Say I have a wallet that contains either a $2 bill or a $20 dollar bill (with equal likelihood) but i dont know which one. I add a 2 dollar bill to my wallet and later without looking reach in and pull out a bill.
its a 2 dollar bill there is one bill remaining in my wallet. what are the chances that its a $2 dollar bill.
State 1:
A1: $2a
A2: $20a
State 2: Add $2
B1: $2a, $2b
B2: $20a, $2b
State 3: Remove 1 bill
B1a: $2a, $2b - $2a: $2b
B1b: $2a, $2b - $2b: $2a
B2a: $20a, $2b - $2b: $20a
B2b: $20a, $2b - $20a: $2b * excluded *
There is a 2/3 chance there’s a $2 bill remaining in the wallet.
Note: If someone else peeks into the wallet and always removes a $2 bill, you have to replace case B2b with:
B2B: $20a, $2b - $2b: = $20a (not excluded)
And then there is a 1/2 chance of a $2 remaining in the wallet.
But, since the bill removal was explicitly stated as blind, we know that case B2b did not occur, therefore we can exclude it from the list of outcomes.
Dr. Crane! Your glockenspiel has come to life!
I think the chances of the remaining (hidden) bill being $2.00 are two out of three.
If you repeated the experiment a hundred times, you’d get 75 twos and 25 twenties. We can throw out the 25 twenties since we already got a two.
That leaves the 75 twos, two-thirds of which are paired with another two, one-third of which are paired with a twenty.
Hopefully, I can convince you to accept “hopefully” as a disjunct adverb.
Frankly, I would be lying if I said I were confident.
Perhaps this subject is simply too complex for me to explain.
Unfortunately, I would be lucky to explain my way out of a paper bag.
Thanks to our host Cecil Adams, I don’t need to even be smart, all I need to do is know how to read!
Please follow the link to his column,