Vsub Segment 3

From Computational Statistics Course Wiki
Jump to navigation Jump to search


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. Do you still have to make an assumption about Monty's preferences (where the slide assumed 1/2)?

Say I always pick door 1. Now, considering all possibilities of the prize and doors being opened (by Monty), we can get the following table:

Considering all probabilities
I pick
Door-1 Door-2 Door-3 Result
Prize Empty Empty When I switch I lose.
Empty Prize Empty I switch and win.
Empty Empty Prize I switch and win.

Thus, it is clear that if I switched doors, after one of the empty doors was opened by Monty, I would win in 2 out of 3 games. If I stayed with the same door, the only case I would win is if I actually picked the door with the prize on my very first turn, which could happen with probability only 1/3.

If Monty did have a preference on which door he would open, it would definitely change the probabilities of the game significantly. Say that, Monty always preferred to open door 2 if it was empty. Then, if Monty opened door 3 (to show that it is empty), then you would most certainly have to switch your choice to 2, since with probability better than 2/3 the prize is likely to be behind door 2.

Thinking about

1. Lawyers are supposed to be able to argue either side of a case. What is the best argument that you can make that switching doors can't possibly make any difference? In other words, how cleverly can you hide some wrong assumption?

The best argument which, according to me, could clearly confuse most humans into the wrong path here is to say that, after the empty door has been opened, you only have an even chance of picking the door with the prize and hence the door you picked first is as good as the one you would pick if you switched.

The main flaw or cleverness in this argument that hides the wrong assumption is that, we hand wave over the door that was opened. i.e. it is very easy to cover the fact that the door that was opened was not chosen purely at random -

a) it is infact chosen after you had made your choice, and

b) also it was chosen to always open an empty door.

2. We stated the problem as requiring the host to offer the contestant a chance to switch. But what if the host can offer that chance, or not, as he sees fit? Then, when offered the chance, should you still switch?

If the host offers the switch at random: Say he flips a coin (biased or fair) and offers you a switch, then it definitely makes sense to switch because you'll still be playing the same game. Where switching would give you a better chance of winning.

However, if you notice a pattern that indicates that the host offers the switch (based on some message from the producer) mostly when participants have picked the correct door in the first time; then you might be better of not choosing to switch. But this requires, you to look for patterns so that you can build a prior. This might depend heavily on watching some game plays to understand if there is a pattern behind when the host offers the switch.

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 use "B" to denote a boy and "G" to denote a girl. There are 2 ways we can consider this problem.

Argument-1: The sample space of 2 children can be given as: BB, BG, GB, GG (where all are equally likely). Since the Smiths have a girl, we can rule out BB. This leaves us with BG, GB, GG - all equally likely again. Hence, the probability that the other child is a girl is .

Argument-2 (incorrect): The order of the child is not important (i.e. Boy,Girl is not different from Girl,Boy - i.e. older and younger doesn't matter) We have 3 possibilities for the children. BB, BG, GG. Since we know that the Smiths have a girl, we can eliminate BB. Thus, the remaining 2 possibilities are BG and GG. Thus, probability is

The reason I think Argument-2 is incorrect is because we do need to distiguish between BG and GB. If this was framed as a coin toss problem (toss 2 coins, 1 is a head, what's the probability the other is a head?, I would have clearly gone with Argument-1. For some reason, making this into children seems to addle my brain :-/ into thinking BG is indistinguishable from GB.

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.

If they don't say anything and only show me the photo, then the probability of the other child being a girl goes up to .

(b) What is the probability that the hidden child is a girl?

If I can't make out anything from the picture, the probability would be .

Comment: If there are hints in the picture, I would go with a prior based on what I see from the picture. e.g. if I see the kid is wearing shorts has short cropped hair I would say with probability 0.75 it's a boy (or 0.25 it's a girl). Othewise it will be 0.5.

(c) If your answers to (a) and (b) are different, explain why there is a difference.

Yes, answers to (a) and (b) are different. In case (a) when the parents "say" they have one girl, we seem to distinguish between the case where the older child is the girl, the younger is a boy and vice-versa. Whereas when we see the picture, we don't distinguish between the 2 cases because, the ordering does not matter any more. The unknown kid has only 2 possibilities - B or G - both equally likely.