Class, 9/29

Today we discussed further the polling example and actually calculated a simple result. I'll try to post a copy of the calculation later. We found that with 6 favoring candidate A out of a sample of 10, the posterior probability that candidate A wins (has over 50% of the vote) is about 71%. The posterior distribution is closely bell-shaped and peaked at r=0.6=6/10. That's a general rule.

We also asked what would happen if we used a more realistic prior that put more weight near 1/2. In response to a question, I pointed out that you can't put it near 0.6 because that would be "cheating," using the same data twice. You have to do it without looking at the data. We found that a prior that rises to a maximum at 0.5 and then falls again will do two things: It will narrow the posterior distribution somewhat, and will also shift the peak closer to 0.5. If there is a whole lot of data, then the effect of the prior will be negligible, but in our example it can be significant.

The following webpage has election predictions with a chart (lower right hand) that shows a similar posterior distribution, based on calculating the electoral vote in a simulation (this is a modern computational technique, even more powerful than the spreadsheet method we discussed). The left-hand chart has other information that summarizes the posterior probability in several ways: Where the maximum of the posterior probability is, what the win probability is, and so forth.

We discussed a situation where two people enter into a consecutive duel: Alan and Bob will take shots at each other in turn. Alan's probability of hitting Bob and putting him out of commission on one shot is 1/3; Bob's probability of putting Alan out of commission is 2/3. We asked, if they keep taking turns until one hits the other, what's the probability of Alan eventually hitting Bob if he goes first? If he goes second?

Although this could be calculated (as one student suggested) by multiplying and adding probabilities until the numbers were very small, I suggested another way.

If Alan goes first, he'll win outright on his first shot 1/3 of the time. He'll also eventually win with probability (2/3)*P(Alan wins eventually | Bob goes first). So

P(Alan wins eventually | Alan goes first)=1/3+(2/3)*P(Alan wins eventually | Bob goes first).

But P(Alan wins eventually | Bob goes first) can be calculated in terms of P(Alan wins eventually | Alan goes first), because the only way that Alan can win eventually if Bob goes first is if Bob misses on his first try (P=1/3). Then Alan will have a second chance, and the probability that he'll eventually win if Bob goes first is therefore

P(Alan wins eventually | Bob goes first)=(1/3)*P(Alan wins eventually | Alan goes first)

Therefore, P(Alan wins eventually | Alan goes first)=1/3+(1/3)*(2/3)*P(Alan wins eventually | Alan goes first). This can be solved for P(Alan wins eventually | Alan goes first). This turns to be 3/7, and P(Alan wins eventually | Bob goes first)=(1/3)*(3/7)=1/7.

Finally, I brought in Charlie, who is a crack shot and never misses. We decided that if Charlie goes first, he'll knock off Bob since Alan is a poorer shot, and if Alan misses, Charlie will knock him off the second time around. Also if Bob went first to be followed by Charlie, Bob will try to knock off Charlie first since if he knocked of Alan, he's a goner. But if Alan goes first, the first guess that he'd go after Charlie seemed to be wrong, as one student pointed out. Actually, Alan has a better chance of survival if he shoots his gun into the air! More on this later.