Envelope Dilemma?

Interesting answer SantoW. Interesting.

This is a poorly defined problem with no mathematical solution.

The answer depends on the probability of the other envelope containing twice as much money - which is completely unknown.

I agree with Danielsong.

This question has sent me in a tizzy. I have no idea what the answer is anymore. If the question simply said MORE money OR LESS money in the 2 envelopes then the answer would be more cut and dry.

I will have to look up the real answer somewhere. Unlike the "lets make a deal" scenario this one has me stumped. Believe it or not the best way to solve the "lets make a deal" situation is to simply think about your probability of being correct inititially and incorrect initially and doing whatever gives greater chance of success, swap or stay.

This question has no such context. I assume you are right initially 50%. After all both envelopes where there for the chosing.

Lets make a deal is NOT a paradox. There is a definite outcome of distinctively higher expectation.

I am stumped now! Thanks David. This is likely a paradox with no definite outcome.

I agree with Fez, Dan and Cutter but now I am going to see what the Wizard has to say.

If you repeat this expirement 10,000 times, 5,000 times you will have more money, and 5,000 times you will have less money. Thus I don't think it makes any difference at all to switch or not to switch.

Now if you are only given the option to switch when you take the envelope with more money, thats a different story...

If your envelope has $50K in it, the only two possibilities are that the other has 100K, or 25K.

Arent you risking 25K to win 50K? Either you wind up with 25K or 100K.

Intuitively, it should make no difference if you swap or not. If you swap, half the time you'd end up with the higher amount and half you'd end up with the smaller amount.

Let X = Lesser Amount, then EV = 0.5*(X) + 0.5*(2X) = 1.5X so it seems that you should ALWAYS swap

However, look at the EV equation this way:

Let X = Larger Amount, then EV = 0.5*(X) + 0.5*(0.5X) = 0.75X so it seems that you should NEVER swap

always swap. Keeping it gives you an average of 50k. swapping an average 62.5k
There's something obviously wrong with my theorizing so please excuse.
I'm painkiller happy.

The higher the dollar amount, the more likely you have the higher paying envelope. The odds truly are not 50/50

It is quite a non intuitive problem..................

No wonder people regularly bet the favorites -- too much angst that they might be wrong. The math that gets you to the EV of the other envelope being $62.5K also applies to the first envelope. The EV of either envelope is the same. Sheesh.

Another way to look at this in economic terms is the bigger the amount of money, the less the marginal utility of the bigger chunk and the greater the marginal utility of the bird in hand. In other words, to exaggerate, if you had $1 million in hand and the 2nd envelope had either $2 million or zero, you'd never switch. That 1st million is worth a whole lot more to you than the 2nd million might be. So, in the paradox case, it could depend on how important the $50K would be to the contestant. Given that there is no +EV advantage, I can't envision a scenario where they'd want to switch except if they needed $100K to pay the ransom and $50K wouldn't do.