University of Minnesota, Twin Cities School of Statistics Charlie Geyer Home Page John Corbett Home Page

Updated Feb 14, 2006.

The *Poisson Distribution* arises in a number of contexts
as the distribution of a random number of points, for example
the number of clicks of a Geiger counter in one second, the number
of raisins in a box of raisin bran, the number of blades of grass
in a randomly chosen square inch of lawn, and so forth.

The formula for the probability of observing `k` of whatever is being
counted when the expected number is `m` is

where `e` is the base of the natural logarithms
and `k` ! indicates the
factorial function (use the `e ^{x}` key on a scientific
calculator to calculate

Theoretically any count between zero and infinity (including zero) is possible, but the probability of large counts falls off very rapidly.

In a lottery, the number of winners cannot have an exact Poisson distribution for two reasons.

- The number of winners cannot be more than the number of tickets sold, whereas the Poisson distribution gives nonzero probability to arbitrarily large numbers of winners.
- The choice of lottery numbers by the players is not completely random. If you choose a popular number, you will have to share with many other winners if you win. Conversely, if you can figure out a number no one else likes and play that, you are guaranteed not to have to share the jackpot if you win. The Poisson distribution assumes every player chooses lottery numbers completely at random.

The first issue is not a serious problem. The Poisson distribution
would be an extremely good approximation if it were not for the other
issue. The second is more serious. Many players (about 70%)
buy quick picks

which
are completely random, but other players choose some number they think is
lucky and that's not random. If every player choose a quick pick

the
Poisson distribution would be an almost perfect approximation. Since
they don't, it is not quite right. However, we will assume the Poisson
distribution is correct to keep things simple.

The reason why the *unconditional* distribution of the number
of winners of the jackpot and the *conditional* distribution of
the number of *other* winners *given* you win are the same
has to do with the assumption of completely random choices of numbers by
all the players, which is required for the correctness of the Poisson
distribution. Then whether you you win or not doesn't change the
probability of anyone else winning. Everyone has the same 1 in 146.1 million
chance of winning, and their ticket choice had nothing to do with yours.

If you win and there are `k` other winners, then the jackpot gets
split `k` + 1 ways, and the amount you win
is `J` ⁄ (`k` + 1), where
`J` is the size of the jackpot.

Your expected winnings are calculated just like any other expectation:
multiply the amount you win in each case,
which is `J` ⁄ (`k` + 1),
by the probability of that case,
which is `m`^{k}
`e`^{&minus m} ⁄ `k` !,
and sum. The sum runs over `k` from zero to infinity, so it
appears to require calculus to sum this infinite series.

Fortunately, there is a trick that allows us to see what the expectation is without doing the infinite sum. The terms in the infinite sum are

Let `W` denote the sum of the `a _{k}`
as

If we multiply each term by `m` we get

where `p`(`k`) is the Poisson probability
defined above.
The probabilities `p`(`k`) must
sum to one as `k` goes from
zero to infinity by the properties of probability. Because of the
`k` + 1 above, the first term is `J` `p`(1).
If we were to add an additional term `J` `p`(0),
the series would
sum to `J` (because the probabilities sum to one).
Thus the series sums to

(Recall that we multiplied by `m` so
the sum is `m W` rather than
`W`). Solving for `W` gives