## HCOL 196, February 16, 2011

We finished the lottery decision tree. The analysis showed that the expected value of a ticket is \$1.28, which after the cost of the ticket is \$0.28; Seems like a positive return, until we recognize that if you want the money right away, you’ll only get about half, and furthermore, Uncle Sam and Peter Shumlin are going to want their share. That makes the expected take negative. Lotteries are not a good way to get rich, and not a good retirement plan.

We had to calculate the probability of 2, 3, 4, … other winners. We know that the probability of an individual winning is 1/80M so the probability that no other people win is (1-p)N=0.0821, where N=200M, the number of tickets. The probability that any particular individual wins and no others win is p(1-p)N-1, but the N versus N-1 doesn’t make a significant difference, so it is 0.0821p. But there are N possibilities, so that makes the probability of one other winner 0.0821(Np)=0.205. That recapitulates the situation for nâ‰¤1, where n is the number of other winners.

Then we tackled the case n=2. One guess was that it should be Npn(1-p)N-n, but this is much too small. It’s the number for one other out of N plus one particular second person. We revised that to Nnpn(1-p)N-n, but this isn’t quite right either because it double-counts people. Another student suggested using the “choose” function, but not everyone was familiar with it. It arises when you expand a sum, like (a+b), to some power N. The calculation is shown on the chart below.

This turned out to be the right way to do it. The “choose” function is written choose(N,n) and it is calculated as

N(N-1)…(N-n+1)/n!

Choose(N,n) is the number of different ways that you can choose n objects (people here) out of a collection of N objects. So if pn(1-p)N-n is the probability of exactly n particular people winning, then choose(N,n)pn(1-p)N-n is the probability of any n winners.

Approximately, since N minus a small integer is for all practical purposes equal to N, just as (1-p)N-n is for all practical purposes equal to (1-p)N for small n. And choose(N,n) is approximately equal to Nn/n! for small n. So, the probability that we want is closely approximated by (Np)n(1-p)N/n!, as illustrated in the following whiteboard shot:

I noted that we can also approximate (1-p)N by the same method that a calculator uses: take the log (natural logarithm), use the fact that log(1-p) is approximately -p if p is small, and then exponentiate the result. Then (1-p)N is approximately , e-(Np) as shown in the final whiteboard shot:

The Poisson distribution is a very important one in statistics. It is used to model random events that happen at an average rate. In our example, the random event is n other people winning a particular draw of the lottery. But it can be used to model the number of telephone calls to be placed at a particular time (enabling us to size our cell phone tower, for example, to handle the expected traffic), or the number of patients that arrive per hour at a hospital, or the number of stars in an area of the sky, or the number of decays in a radioactive sample per second, and similar types of situations.