Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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?

share|improve this answer

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.

share|improve this answer
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.