HCOL 196, February 14, 2011

Today we agreed to have the first quiz on Monday, February 28. I will give you a study list and we will go over it next week instead of a problem set. There will be no problem set this week. A Journal is due this Friday, but there will be no Journal due on Friday, Feb. 25

We introduced the idea of a decision tree by expanding probability trees with a new kind of node, a decision node, indicated by a square instead of a circle. Unlike a probability node, where we have no control over which branch is realized, the choice of node at a decision node is entirely up to us. Otherwise the decision tree looks very similar. We considered the problem of bringing an umbrella. If it rains and we have the umbrella, or if it doesn’t rain and we don’t have it, all is OK, and we can indicate this by putting a loss of 0 at the end of those branches. But if it rains and we don’t have the umbrella, we’ll get wet, which most people don’t like. We assigned a loss of 2 to that branch. If it doesn’t rain and we’ve been carrying the umbrella around, we might look like a dork, but that may not be as bad, and we assigned a loss of 1 to that branch.

Then we evaluated the tree. The rules are: Multiply the loss at any point by the probability of the branch it is on, and add the products to get the expected loss of the (probability) node that the branches are emanating from. Then, if there are additional probability nodes, do the same thing as you go back towards the root of the tree. If you have a decision node, on the other hand, you should choose the branch that has the least expected loss, and cut off the other branches. We decided that the probability of rain was 0.3 (get that from the web or TV). The evaluated tree is below; we chose not to bring our umbrella. Had the loss for getting wet been 3 instead of 2, we would have brought the umbrella.

Someone asked if we could do the same thing with gains instead of losses. Yes, you can. We call the gains “utilities”, and if we were using gains we would choose branches so as to maximize the expected utility.

You can’t mix loss and utilities. Pick one or the other, whichever is best suited to the problem.

Then we looked at the problem of the lottery. A particular example took place some years ago. The jackpot was $280M dollars (‘M’ means “million”), a ticket cost $1, the probability of winning was 1/80M since there were 80M distinct tickets, and 200M tickets had been sold. Should we buy a ticket?

We started a tree. Again, a decision node sits at the root of the tree (buy a ticket or don’t buy it). If we don’t buy, the gain is 0 (it’s better to use gains than losses here). But we recognized that in the case of buying the ticket, there is a possibility of more than one winning ticket. In fact, we might expect 200M/80M winning tickets, or 2.5 on average. However, we might get lucky and have the only ticket, or our luck could be very bad and there might be 5 or 6 or even more winners. We’ll have to divide the jackpot with an unknown number of others if we win. So our tree has an additional probability node, for the possible other winners, labelled 0, 1, 2, …, and gains (utilities) of $280M, $280M/2, $280M/3, and so on.

So we need to compute the probability of no additonal winners, one, two, and so forth. For no additional winners, with p=1/80M, N=200M, the probability of a particular ticket not winning is (1-p), and since the tickets are independent, the probability that they all lose is (1-p)N. We calculated this number to be 0.0821 on a calculator.

Similarly, the probability that a particular ticket wins all all the others lose is p(1-p)N-1. However, there are N tickets, and any of them could be a winner, so the probability of exactly one additional winner is Np(1-p)N-1. For all practical purposes, (1-p)N=(1-p)N-1, because (1-p) is so close to 1 that one fewer factor (or even quite a number of fewer factors) won’t make a difference, at the level of accuracy that we are computing. So the probability of one additional winner is 2.5×0.0821=0.2052.

I asked you to think about the other probabilities we need: exactly 2 other winners, 3 other winners, and so forth, to discuss in class on Wednesday.

See you on Wednesday!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: