# Lottery Probability

• Mar 18th 2008, 12:46 AM
reyup
Lottery Probability
I'm having a tough time with this problem and was hoping someone coule help.

It involves the Mega Millions lotter.

5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

I know the probability of winning with 1 ticket is
1/175711536

The question is, if exactly $200,000,000 is wagered (at$1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

Any help would be greatly appreciated.
• Mar 18th 2008, 05:09 AM
colby2152
Quote:

Originally Posted by reyup
I'm having a tough time with this problem and was hoping someone coule help.

It involves the Mega Millions lotter.

5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

I know the probability of winning with 1 ticket is
1/175711536

The question is, if exactly $200,000,000 is wagered (at$1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

Any help would be greatly appreciated.

BTW, I believe your probability of winning one ticket is wrong. It should be equal to the fraction of one over $\displaystyle 56*55*54*53*52*46 = 21,085,384,320$. Let that be equal to $\displaystyle p$.

Probability of two or fewer can be broken down as P(2), P(1) and P(0).

$\displaystyle P(2)= \frac{200,000,000!}{2!*(200,000,000-2)!}*p^2 * (1-p)^{200,000,000-2}$

$\displaystyle P(1) = \frac{200,000,000!}{1!*(200,000,000-1)!}*p \Rightarrow 200,000,000p*(1-p)^{200,000,000-1}$

$\displaystyle P(0) = \frac{200,000,000!}{0!*(200,000,000-0)!}*p \Rightarrow p(1-p)^{200,000,000-0}$

The answer to your question is the sum of those three probabilities.
• Mar 18th 2008, 06:51 AM
CaptainBlack
Quote:

Originally Posted by reyup
I'm having a tough time with this problem and was hoping someone coule help.

It involves the Mega Millions lotter.

5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

I know the probability of winning with 1 ticket is
1/175711536

The question is, if exactly $200,000,000 is wagered (at$1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

Any help would be greatly appreciated.

You are expected to use the Poisson approximation to the Binomial distribution for this problem.

RonL