# Probability: independant lottery ticket

I have one chance over 1,000,000 to win a lottery. If I buy the same number of ticket with different combination on each, I have 100% chance of winning. What if my ticket number are randomly choosen ?

What is the probability of winning a lottery (that has 1,000,000 possible combination) if I choose randomly the 1,000,000 ticket's number ?

Thank you

As the answer of Zubin Mukerjee has already pointed out the probability is

$$1-\left(\frac{999999}{1000000}\right)^{1000000}$$

You can use the relation $\lim\limits_{n\to \infty }(1-\frac{x}{n})^n=e^{-x}$ to get an appropiate approximation.

$\frac{999999}{1000000}=1-\frac{1}{1000000}$. Therefore we can write

$1-\left(\frac{999999}{1000000}\right)^{1000000}=1-\left( 1-\frac{1}{1000000}\right)^{1000000}\approx \boxed{1-e^{-1}}=63.21\%$

I'll assume that the ticket numbers are chosen independently and at random, and each has a $1/1000000$ chance of winning.

In this case, each ticket has a $999999/1000000$ chance of losing. If you buy $1000000$ tickets, the probability that all of them lose is $$\left(\frac{999999}{1000000}\right)^{1000000} \, \approx \, 0.367879$$

The probability that at least one of them wins is equal to $1$ minus the probability that all of them lose:

$$1 - \left(\frac{999999}{1000000}\right)^{1000000} \,\approx \,0.63212 \,\approx \,\boxed{\,63.2\%\,}\,$$