Class, 9/22

In class today we talked about the value of a human life. One student proposed that we could multiply the amount that a person would earn per year by the number of years worked. We decided that a kind of "average" amount would be appropriate, chose $50,000/year and 40 years to get a value of $2 million. There was some discussion about the costs of an individual to society, but I pointed out that people do pay taxes so maybe that's not something that enters into the equation. I also stated that although the exact number varies from government agency to agency, one such number I had recently read had the transportation department valuing a human life at around $5 million. In our philosophy of Fermi Problems, $2 million and $5 million are not that far removed, so our estimate wasn't bad.

We then asked whether air bags would be something that the government should require, given the cost of an air bag and the number of people that would be saved per year if air bags were required. This required estimating several numbers. One is the number of highway fatalities per year. Several numbers were raised, of the order of several hundred thousand per year. I knew that the actual number is smaller, about 50,000 per year. OK, that's only a factor of 10 (significant, but not horrible). Armed with this number and the population of the U.S., which everyone thought was about 300 million, we figured that the incidence of highway deaths is about 1 per 10,000 per year. We also estimated that maybe 1/3 (and the person who made this estimate thought it was high, which it is) of the deaths could be saved if air bags were installed in every car. The actual number of saved lives is estimated at about 10,000 per year.

So now we set up a very simple decision tree. On one branch, if air bags are installed, that saves 10,000 lives per year, each worth an estimated 5 million dollars, for a total savings of 50 billion dollars per year. On the other hand, one student said that to replace the air bags in a car costs about $700, but that's more than it costs to install them initially, which may be about $300. We also estimated that maybe 20 million cars go on the road each year. That adds up to an additional $6 billion per year. So, by investing $6 billion in air bags each year, we'll save lives estimated in value at $50 billion. This cost-benefit analysis says that the government should mandate air bags, as the net savings to society is only one-tenth of the cost of mandating them.

Class 9/19

Today you all took a "quiz" (two forms) which asked various questions for you to express your opinion on. The point of the "quiz" was to point out various ways in which our decisions may be affected by the language used to pose the problems, and other factors that seem on their face to be irrelevant. When the risks are posed in a negative way, the reaction we have and the decision we make may well be different from the same problem when the risks are posed in a positive way (lives saved versus lives lost, for example). We talked about the trolley problem and saw that, even though the decisions seemed to be the same when you just look at the outcomes (one person versus five people dead), the reluctance of most people to push the fat man off the bridge contrasts with their greater willingness to simply throw a switch, may have evolutionary reasons. I pointed out that when this question is posed to a person having their brain scanned in an MRI machine, different parts of the brain are active when the two different questions are posed.

Problem set comments

We talked about the problem set today. In particular we discussed the King and Brother problem and the ordinary dice problem.

In the King and Brother problem, the easiest way is to list all of the (assumed equally probable) ways that a couple can have two children:

BB p=1/4
BG p=1/4
GB p=1/4
GG p=1/4

The last of these didn't happen, so only the remaining three remain. Their probabilities are equal, so by inspection we see that in one of the three cases, the king has a brother, and in two of them, he has a sister. So, the probability that the king has a brother is 1/3. In symbols,

P(brother|king)=1/3

Again, looking at the list above, we can see that considering all kinds of families, the probability that there will be a king if the royal family has two children is 3/4, that is, there will be a king in all cases except GG, in which case the oldest girl will become Queen. We also see that the probability that there is one brother in addition to the king is P(brother,king)=1/4, since that only happens in the BB case. So, using the formula for conditional probability, we can calculate

P(brother|king)=P(brother,king)/P(king)=(1/4)/(3/4)=1/3, same as above.

We also got this using a tree, but I haven't figured out how to draw one and put it in the blog yet.

We found that in the case where there are three siblings, that

P(two brothers|king)=1/7
P(one brother|king)=3/7.

We got this both by listing all cases and using the formula.

We also did the ordinary dice problem by listing all of the possibilities in a square array, and putting the total in each entry:

| 1 2 3 4 5 6
------------------------------------
1 | 2 3 4 5 6 7
2 | 3 4 5 6 7 8
3 | 4 5 6 7 8* 9
4 | 5 6 7 8 9 10
5 | 6 7 8* 9 10 11
6 | 7 8 9 10 11 12

(Sorry, this isn't coming out formatted the way I expected, I apologize).

We noted that there are 36 entries in the table, and five of them total 8 (marked in red), so

P(total 8)=5/36

We discussed why (4,4) should not be repeated twice. By coloring the dice red and green, and tossing first the red one and then the green one, we get only one (4,4) amongst the examples that total to 8.

We see that there are 11 total cases that have a five showing (row 5 and column 5). Of these, two total 8 (starred). So,

P(total 8|5 shows)=2/11

Also, we just saw that there are 5 cases that total 8 (red), and of these two have a '5' showing (starred). So,

P(5 shows|total 8)=2/5

We also got the last two results by using the formula for conditional probability.

Class 9/15

Today we did several things. I told you about the problem in this week's "Cartalk" program on NPR. The problem involves two people who want to hire themselves out. Their employer insists that there be a 90% chance that they will show up for work (they have to drive), but one of them has a car that is only 70% reliable and the other's car is only 80% reliable. We were able to show that between them they would be able to get to work with better than 90% reliablility, using two different arguments.

(I still haven't figured out how to get pictures into here, so bear with me).

The first solution went this way: They first try to start the car that is 80% reliable. If it starts, they go in it. Otherwise, they try to start the car that is 70% reliable. That will happen on 20% of the days, so the probability that they will use the second car is 0.2*0.7=0.14. Add that to the 0.8 if they use the first car, and the result is 0.94, or a 94% probability that they will get to work.

Put in terms of conditional probability, we have

P(car 1 starts)=0.8,
P(car 1 doesn't start)=0.2,
P(car 2 starts|car 1 doesn't start)=0.7,
P(car 2 starts, car 1 doesn't start)=0.7*0.2=0.14 (by conditional probability formula)
P(they get to work)= 0.8+0.14

The other idea was even simpler. The probability that neither car will start is 0.2*0.3=0.06. Subtract that from 1 and you get 0.94, same as before.

Here we use the fact that the probability that car 2 will not start is independent of whether car 1 starts or not, so

P(car 2 doesn't)=P(car 2 doesn't|car 1 doesn't star2)
=0.3,
P(car 1 and car 2 both don't start)
=P(car 1 doesn't start)*P(car 2 doesn't start)
=0.2*0.3
=0.06

Both solutions are correct.

We used the opportunity to explain independence and dependence.

A and B are independent if P(A|B)=P(A). You will work on two problems involving independence for next Monday, including showing that several definitions of independence are entirely equivalent. We also wrote down a table of joint probabilities of A and B, identified the marginal probabilities that are obtained by summing across the rows and down the columns of the table. There's another problem for Monday that uses tables like this.

We spent the rest of the period discussing various flavors of the Monty Hall problem (mentioned below). We found just by our discussion that it always pays to switch in the Angelic Monty problem, it never pays in the Monty from Hell problem, and it doesn't matter in the Ignorant Monty problem. Finally, we analyzed the Mixture Monty problem by recognizing that there are six states of nature, 3 doors x 2 possible Montys. We set up a table, and used the fact that our evidence was that Monty opened door 2 and we saw a goat, to figure out the likelihood. There was some confusion, for example we found that

P(Open 2 & see goat|Door 3)= 1 for Angelic Monty and 0 for Monty from Hell

(because if you've chosen the wrong door, Angelic Monty will in this case show the only door that he can that has a goat, door #2, whereas Monty from Hell will open the door with the prize and will not open door #2).

We found that it is better to switch if we are facing Mixture Monty.

The easiest way to see this intuitively is that Angelic Monty will offer you a chance to switch twice as often as Monty from Hell will. That's because you'll choose the wrong door 2/3 of the time, so 2/3 of the time Angelic Monty will offer you a chance to switch, while Monty from Hell will offer you a switch in only that 1/3 of the cases where you have picked the right door. If you're offered a chance to switch, it's twice as likely that you're facing Angelic Monty than Monty from Hell.

Class 9/12

Today in class we mostly pursued the Monty Hall problem, in various disguises. In the process we learned about Bayes' theorem:

P(A,B)=P(A|B)P(B)=P(B|A)P(A),

and as long as P(B) is not zero, we can divide to get the usual form of Bayes' theorem:

P(A|B)=P(B|A)P(A)/P(B).

We identified P(A) as the prior probability of A (before we observed B), P(B|A) as the likelihood, P(B|A)P(A) as the joint probability of A and B, P(B) as the probability of observing the evidence B that we actually observed (that Monty opens the second door), and P(A|B) as the posterior probability of A being true, given that we've observed evidence B.

We also learned that the denominator, P(B), is gotten by summing P(A,B)=P(B|A)P(A) over all possible values of B (not A!).

We solved Monty Hall first with a probability tree, then with a "natural frequencies" approach like that in Calculated Risks, and finally with a spreadsheet-like calculation in which we had columns. We put the possible states of nature (here the different places the prize might be) in the first column, the prior probabilities (1/3) of each state of nature in the second, the likelihood (probability of the data that Monty opens door 2 given that you choose door 1 and the prize is behind the door for that row) in column 3, the product of columns 2 and 3 in column 4 (the joint probability), the sum of column 4 under that column, and then the posterior probability in column 5, which is the entry in each row column 4 divided by the sum under that column.

I haven't figured out yet how to produce such a table in the blog, when I do I'll add it.

Journals

I was really pleased with the journals. Two were short of three full pages, please watch this. But the journals this week were really thoughtful and had some good ideas. Keep it up!

HCOL 195 Class Members, Post Your Comments/Questions

OK, folks, the blog is started.

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Please note that entries prior to September 1 are for a graduate class I taught a year ago, and have no relevance to our class. You are welcome to read them and comment on them, but they are quite advanced!

Bill