Interesting... here's my "gut reaction" as a non-statistician. The problem is that 'A' has a probabilistic component as well. If you let X be the "lesser of t

Interesting... here's my "gut reaction" as a non-statistician. The problem is that 'A' has a probabilistic component as well. If you let X be the "lesser of the 2 dollar amounts" and let A = X/2 + 2X, then you come out with equal expected values for switching and non-switching.