In a box there are two envelopes. It is known that with probability 1/2, one envelope contains $1 and the other one $10; with probability 1/4, one envelope contains $10 and the other one $100; with probability 1/8, one envelope contains $100 and the other one $1000; and so on.

You open one envelope and find x dollars in it. Now you can keep the money or take instead the other envelope. What do you do?

## Friday, January 04, 2008

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I guess you keep whatever you got. If you get, say, $100, then you know there's a greater chance that it's part of the $10/100 pairing than the $100/1000 pairing, and therefore if you switch envelopes you are more likely to get a lesser amount.

ReplyDeleteI haven't posted on here in forever!

well i guess that's not right...the odds will favor 2:1 that it's a smaller bill, but the small bill is 1:100 smaller than the bigger bill.

ReplyDeleteYou should always switch out. Note that this is a theoretical problem and can't happen in real life. This is because there is no limit on the amount of money that could end up in the envelopes (and who has infinite money to play with?)

ReplyDeleteBut it turns out that the probability of the 'other' envelope, the one you didn't pick, is most likely going to contain more money in it.

On your first choice the probability of the envelope you choosing having $1 in it is 1/4. The probability of getting $10 is 1/4 + 1/8 = 3/8. The probability of getting $100 is 1/8 + 1/16 = 3/16. And so on...

After choosing the first envelope, the probability of the second envelope containing $x/10 is 1/3 while the probability of it containing $10*x is 2/3. So, you have a very good reason for switching, no matter what the envelope contains.

If my first envelope is the 1000, im not switching in that case.

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