*Originally posted by Palynka*

**I'll contend that the strategies are indifferent, and try to show why, but I'll wait to see if someone takes a crack at it.
**

If you're curious, try writing down a MATLAB program where first A and B quantities are set and then you play a strategy of "always open and switch". Then compare it with a strategy of "choose randomly and stick with it". After thin ...[text shortened]... ng about it in this way, does the strategy of switching give you a higher expected value?

i understand your point that the final outcome SEEMS to be irrelevant in terms of the choice you make. since the dollar amounts are predestined, you're making a "random" choice either way, why wouldn't it be the same total outcome regardless of what you do?

but consider for a moment the fact that given any finite dollar amount that you see, when you are confronted with the choice between possibly doubling it, or possibly cutting it in half, in terms of dollar amounts you have more to gain than to lose. say you open the envelope and there is $1024 in it. now your choice is a 50/50 chance of either winning $1024 more dollars, OR losing $512. and you are only allowed to make the decision once - no "double or nothing" choices afterwards, etc. i haven't done the MATLAB program as you suggested, but i believe that it will follow my predicted results, PROVIDED you keep track of two very important sums along the way. the running total sum of the envelopes you see, and the running total sum of the envelopes you don't see. that is somewhat logically different than the program you suggested, but i think gets at what i'm trying to describe as the reason why switching is advantageous.

for instance, during the simulation, imagine $16 is chosen 100 times. half those times there will be $32 in the other envelope, half the time there will be $8 in the other envelope. so we see a total of $1600, but if we add up all the money in the envelopes we don't see, it's $32*50 + $8*50 = $1600 + $400 = $2000 in the unseen envelopes. so why didn't we switch?

it seems that when we pin down a finite dollar amount on one of the envelopes, the choice favors switching... and i think the difference in our reasoning may have something to do with the fact that in your setup of the problem the set of possible dollar values is only

**semi**-infinite? we are dealing with a set that has a smallest value, but then are disallowed from having the smallest value be the envelope we choose. or perhaps the infinitude of the set (lack of a highest value) is causing a logical error on MY part. i'm not certain, and want to assure you i am not a person who is unwilling to listen or change - in fact i will be happy to be proven wrong!

i'm really just musing here and will continue to dwell on this question, and perhaps will find the time to run a simulation for future truth-seeking.