Quote:
Originally Posted by mr fantastic  Consider the expected value of the prize that can be selected from behind the curtain:
(1/3)(0.3) + (1/3)(30) + (1/3)(120) = ......
Is this greater or less than $50 .....? |
Sorry for double posting!
Since the expected value of the prize is greater than $50, then he is willing to trade the $50 for a chance to pick a curtain at random.. is that correct?
What I had initially done was calculate the loss and gain of each possibility:
(1/3)(-$49.7) + (1/3)(-20) + (1/3)(70) = 1
Would it also be right if I did it like that? I assumed that an answer of 1, as opposed to 0 for neutral and -1 for a loss, would mean a gain.