Class, 10/17

Today we spent most of the class discussing the lottery problem I left you with last time.

What we need to compute is the probability that no one wins the lottery, the probability that exactly one person wins the lottery, exactly two people, and so forth.

After some discussion we decided that the probability that no one wins is the probability that the first person loses AND that the second person loses AND....AND that the last person loses. Since these are independent events, we need to multiply the probabilities of each of these events (AND always means multiply the probabilities). If I write w=1/80,000,000, the probability that a given person wins the lottery, then the probability that that person loses is (1-w). The probability that everyon loses is (1-w)^N, where N=200,000,000 is the number of tickets sold. Although that looks terrible to compute, actually a hand calculator correctly computed this number to be 0.082.

To get the probability that exactly one person wins, we decided that it is equal to the probability that (the first person wins AND all the others lose) OR (the second person wins AND all the others lose) OR...OR the last person wins ane all the others lose). The AND means multiplication, and the OR means adding probabilities. To get the probability that one specified person wins AND all the others lose, this is equal to w*(1-w)^N-1 which is hardly different from w*(1-w)^N since the extra factor of (1-w) is very, very clost to 1. But there are N tickets, so the OR means we add this number to itself N times and the probability that exactly one person wins, one of the N tickets, is (Nw)*0.082 or 0.205.

For two people we follow the same principle: We compute probability that (the first person wins AND the probability that exactly one of the other people wins AND that all the other people lose) OR (the second person wins AND exactly one of the other people wins AND that all the others lose) OR...OR (the last person wins AND exactly one of the other people wins AND that all the others lose). We just computed the probability that one of the other people wins AND that all the others lose, it's 0.208. And the probability that a particular person wins is still w. And, there are N identical numbers that are OR'ed together, so we have to multiply by N again, getting (Nw)*0.208. However, there is a little complication, because if you look at the first two numbers above, in both of them there is a piece that comes from the first person winning AND the second person winning. And a similar thing can be said about any pair of items above. So what has happened is that each pair of people appears twice in the sum and is therefore counted twice as much as it should be. So the probability we want has to be divided by 2, and the answer we need is: The probability that exactly two people win is (Nw)*0.208/2=0.257.

In a similar way, the probability that exactly three people win can be computed as (Nw)*0.257/3=0.214; similarly to the case of two people, we note that each triple of tickets gets counted three times, so we have to divide by 3. And so forth for the case of exactly four, five, six and so on. Once you get to seven people, there's less than a 1% chance that that many people will win.

Now we can compute the expected value of a ticket. Your probability of winning is w. If no one else wins (probability 0.082) then you would win $280M. If exactly one other person wins (probability 0.205) then you would win $140M. And so forth. Adding it all up, the expected value of a ticket is $1.29. We calculated $1.12 by dividing $280M by 2.5=200M tickets/80M probability of winning.

But still, taxes and the fact that you can't take home the entire jackpot if you want it all at once means that it is not worth it (from an expected value point of view). The only reason to buy a ticket is for the fun of it.

I gave out some worksheets for estimating your utility function for money. You should work them out this weekend. We'll discuss them on Monday.