The Poisson Variation of the Binomial

My recent post about Why the Other Line Moves Faster reminded me of another variation of the binomial distribution that is not covered in AP Statistics, the Poisson Distributions. (Ok, the Poisson is lots more than just a special case of the binomial, more about that later, but in the days before hand-held calculators it was in that sense that we met it.)

The distribution is named for the French Mathematician Simeon Denis Poisson (hey, his name is on the Eiffel Tower.. 2nd from the right on the South East side).




Suppose you had a very rare binomial event, let's say something that happened only one in 200 trials (p=.005) and we wanted to know how probable it was to happen four times (x=4) in three-hundred trials (n=300). The calculation of the binomial probability is now as easy to do with modern calculators as with the Poisson approximation to it, but in my youth a calculation of looked nearly impossible.

The Poisson frequently uses the Greek letter lambda, , for the mean or expected value np. In this case NP=300(.005) = 3/2 indicates that, on average, we would expect only 1.5 successes in 300 trials. We want to calculate the probability of getting four successes. The Poisson probability is given by . The difference between the two calculations is less than .0002 on my Ti-84 calculator.

The Poisson is not limited to binomial events. More often it is applied to events which are distributed randomly across time. The same calculation above could be used to calculate the probability of four people entering a bank in a 15 minute period when only and average of 1.5 visitors would be expected. If a manufacturing process produces an average of 1.5 failures per day and you wanted to calculate the probability of four failures, you use the same calculation again.

And if you wanted to know how many cashiers to keep open at the market, how many mutations in a string of DNA exposed to radiation, the number of deaths from a rare side effect of a drug, or other similar situations which involve a very rare event with a very large number of possible incidences, the Poisson will often suffice as long as conditions of independence are met.

The Poisson distribution is sometimes called the Law of Small Numbers after a book by the same name by Ladislaus Bortkiewicz published in 1898 on the Poisson distribution.

For AP Statistics Students, it is often possible to approximate the Poisson with the Normal Distribution as long as the value of lambda is about ten or more. For the Poisson, the mean and variance are both the same, so if you approximate it with a normal, use a mean of lambda, and a standard deviation that is the square root of lambda. Do remember to do the continuity correction for the fact that the Poisson is discrete and the Normal is continuous.

A Different Look at an Old Puzzle

Reading through a paper on Public Key Cryptography (yeah, I know how to live), by Shai Simonson of Stonehill College, I was reintroduced to an old problem from my youth. "In the 1995 movie Die Hard: With a Vengeance (aka Die Hard III), Bruce Willis and Samuel L.Jackson play a bomber's deadly game as they race around New York trying to prevent explosions.In order to stop one explosion, they need to solve the following puzzle:

Provided with an unlimited water supply, a 5-gallon jug, and a 3-gallon jug, measure outprecisely 4 gallons, by filling and emptying the jugs." If you have never tried this problem, you might pull out the pencil and paper and have a go before you read on.

In a graph theory variant of this problem, the unlimited supply is often replaced with an eight gallon jug, but the problems are equivalent. I had just hours before read the same problem in reviewing Nets, Puzzles, and Postmen by Peter M. Higgins.

Simonson goes on to describe a little history of the problem, "Of course this puzzle was not invented for this movie. Most references attribute the puzzle and its variations to Tartaglia [Tweedie, M.C.K., A Graphical method of Solving Tartaglian Measuring Puzzles, Mathematical Gazette, 23, (1939) 278-282.], but the earliest known version of the problem occurs in the "Annales Stadenses" compiled by Abbot Albert of the convent of the Blessed Virgin Mary in Stade. Stade is a small city on the west side of the Elbe estuary a bit downriver from Hamburg. The date of compilation is uncertain, but seems to be 1240."

"The puzzle is also discussed in detail at the wonderful educational mathematics site cut the knot created by Alex Bogolmony ".

Knowing you would be interested, I copied a short clip from Bogolmony's site where he retells the story of how, “Siméon Denis Poisson (1781-1840) was a famous French mathematician of the last century. He worked in celestial mechanics, probability, calculus, electricity and magnetism. He is known to be designated for a medical carreer by his family. Edward Kasner and James Newman Mathematics and the Imagination refer to Poisson's biographer, Arago:
“Siméon Poisson's family tried to make him everything from a surgeon to a lawyer, the last on the theory that he was fit for nothing better. One or two of these professions he tackled with singular ineptitude, but at last he found his métier. It was on a journey that someone posed to him a problem similar to the one below (above on this page - AB). Solving it immediately, he realized his true calling and thereafter devoted himself to mathematics, becoming one of the greatest mathematicians of the nineteenth century." ahhhh now you wonder how many young people discovered their love for mathematics in a darkened theater as they shouted out help to Willis and Jackson, "No, Fill the three and dump it into the five!".

The graphic NET for the problem is in the photo at the top. Imagine the points x,y as giving the amount in each of the two jugs. The x-coordinate shows how much is in the five-gallon jug and the y-coordinate shows how much is in the three-gallon jug. Each of the paths on the graph shows a possible move. For instance, if we start at (0,0) nothing in the two jugs, there are two paths we may follow, rigth out the x-axis to the point (5,0), or vertically up the y-axis to (0,3). The object is to end up at (4,0) or (4,1) or (4,2) or (4,3) since each of these will have a jug containing four gallons. To solve it, just move along the lines from point to point and if you get stuck, go back a point and take a different path. In the language of the graph-theory people, this is called a breadth first search (I think).

Have fun, but keep in mind you may be learning important math along the way. Simonson points out the reason for his including the puzzle in his work, "This puzzle is useful for studying public-key cryptography, because the solution embodies the two major related number-theoretic results: Euclid’s algorithm [I just realized I have never written up anything on this, perhaps because so much has been done so well in other places, like this page ] and Fermat’s Little Theorem.

I found a post by David Singmaster, one of the foremost experts on Recreational Mathematics, who provides some additional history of the problem, and an incentive for young mathematicians (see the bold below)...
" Measuring With Jugs.... I have reorganised the notation of these problems. Tweedie, MG 23 (1939) 278-282, is the source of the triangular graphical method of solution. Halving 8 using 5 and 3 is in Abbot Albert which seems to be the earliest version. Pacioli, De Viribus,
seems to be the first to use any other values, e.g. halving 12 using 7 and 5. Tartaglia seems to be the first to divide in thirds, e.g. divide 24 in thirds using 5, 11, 13. The general problem of what can be obtained from a, b, c with c full to start seems to be first treated by A. Labosne is his 5th ed. of Bachet's Problemes and appears to still be unsolved when c < a+b." So have a go children!!!