# Probability

This is a lottery odds question. It involves the Mega Millions, where 5 numbers are selected from the intergers 1-56 (without replacement) then a special 6th ball is selected from a new set of intergers 1-46. The odds of winning are 1/175711536. If exactly 200,000,000 random numbers are entered, what is the probability of 2 or fewer winners. I know how to find the prob of 0 winners, but I don't know how to find exactly 1 or exactly 2. Any help would be appreciated.

asked by Rey
1. Let P be the probability of winning = (1/175...). Let n be the number of tickets = 200M. Let x be the exact number of winners.

First, calculate z, the number of ways you could have exactly x winners. It is n-choose-x; the formula is:
z=n!/(n-x)!x!
(where ! means factorial n! 1*2*3...*n. Also, zero!=1)
So if x=0 then z=1. if x=1 then z=n, if x=2 then z=n*(n-1)/2

Probability of exactly x winners is:
z * P^x * (1-P)^(n-x)

posted by economyst
2. I will let a = 1/175,711,536= 5.591145*10^-9 , to save typing.
The expected number, among 200,000,000, is L = 2*10^8 a = 1.13823

The probability of P(k) of k winners is given by Poisson statistics:
P(k) = L^k*e^-L/k!

Probability of 0 winners = 0.32039
Probability of 1 winner = 0.36467
probability of 2 winners = 0.20754
Probability of 3 winners = 0.07874
Probability of 4 winners = 0.02241
Probability of 2 or fewer winners = 89.26%

posted by drwls
3. The Poisson formula I used is different from Economyst's formula, which may be the more exact. The numbers I calculate with Poisson differ by less that 1% in a relative sense. The Poisson formula is easier to use. Try them both and see.

posted by drwls
4. Thanks drwls, and excellent suggestion to use the Poisson. Just for grins, I checked your results with my methodology; they are the same in all cases (at least up to the 5th significant digit)

posted by economyst

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