
AMOUNT YOU WIN | PROBABILITY |
---|---|
20 | 12/38 |
-10 | 26/38 |
To summarize this probability distribution, we compute an average value, which is often called the mean . We compute this average in two steps:
We illustrate this computation for the roulette winnings in the table below. In the PRODUCT column we multiply each winning by its probability. The value at the bottom of the PRODUCT column is the average.
AMOUNT YOU WIN | PROBABILITY | PRODUCT |
---|---|---|
20 | 12/38 | (20)(12/38) = 240/38 |
-10 | 26/38 | (-10)(26/38) = -260/38 |
SUM | -20/38 = -.53 |
Here we compute the average value to be $-.53 or 53 cents. What does this mean?
Let's illustrate this using a computer to simulate many games. I play the game 100 times one day and I record the winnings (in dollars) for the games in the table below.
20 -10 -10 -10 20 -10 -10 20 -10 -10 -10 -10 20 20 20 20 20 -10 20 -10 20 -10 -10 -10 20 20 -10 -10 20 -10 -10 -10 -10 -10 -10 20 20 -10 -10 -10 -10 -10 20 -10 -10 20 20 20 -10 -10 20 -10 -10 -10 20 -10 -10 -10 -10 20 -10 20 -10 -10 20 -10 -10 -10 -10 -10 -10 -10 -10 -10 20 -10 -10 20 -10 -10 -10 -10 -10 20 -10 -10 20 -10 -10 20 -10 20 -10 -10 -10 20 -10 -10 20 -10
How have I done at the end of the day? It turns out that I am $40 in the hole after these 100 games. In other words, I have lost an average amount of $40/100 = $.40 or 40 cents on each game. This is close to the average value .53 that I computed from the probability distribution above. I would observe an average loss closer to .53 if I played this game a much larger number of times.
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