two envelopes are placed in front of you and you are told one contains twice As much money as the other, but the absolute quantities are unknown. So u pick one and open it and see $10. You are then asked if you wish to switch.

The EV of switching is 12.5 (.5)(20)+(.5)(5)

why is the ev not 2.5; where you subtract the $10 you have to give up in order to switch?

This is a famous problem in probability. I suggest you to read the wiki link I provided to see how many interpretation this problem can have and which difficulties it arises. It is a very interesting problem.

Regarding your question, the EV indicates how much value you make on average. If you don't switch you make 10$, so your EV is 10$. If you switch, assuming that your reasoning is correct (which is not), you make on average 12.5$, so your EV = 12.5$. 2.5$ is just the difference in EV between the two choices (again, assuming the calculations are good).

Thanks for the link. It presents an interesting scenario. Learn something every day

The correct way to calculate the EV is different.

In reality it is like this; The value revealed is x. The real maximum is a and the minimum is therefore a/2. But we do not know what a is, we only know x that we see with whatever value such information has in general.

Changing envelopes say has then this Value change as a strategy;


(Probability of first holding the maximum given that it is is revealed as x)*Change in EV then+(probability of first holding the minimum given that it is x)*Change in EV then=

(Probability of first holding the maximum given that it is is x)*(a/2-a = lose a/2 because x=a then )+(probability of first holding the minimum given that it is x)*(a-a/2 because x=a/2 then).

In other words the maximum must remain unknown and called "a" in the calculation because it is exactly that an unknown thing.

Now if we know that x belongs in a distribution and the very fact it is x when we see it means something that can impact eg Probability of first holding the maximum given that it is is x ie a Bayesian probability, then this can improve the evaluation of the strategy of changing envelopes and lead to one choice being better than the other.

For example if the money is coming from a bank that has at most 1 mil bankroll and the minimum denomination is 1$ and we see inside the envelope (or make it box to have room for a a lot up to 1mil in bills) then we know that switching is 100% the best choice because 1 is the minimum possible so the other box/envelope has definitely 2 and nothing else.

If the envelope had 590000 though we also know that we have to keep it because the bank cant possibly have 590000*2=1180000.

So as you see if we knew the distribution of the money that is placed in these envelopes the actual value revealed can impact the probability.

Now if we knew therefore that the distribution was eg some uniform from $1 to 1 mil there are values of the money revealed in the first envelope that can impact the decision in a meaningful way. If it was another distribution eg lognormal the decision may also change for some values as compared to others.

If the money revealed was $350 you wouldnt be able to say a lot though regarding the conditional probability its the maximum eg in the uniform case.


The usual argument then that

Probability of first holding the maximum given that it is is revealed as x =50% is not accurate in general and only correct or approximate in some distributions.

So the final correct EV of changing is;


(Probability of first holding the maximum given that it is is x)*(-a/2)+probability of first holding the minimum given that it is x)*(a/2) and when you can see these probabilities as roughly 50% then changing has EV 0 as expected.




The problem here starts when one tries to perform the EV calculation this way instead;


(Probability of first holding the maximum given that it is is x)*(x/2-x) +probability of first holding the minimum given that it is x)*(2x-x) and then proceeding additionally to call these 2 conditional probabilities 1/2 to obtain

-1/2*x/2+1/2*x=1/4*x>0 (for changing)

The problem here is that we have assumed in the same expression for the total EV change that the envelope can have x and 2x and x/2 at the same time as possibilities of one world. But we do have 2 worlds instead. One in which its x and x/2 and another in which it is x and 2x and we cant do the EV as if this was one world with possible values x, x/2 and 2x at the same time.

Or maybe its easier to see the problem if examined this way. Your envelope has x. The other envelop therefore has

EV2=(probability x is maximum)*x/2+(probability x is the minimum)*2x=> (in the 50% assumption case)
EV2=x/4+x=5/4x>x (suggesting a switch)

The problem now is seen better that you imagine a world that you can have x , x/2 and 2x with some probability in these 2 envelopes. But its either x, x/2 or x and 2x.


There is also this old thread about it

http://forumserver.twoplustwo.com/47...aradox-832194/