2 envelopes

[huy.duong@ctsu.ox.ac.uk]

Dear anh Paul,

>choose other one because if you're to lose, you lose
>50, otherwise you win 100.

If you generalise the case above, you get:

`Whatever amount in this envelope, without wasting time to
open it, I'll choose the other one because if I loose, I'll loose
half and if I win, I'll win another whole of this envelope'. So you 
choose the other evelope and say the same thing again...forever.

Instinct: it can't make any difference.

Attempting to justify my prejudice:

The envelopes contain x and 2x. First choice is probabilistic: p(x)=p(2x)=0.5. 
Swap is deterministic: prob(x->2x)=prob(2x->x)=1, prob(x->x)=prob(2x->x)=0, ie 
always get what you didn't get first time.

If no swap: total P(x)=p(x)=0.5, P(2x)=0.5. 
If swap: total P(x)=p(x)*p(x->x) + p(2x)*p(2x->x)
                   =0.5*0        + 0.5*1
                   =0.5.
Therefore total P(2x)=0.5.

So it doesn't matter.

Huy




If you swap, the total prob of getting 2x is therefore 
prob(x)*prob_swap(x->2x)=0.5*1.0=0.5, same as total prob of getting x, same as 
not swapping.

Huy