At the end of the program “What Were You Thinking,” Professor Debbie Bennett left listeners with a brain teaser, the answer supplied here.Dr. Bennett writes: “The problem involved two identical envelopes--one with two $1 bills in it and the other with a $1 bill and a $100 bill. The envelopes are mixed and you choose one to keep. But before you get to keep it, I randomly remove a bill from your envelope and show it to you. It is a $1 bill. I return it to your envelope and offer to switch envelopes with you. Do you keep your envelope or switch with me? In other words, is your envelope more likely, less likely, or equally-likely to contain the $100 bill?
ANSWER: You should switch. My envelope is twice as likely to contain the $100 bill. If your envelope has the two $1 bills, there is a 100% chance that when I remove a bill, it is a $1 bill. If your envelope has the $1 bill and the $100 bill, I have only a 50% chance of removing a $1 bill.
Since I did remove a $1 bill, it is more likely to have come from the envelope with the two $1 bills. Considered another way: Since you saw me remove a $1 bill, there are three possible scenarios:
1) You have the $1 and $100 and I removed the $1.
2) You have the two $1 bills and I removed the first one (the second remained hidden).
3) You have the two $1 bills and I removed the second one (the first remained hidden).
In 1 out of 3 equally-likely scenarios, you have the $100 bill and in 2 out of the 3 scenarios, I have the $100.”
Any questions? Send a $100 to “Are We Alone?” and we’ll answer it for you (or, leave a comment and Dr. Bennett will)!
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6 comments:
Professor Bennett is eloquent and clear in her explanation of the problem. Her colleagues must be in awe of her intelligence!
I think it is very uncouth for Prof. Debbie Bennett to anonymously comment on herself like that!
In all seriousness I thought it was a good interview, but I can't say that I know of anyone who has the kinds of problems with 'if, then' statements that she described. Seriously, do people get confused?
PS Why isn't there an Are We Alone? Twitter feed?
I've heard and read about the Monty Hall problem and its solution. Though I consider myself reasonably intelligent, this is the first time it ever really made sense. And it fact, it wasn't even Dr. Bennett's explanation, but Seth's restatement. Thank you much!
Hope this is correct.
Let Envelope A be the envelope with the 2 $1 bills, and Envelope B be the envelope with a $1 bill and the $100 bill. So, the conditional probabilities would be:
P( $100 | A ) = 0
P( $100 | B ) = 0.50
P( $1 | A ) = 1
P( $1 | B ) = 0.50
Therefore, by Bayes theorem,
P( A | $1 ) = [P( $1 | A )P( A )] / P( $1 ) = ( 1 * 0.5 ) / 0.75 = 0.667
P( B | $1 ) = [P( $1 | B )P( B )] / P( $1 ) = ( 0.5 * 0.5 ) / 0.75 = 0.333
Given that Professor Bennett drew a $1 from your envelope, you should switch, since there is a two-thirds chance that a $1 drawn from one of the two envelopes comes from Envelope A.
Man I hate these things.
Here's my logic for the monty hall problem.
If you have a door with a goat and you switch you have 100% chance of winning because Monty has revealed the other goat. If you already have the car and you switch you have a 100% chance of losing. Since 2/3 of the time you will have the goat to start with then switching gives you a 2/3 chance of winning while staying gives you 1/3.
QED - the correct answer.
Now applying similar logic to the Dr Bennett problem.
If you have the $100 bill and you switch you have a 0% chance of winning. If you do not have the $100 bill and you switch you have a 100% chance of winning. At the start you have a 50% chance of having the $100. So switching still gives you a 50% chance of winning. ie it makes no difference.
When $1 is removed from your envelop the choices for you are not reduced so it does not change the probabilities
Why is this wrong?
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