# Odds of winning the lottery if I buy 10,000,000 tickets

In a lottery whose numbers consist of 7 digits, 0–9, there are $10^7$ possible winning combinations. What would be the odds of winning if I purchase 10,000,000 "quick pick" tickets, where the numbers are randomly picked on my behalf?

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Well, the probability of one of the tickets not being the winning combination is $$\frac{10^7-1}{10^7}.$$ If all the tickets were truly randomly chosen (so that one or more tickets may be identical), then the probability that none of them is the winning ticket is $$\left(\frac{10^7-1}{10^7}\right)^{10^7}\approx0.367879.$$ Can you get the rest of the way?

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Let $N=10^7$. I will assume that the "quick pick" selections may repeat. The probability that one particular ticket is a non-winning one is $1-\frac{1}{N}$, so the probability they are all non-winning is $\left(1-\frac{1}{N}\right)^N$. Thus the probability of at least one win is $1-\left(1-\frac{1}{N}\right)^N$.

The number $\left(1-\frac{1}{N}\right)^N$ is very close to $\frac{1}{e}$, where $e$ is the base for the natural logarithms. This is because $e^x = \lim_{n\to\infty}\left( 1 + \frac x n \right)^n$ and $\frac 1 e = e^{-1}$.

Numerically, the probability of at least one winning ticket is approximately $0.63212$.

You may get more than one winning ticket, in which case you have the pleasure of sharing the grand prize with yourself, and perhaps others. On average, the more you play, the more you lose.

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@user84751: format that as $a^n-a^{n+1}<a^{n-1}-a^n$ if you want it to look like $a^n-a^{n+1}<a^{n-1}-a^n$. As it is, it's not even parenthesized as it should be... –  dfeuer Aug 15 '13 at 16:34
Details depend on the structure of the lottery. Usually there is a "pot" to be shared by all grand prize winners. It is conceivable that this prize is so huge that a ticket has positive expectation. (Never has happened, probably never will since if pot is huge more people buy tickets.) But ordinarily a ticket has negative expectation, the cost is substantially greater than the mean win, usually at least twice as large. Then "the more you play the more on average you lose" has to do with the fact that the expectation of a sum is the sum of the expectations. –  André Nicolas Aug 15 '13 at 16:39
But on reasonable assumptions, your inequality will do it. –  André Nicolas Aug 15 '13 at 16:41