The first problem we discussed is the second half of the drug company’s decision.

Basic decisions are to continue research on marketing the drug, or to stop. Since the “sunk costs” of the research so far are the same regardless of what we do now, we can set the loss or gain at zero (remember, the gain/loss can always be added to by an arbitrary constant, or multiplied by an arbitrary scale). Also, since the goal is to make a gain, it’s probably best to frame this decision in terms of gain (utility) rather than loss. So, “do nothing” has a gain of zero.

If we were to decide to continue the research, from the data we already have there is a probability p that the drug is better than the old one, and (1-p) that it is not. That is a probability node (we could use as our p some higher criterion, such as that the new drug is twice as good as the old one.) To test the new drug will cost $30 M (probably low, by the way). If we test it, there is a possibility that the early testing may not pan out. The probability of cure rate may end up at q, which might even be less than the p that we got in the early tests. That has to be folded into the costs of marketing, bringing the drug to market, and the possible rewards of pricing the drug so that the expected number of doses sold will (over the 20-year lifetime of the drug) handsomely reward our company and our stockholders. That is illustrated in the above chart in a very sketchy way.

Our next problem was to consider the claim of an astrologer that he has powers that allow him to predict the future with 85% acccuracy. He makes 11 predictions, 4 of which are correct. What is the probability, given that he has some powers, that he can predict the future with 85% accuracy? That leads to our usual spreadsheet (where the division into 0.05, 0.15, …., 0.95 is for illustration and is adequate for the exam):

As usual, the likelihood is p^{4}(1-p)^{7} where p takes on the values that we put into our spreadsheet. We complete the spreadsheet in the usual way, and then to decide the probability that the astrologer has proven his case, given that he has the powers he claims, we add the probabilities for the states of nature ≥ 0.85 (that is, for this spreadsheet, 0.85 and 0.95).

But that doesn’t take into account our own experience. When we discussed this in class, people seemed to be skeptical about whether some people could actually make such predictions. To focus things, I asked whether people would think that I had abilities to predict things if, you given a fair coin and tossing it yourself out of my view, I were to predict that it would come up heads or tails correctly, no one would be impressed. If I did it ten times in a row, (one chance in 1000), some would pay attention. And if I did it 100 times in a row, (one chance in 30 million) most would think that something (maybe some sort of cheating) was going on. So that leads to a short spreadsheet like the one below, which puts a very small prior probability on the hypothesis that the astrologer really has the claimed powers.

We then considered the problem of polling people when there are controversial issues that some people might lie about when polled, as for their personal drug use, or their opinions on controversial subjects. Polls can be skewed when people lie.

So, for example, the poll is constructed so that the person being polled tosses a coin. If the coin comes up “heads”, then he is instructed to answer one way always (i.e., “yes, I used drugs last week.”) If it comes up “tails”, then he is instructed to tell the truth. The idea here is the the pollster has no idea whether the person being polled used drugs or not, so that protects the privacy of that person. But nonetheless, the pollster can back out the desired information about the group. Here’s what I wrote on the board:

The point is this: We need to know the probability that a person will say “yes,” given any probability p that the person has used drugs (or whatever the question is). That’s not hard to figure out.

P(“yes”|p)=P(“yes”,H|p)+P(“yes”,T|p)

=P(“yes”|H, p)P(H|p) + P(“yes”|T,p)P(T|p)

=1*(1/2)+p*(1/2)

=(1+p)/2

I didn’t do the calculation this carefully, but this is the result. For the probability of “something happens,” the probability of “says no”|p is (1-p)/2.

So, if 57% of the respondents answered “yes,” then the naive calculation is that

(1+p)/2=0.57, or p=0.14, the proportion of respondents that used drugs (or whatever the question is).

But we can do better. The quantities we’ve computed above are the likelihood function, and we can put them into a spreadsheeet:

The spreadsheet can have 10, 100, 1000, however many rows we need. The more we take, the more accurate the calculation. Or, we could, if we knew about calculus, do a fancy calculation that uses that skill. (This would not lead to better insight into this problem, only to greater accuracy.)

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