Permalink Submitted by Chen Weidong on October 4, 2017

write a and 2a for the amount of money in those two envelope, write x for the amount of money that is in your chosen envelope. The x can be a or 2a in both cases with the probability of 1/2. When the x is a, the amount of another envelope must be 2a, vice versa. If you change your mind to choose another envelope, in fact the expected amount you will get is:
"the expected value when your chosen envelope is a " plus "the expected value when your chosen envelope is 2a", which is 1/2*(2x) (when x equals a ) plus 1/2*x/2 (when x equals 2a).

so the expected amount you will get is 1/2*(2x(x=a)+x/2(x=2a))=1/2(2a+a)=3/2*a.

At the same time, the expected value of the first envelope is 1/2*a+1/2*2a=3/2*a.

In conclusion, you do not have to change your mind if you choose one envelope.

The key point is that the " x " in the chosen envelope is a random variable, the x in 2x or x/2 for the amount of other envelope is two different value .

## the value of x in 2x and x/2 is different

write a and 2a for the amount of money in those two envelope, write x for the amount of money that is in your chosen envelope. The x can be a or 2a in both cases with the probability of 1/2. When the x is a, the amount of another envelope must be 2a, vice versa. If you change your mind to choose another envelope, in fact the expected amount you will get is:

"the expected value when your chosen envelope is a " plus "the expected value when your chosen envelope is 2a", which is 1/2*(2x) (when x equals a ) plus 1/2*x/2 (when x equals 2a).

so the expected amount you will get is 1/2*(2x(x=a)+x/2(x=2a))=1/2(2a+a)=3/2*a.

At the same time, the expected value of the first envelope is 1/2*a+1/2*2a=3/2*a.

In conclusion, you do not have to change your mind if you choose one envelope.

The key point is that the " x " in the chosen envelope is a random variable, the x in 2x or x/2 for the amount of other envelope is two different value .