# Probability and Measure

1.1.1. Introduction

Texans used to play the lotto by selecting six different numbers between 1 and 50, which cost $1 for each combination.[1] Twice a week, on Wednesday and Saturday at 10 p. m., six ping-pong balls were released without replacement from a rotating plastic ball containing 50 ping-pong balls numbered 1 through 50. The winner of the jackpot (which has occasionally accumulated to 60 or more million dollars!) was the one who had all six drawn numbers correct, where the order in which the numbers were drawn did not matter. If these conditions were still being observed, what would the odds of winning by playing one set of six numbers only?

To answer this question, suppose first that the order of the numbers does matter. Then the number of ordered sets of 6 out of 50 numbers is 50 possibilities for the first drawn number times 49 possibilities for the second drawn number, times 48 possibilities for the third drawn number, times 47 possibilities for the fourth drawn number, times 46 possibilities for the fifth drawn number, times 45 possibilities for the sixth drawn number:

The notation n!, read “n factorial,” stands for the product of the natural numbers 1 through n:

n! = 1 x 2 x ■ ■ ■ x (n — 1) x n if n > 0, 0! = 1.

The reason for defining 0! = 1 will be explained in the next section.

Because a set of six given numbers can be permutated in 6! ways, we need to correct the preceding number for the 6! replications of each unordered set of six given numbers. Therefore, the number of sets of six unordered numbers out of 50 is

/50 def. 50!

II 6) = 61(50 — 6)!

Thus, the probability of winning such a lotto by playing only one combination of six numbers is 1/15,890,700.[2]

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