Odds on the Number of PowerBall Lottery Winners

There is a simple formula for the average number of lottery winners, and for the probability of various number of winners or the distribution of winners.  It of course depends on the ratio of the number of tickets bought N to the number of possible number outcomes, which is the same as the product of the the numbers of tickets bought time the probability of winning p.  This ratio or product we call <n> = N p.  <n> is called the expected number of winners, even though in general it is not an integer.  For the current Powerball jackpot, p = 1/292,201,338.  If N = 292 million Powerball tickets are bought, then the expected number of winners is <n> = N p = 1.  If twice 292 million tickets is bought, then <n> = 2 X 292 million / 292 million =2.

The formula is the Poisson distribution giving the probability P (n) for a given number of winners n, and integer, when the expected number of winners is <n>.

P (n) = exp(-<n>)  <n>^n / n!

The n! is called n factorial, and is the product:

n! = n (n-1) (n-2) … 1.

For example,

0! = 1, 1! = 1, 2! = 2 x 1 = 2, 3! = 3 x 2 x 1 = 6, 4! = 4 x 3 x 2 x 1 = 24, 5! = 5 x 4 x 3 x 2 x 1 = 120, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.

<n>^n means the expected number <n> is raised to the power n, I.e. <n> is multiplied by itself n times.

The exp(-<n>) means the exponential number e = 2.718… Is raised to the power -<n>, which is the same as the reciprocal (1/e) being raised to the power <n>.

Table of the distribution of P(n) for various number of winners n, for a given <n>.

<n> = N p is the column label.

n is the row label.

n <n> =0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
0 0.607 0.368 0.223 0.135 0.0821 0.0498 0.0302 0.0183
1 0.306 0.368 0.335 0.270 0.205 0.149 0.106 0.0732
2 0.0759 0.184 0.251 0.270 0.257 0.224 0.185 0.146
3 0.0126 0.0613 0.125 0.180 0.214 0.224 0.216 0.195
4 0.00158 0.0153 0.0470 0.0900 0.134 0.168 0.189 0.195
5 0.00016 0.00307 0.0141 0.0360 0.0668 0.101 0.132 0.156
6 0.00001 0.00051 0.00352 0.0120 0.0278 0.0504 0.0771 0.104
7 0.00007 0.00076 0.00343 0.00994 0.0216 0.0386 0.0595

Reading the table, for <n> = N p = 4.0, the greater probability for the number of winners is 3 or 4, with 5, 2, and 6 next.

For <n> = N p = 2.0, the greater probability is for 1 or 2 winners, with 3 or 0 winners next.

For <n> = N p = 1.0, the greater probability is for 0 or 1 winners, with 2 winners next.

For <n> = N p = 0.5, the greater probability is for 0, and then for 1.  Those two numbers take up 91% of the winning possibilities.  There is only a 9% probability of 2 or more winners.

For the recent $1.6 billion PowerBall lottery, there were 3 winners. For that lottery, there were N = 635,103,137 tickets sold, for a sales of $1,270,206,274.

<n> = N p = 635 million / 292 million = 2.1735.  Approximating from the table for <n> =2, the most likely number of winners for <n> = 2 would have been 1 or 2 at 27% probability each , 3 winners at 18% probability is still sizable.

Here we give the exact result from the <n> = 2.1735 giant PowerBall drawing:

n = 0:  0.1138, n = 1:  0.247, n = 2:  0.269, n = 3:  0.195, n = 4:  0.106, n = 5:  0.0460, n = 6:  0.0167, n = 7:  0.0052.

So the exact probability for 3 winners for the <n> = 2.1735 PowerBall was 19.5%.

While the $1,270 million sales would seem to not cover the jackpot and other prizes, the amount invested in previous weeks should be added to the total from which the prize was given.

For <n> = N p = 5, the table values would be: n = 0:  0.00674, n = 1:  0.0337, n = 2:  0.0843, n = 3:  0.140, n = 4:  0.176, n = 5:  0.176, n = 6:  0.146, n = 7:  0.104, n = 8:  0.0653, n = 9:  0.0363, n = 10: 0.0181.

The <n> = N p = 5 would favor 4 or 5 winners, and next 6, 3 or 7 winners.

About Dennis SILVERMAN

I am a retired Professor of Physics and Astronomy at U C Irvine. For two decades I have been active in learning about energy and the environment, and in reporting on those topics for a decade. For the last four years I have added science policy. Lately, I have been reporting on the Covid-19 pandemic of our times.
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