Todd - Segment 3
1. The slides used a symmetry argument ("relabeling") to simplify the calculation. Redo the calculation without any such relabeling. Assume that the doors have big numbers "1", "2", and "3" nailed onto them, and consider all possibilities.
There are 6 possibilities, because the contestant can pick one of 3 doors, and the host can open one of 2 remaining doors. These calculations show that relabeling is just a convenience, but does not change the fact that the contestant should always switch.
Contestant picks door 1; host opens door 2.
Contestant picks door 1; host opens door 3.
Contestant picks door 2; host opens door 1.
Contestant picks door 2; host opens door 3.
Contestant picks door 3; host opens door 1.
Contestant picks door 3; host opens door 2.
Do you still have to make an assumption about Monty's preferences (where the slide assumed 1/2)?
Monty could choose a door (any door that does not contain the car) with probability less than 1 and the contestant should still switch. If Monty always selects the same door (probability 1), then the posterior probabilities are equal between switching or not switching.
WLOG the contestant picks door 2 and Monty opens door 3:
Assume Monty's preference is less than 1 (say 0.99). Then the contestant should switch to door 1.
However, if Monty always opens door 3 when door 2 is selected and contains the car, then the probabilities are equal between the two remaining doors:
To Think About
3. Mr. and Mrs. Smith tell you that they have two children, one of whom is a girl. (a) What is the probability that the other child is a girl?
We know that there are 4 possible outcomes: .
However, we know that BB is not possible, because the Smiths said (at least) one of their children is a girl. This leaves 3 possible outcomes, and only 1 of which would mean that both children are girls. So, the probability of the other child also being a girl is .
Mr. Smith then shows you a photo of his children on his iPhone. One is clearly a girl, but the other one's face is hidden behind the family dog, and you can't tell their gender. (b) What is the probability that the hidden child is a girl?
Assume each child has an ordering, and WLOG the Smiths say that Child 1 is a girl. Then we are looking for the probability that Child 2 is also a girl. Because Child 1 is a girl, there are two possible outcomes: . Only GG has the result that Child 2 is also a girl, so the probability is .
(c) If your answers to (a) and (b) are different, explain why there is a difference.
In part (a), there is no ordering of the children, but in (b) we can assign an ordering, which changes the event space.
We can also think of it like this: Suppose the parent randomly choose a child to talk about, and that child is a girl. Then when we see the photo and the only child clearly shown is a girl, the chances of both children being girls becomes higher (assuming the probability of the clearly visible child and the choice of who to talk about are independent).
Having both pieces of data (hearing the parent mention one of the child's gender and seeing one of the children in a photo) gives more evidence to support the number of girls the family has.