Segment 3: Daniel Shepard

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Problems

To Calculate

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)?

Solution:

Suppose that you choose door the th, which will be indicated as , where is an element of the set . Then Monty opens the th door revealing an undesirable prize, which will be indicated as , where is an element of the set such that . We would like to determine where is an element of the set . This problem results in the following relations:

    for all valid  and  (i.e., choosing a door does not effect the probabilities)
   
    for all valid , , and 
   
    for all valid , , and 

This gives the same result presented in the slides that

    for all valid , , and 

Note that the assumption about Monty's preference is still required. Although this preference does not need to be 50:50, a systematic way of choosing which door to open cannot be used and still preserve the problem. Say, for example, that he always chose the lowest numbered door that you did not choose and did not contain the prize. This would result in a 50:50 probability between the two remaining doors in some cases and a certainty in which door the car was behind in other cases. Suppose that under this scenario you chose door number 3. If Monty opened door 2, then the prize must be behind door 1. However, if Monty opened door 1, then it would be equiprobable that the prize was behind door 2 or door 3.


To Think 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?

By opening a door and revealing that the prize was not behind it, Monty simply removes that door from consideration. Now you are left to choose between the two remaining doors and, therefore, there is a 50:50 chance that the prize is behind either door. There is no benefit to switching doors because no matter what door you chose, there is always a door that Monty can open and reveal that the prize is not behind that door. This means that there are no clues given as to which of the remaining two doors you should prefer.


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? (Spoiler alert: see this New York Times interview with Monte Hall.)

If the probability that the host offers a chance to switch doors is independent of whether or not the prize is behind the door you chose, then you should still always switch doors because the probabilities derived in the problem stated above are preserved. If the host is more likely to give you the offer to switch if the door you chose is not correct, then this only improves your chances by switching and you should still switch. If the host is more likely to give you the offer to switch if the door you chose is correct, then the choice to switch depends on the exact likelihood ratios for the host giving you the offer to switch.


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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?

What makes this question interesting is that order doesn't matter. There is no child 1 and child 2 where you know either that child 1 is a girl or child 2 is a girl. You only know that one of the children is a girl. I will therefore disregard ordering and represent a girl by and a boy by . Without any knowledge of the children's genders, assuming that both genders are equiprobable, and that the gender of one child is independent of the other, the prior probabilities are

   

Consider that the knowledge that one of the children is a girl comes from mention of one of the children, which will be denoted as . Assume that there is equal probability that he mentioned either child regardless of order or gender. Applying Bayes rule results in

   


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?

This photo provides a view of one of the children that is a girl which is independent of the mention of one of the children being a girl. There is no reason to assume that this must be the same child that was mentioned before. I will denote this photo of the girl as . The addition of this new evidence can again be performed using Bayes rule

   


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

The answers for (a) and (b) are different because the addition of the photo in (b) provides additional evidence independent of the mention of a girl that was included in (a). This increases the likely-hood that both children are girls because a randomly chosen child will always be a girl if both children are girls, but only has a 50% chance of being a girl if one child is a girl and the other is a boy.