Lottery ticket sales and ‘coverage’

Question

Take a small 6/40 lottery with 3,838,380 possible combinations. Prior to the draw, the operator sells only 3 million randomly drawn tickets (for simplicity’s sake, I’m completely ignoring the fact that many thousands of players use combinations like 1,2,3,4,5,6 etc.).

Since we can safely assume that some of these random tickets will be duplicated, how can we estimate the probability that the winning combination will be found among the 3 million sold? I’ve heard lottery operators using the term ‘coverage’ for this.

Answer 1

The probability that any given ticket is not the winning combination is $$ p = \frac{3838379}{3838380}$$ So the probability that none of the three million tickets has the winning combination is given by ... $$ P_\text{no winner}=p^{3000000}\approx 0.458 $$

Answer 2

You can estimate by using the well-known limit

$$\lim_{n\rightarrow \infty} (1-\frac{1}{n})^n=\frac{1}{e}$$

Except here you have the probability $1-(1-\frac{1}{3838380})^{3000000}$ which is essentially

$$1-\lim_{n\rightarrow \infty} (1-\frac{1}{n})^{\frac{3000000}{3838380}n}$$ $$=1-\lim_{n\rightarrow \infty} [(1-\frac{1}{n})^n]^{\frac{3000000}{3838380}}$$ $$=1-(1/e)^\frac{3000000}{3838380}$$ $$\approx 1-(\frac{1}{e})^{0.78158}$$ $$\approx 0.5423$$ which is the probability you will have a winning ticket.