1

Assuming there's a lottery with 800,000 tickets and 24 of these tickets contain a win, my chance to win (if I buy only one ticket) is $\frac{24}{800,000}$ or $\frac{1}{33,333.\overline{3}}$, right?

But what are my chances to win if I buy two tickets?

4242
  • 13

1 Answers1

1

Your chances to win at least once are roughly twice as high if you buy two tickets. Not exactly twice, because there is a very small chance both tickets will win, but this is small enough to ignore in an approximation.

vadim123
  • 83,937
  • So that would mean the chance that I will win is now roughly $\frac{1}{16,666}$? And if I buy 10 tickets, will it be $\frac{1}{3,333}$? That doesn't sound right to me. – 4242 Dec 01 '16 at 16:30
  • That is, roughly, correct. However, if you buy 10 tickets, the probability of more than one victory from those 10 becomes much larger, so it's actually significantly less than $\frac{1}{3,333}$. – vadim123 Dec 01 '16 at 16:35
  • @4242 the exact calculation can be done via routine counting methods. If we want the probability of not winning after having bought $k$ different lottery tickets it will be $\frac{\binom{800000-24}{k}}{\binom{800000}{k}}$ where $\binom{n}{r}$ is the binomial coefficient $\frac{n!}{r!(n-r)!}$. The probability of winning at least once will be one minus that amount. For ten tickets that evaluates to approximately $0.0003$, similar to $1/3333$ but not exactly. – JMoravitz Dec 01 '16 at 16:44