(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)Both solutions are correct.
=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
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.
2 comments:
http://edge.org/3rd_culture/taleb08/taleb08_index.html
I came across this article, it's an interesting take on the risk of certainty and makes a case for setting a limit on the use of statistical probablility
This is an interesting article, and well worth reading.
Taleb has for a long time warned against failing to take into account extremely rare or unexpected events (the subject of his book, "The Black Swan").
It is not so much an indictment of the use of statistical probability as a tool for making decisions, but rather a warning that, as a famous statistician, George Box made, "All models are wrong, but some models are useful."
So, a wise person using the tools we are learning about will always know and understand that the models we are using are just models, that is, they are our way of trying to capture the way the world is working (like assuming that the probability of a child being a boy is equal to its being a girl, or the probability of the second child's gender being independent of the gender of the first child). But these models are probably both wrong. The question is, can you make good decisions, even if your models are wrong?
Sometimes you can, as Taleb points out. He points out the "Mediocristan" decisions (p. 8 of the article), and this is the sort of thing we will be discussing in this course, most of the time.
Sometimes the model you are using (and remember, it is wrong, if George Box is right), is so badly wrong that the decisions made are also badly wrong (in Taleb's article, "Extremistan").
So this is a useful article in pointing to the limits of applied decision theory. They are:
1) All models are wrong,
2) Some models are useful,
3) Models can be wrong in ways that will not badly affect the decisions. This is the usual situation.
4) Models can be wrong in ways that make decisions made with them very bad. These situations are rare, but when they happen, the outcomes can be very, very bad. The problem is that the rarity of the situations that give rise to these bad outcomes is so great that we have no good experience to include them in our models. It's like large asteroids hitting the Earth and wiping out a large fraction of the Earth's biota, as seems to have happened to the dinosaurs some 40+ million years ago (and several times earlier).
There's a lot more to the article, in particular Taleb's assertion (which I agree with) that increasing the complexity of things may appear to make situations more stable but increases the risk that extreme events can make the whole system much more unstable, as the current financial turmoil with subprime mortgages, Fannie Mae and Freddie Mac, and now the failure of the 150-year-old Lehman Brothers and the takeover of Merrill Lynch and AIG demonstrate.
Stay tuned!
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