STAT 330 September 4, 2012

Today we looked at some more examples…a more complex version of Bertrand’s Box, the famous Monty Hall problem (see the video that Anna found), examples from my experience with the Hubble Telescope, coin tossing examples, and other examples that illustrated conditionalization and marginalization. We finished with the definition of independence.

I did not mention several other Monty Hall variations. We discussed the standard one, where it pays to switch, and I mentioned a version where Monty randomly opens another door (I call this one “Ignorant Monty” since he doesn’t know where the prize is). There’s also “Angelic Monty”, where Monty only opens a door if you have chosen the wrong door, and shows you that you are a winner if you chose the right door, and “Monty from Hell”, where Monty only opens a door and asks if you want to switch if you have chosen the right door, and opens the door with the prize to show you that you’ve lost if you chose the wrong door. There’s also “Mixture Monty,” where Monty flips a fair coin in advance, and if it is heads, behaves like “Angelic Monty”, and if it is tails, like “Monty from Hell”. Think about these variations. In which ones does it pay to switch if you choose Door #1 and he shows you a goat behind Door #2? In which ones does it not pay?

Advertisements

2 Responses to “STAT 330 September 4, 2012”

  1. Cathy Says:

    Hi everyone,
    I was looking ahead at slide 47 “Independence” and the claim that if P(A|B and H) does not equal P(A|H) then P(B|A and H) does not equal P(B|H). I’m thinking that we may need to stipulate that P(B|H) is not equal to zero.

    For example, if we do not stipulate that P(B|H) be nonzero, then I think we could have P(B|H)=0, P(B|A and H)=0, P(A|H)=1/3 and P(A|B and H)=1/2 as a counterexample. Notice that for these values, P(A|B and H) does not P(A|H), but this does not imply that inequality of P(B|H) and P(B|A and H) (because these last two quantities are both zero).

    I ran into a similar issue on slide 50.

    I was just wondering what others thought about this.:)

    • bayesrules Says:

      Yes, you are right. I was going to point this out in class, but it is absolutely correct that any possibility of dividing by zero, or (as you say) multiplying something by zero and then equating both sides without that multiplier, will lead to nonsense.

      The important point to remember is that in these calculations, we do not assume that B is false (so that P(B|H)=0). Rather, we do not know whether B is false at the outset, so that there is some possibility that B is true, so P(B|H)≠0), and so this problem doesn’t arise since we are not dividing by zero.

      Thank you, Cathy, for sharpening this example. This is an important point.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: