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## The Box Model

I had never heard of the Box Model until I used Statistics by David Freedman, Robert Pisani, and Roger Purves as a course textbook a few years into the 21st century. It simplifies a great many problems by seeing them as a process of pulling tickets from a box. So, for example, suppose you bet a dollar on red while playing roulette. There are 38 slots on the roulette wheel, 18 red, 18 black, and two green. Thus, you can view the game as one of drawing a ticket from a box of 38 tickets. Eighteen of them have \$1.00 on them, and 20 of them have -\$1.00 on them (because you will either win a dollar or lose a dollar.)

The advantage of converting problems to this form is that it allows us to analyze problems with a very simple probability distribution, the uniform distribution, even when they do not initially appear to be uniform problems. Each ticket has the same probability, so to find the expected value or mean of the distribution, we simply add up the sums on the tickets and divide by the number of tickets. If we do this in the case of the roulette wheel, we see that the expected value is about a negative five cents; the odds favor the casino. We can also compute the standard deviation fairly simply. We use the formula that we learned near the beginning, but instead of dividing by n-1, we divide by n. (Check back to discrete probability distributions to figure out why.)

For some boxes there is a short-cut formula for the standard deviation that is useful. If the box contains only two kinds of tickets, we can multiply the fraction of tickets with the big number by the fraction with the small number. Then we take the square root of that product, and multiply the result by the difference between the big number and the small number. In the roulette case, we multiply 18/38 times 20/38 to get .24930747922. Taking the square root gives .499307, or approximately .5. The difference between the big number and the small number is \$2.00, so the standard deviation is \$.998614 or approximately \$1.00.

The power of this model comes when we use it with the Central Limit Theorem. If we play roulette 100 times, it would be like pulling 100 tickets from the box with the 38 tickets described above. The expected sum would be 100 times the expected value of the box, or a negative \$5.26. However, there will be variation each time we draw those 100 tickets, and the expected error of the sum, which is the square root of 100 multiplied by the standard deviation of the box, tells us how much variation we can expect in the sums. Ten times .998614 is \$9.99. So if we play roulette 100 times, always betting one dollar, we would expect to lose about \$5.26 give or take about \$10.00.

To compute our chances of breaking even or better, we use the normal curve. It is centered at -5.26 and we want to know what the probability of getting \$0 or above is. We need to convert to standardized or z-scores. To convert the \$0 to a z-score, we subtract the mean and divide by the standard error of the sum, or (\$0-(-\$5.26))/\$9.99 = .53. Checking this on a table shows that the player has a 30% chance of walking out of the casino with more than the started with (and, obviously, a 70% chance of leaving with less.)

In the long run the casino will win. A player who makes this bet 10,000 times will have an expected loss of \$526.53 give or take \$99.86. The chance of leaving the casino ahead after this many plays is less than one in a thousand.

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