You are told there are two envelopes. One contains twice as much money as the other one. You pick one but are allowed to change your mind after picking it. (You are equally likely to pick the one with less money as the one with more money.)

To figure out how much on average the other envelope should contain, one might average x/2 and 2x because one is equally likely to pick one as picking the other. Since this comes out to 5x/4, one might always change his or her mind. But wouldn't this end up with one never making up his or her mind?

Let's suppose someone comes to you with a proposal: Give me a 100 bucks, and any number, N. I will flip a coin N times. For each time the coin falls on heads, I will double your money, and for each time it falls on tails, I will take a half for myself. In that case, what is the profit's expectancy?

the answer, surprisingly, is - statistically, the purposal is very worthwhile. Take a random variant X, which has 50% of having to value 1 and 50% having to value -1. let S(X, N) be the sum derived by summing up random X's (1's or -1's) N times. then S is a function with expectancy of 0, with normal distribution around 0. it is easy to see that is the man's purposal would have been "I will give you a dollar every time it's tails and take a dollar every time heads is out" - the expectancy of profit would be zero.

but now - let's look at 100*2^S(X, N). I'll omit that actual calculation, but the expectancy of this function is much higer than 100 - to satisfy intuition - when S sums up to a negative number, 2^S is a number between 0 and 2. when S sums up to a negative integer, 2^S can be a lot higher than 4.

 

Posted by ronen on 2004-05-27 19:45:50