 # The two envelopes - Revisited Palynka Posers and Puzzles 19 Mar '09 11:18
1. 09 Apr '09 02:56
I believe the crux of the paradox lies in the fact that in the original problem, our high value is unbounded. It is only after we know what is in our envelop that we encounter any limitation to it.

How this generates the paradox appears to be a bit difficult to grasp precisely, but it does seem to have something to do with uniform distribution of random numbers over an infinite range, although this lack of upper bounds may play with the statistical math in other way, by failing to provide enough of a infrastructure framework to set our statistical expectations properly.

However, I am curious how the math works out if one chooses a minimal discreet step for the currency and uses odd/even analysis. If the amount in the envelop is an odd number of this units, it clearly cannot be the doubled envelop, and thus switching is a guaranteed doubling of our amount.

What if it is even though? Assuming an even distribution of possibilities for the lower envelop in the range which it could be (assuming that is finite, even if it is indefinite), it has a 50% to be even and a 50% of being odd. I thus have a 50% chance of picking the high envelop, which is always even, but only 25% of picking an even low envelop, making the odds that the envelop is the smaller envelop 1-in-3 by Baynesian analysis.

Now plug in the expected net exchange based on these odds. I arrive at the following numbers (2 x 1/3 + 1/2 x 2/3) = 2/3 + 1/3 = 1.

So that account for odd and even, if I receive an even envelop, neither keeping it nor swapping it has any advantage statistically. However, if the 1/4 of the time it is odd gets added in, I find this boosts the return for swapping by 25%, so I arrive at the same 125% return we got with simpler analysis..
2. 14 Apr '09 14:57
H: High
EE: both even
EO one even, one odd

That doesn't seem right. The odds are higher that when you open and find an even sum, then you are in the EE case, but not necessarily with the higher envelope. Let's see the math:

We want to find P(H|Even).

P(H|Even) = P(H|Even & EE)*p(EE)+P(H|Even & EO)*p(EO)

Also:
P(Even & EE) = P(Even|EE)*P(EE) = 1*1/2=1/2
P(Even & EO) = P(Even|EO)*P(EO) = 1/2 * 1/2 = 1/4

Moreover:
P(H|even & EE) = 1/2
P(H|even & EO) = 1

Repeat from the first line:
P(H|even) = P(H|Even & EE)*p(EE)+P(H|Even & EO)*p(EO)
P(H|even) = 1/2*1/2 + 1*1/4 = 1/2

Note: I'm abstracting from the bound criticism here (which for me is the source of the paradox), so all this math should be seen with the caveats I mentioned before.
3. 23 Apr '09 22:24
Originally posted by Palynka
H: High
EE: both even
EO one even, one odd

That doesn't seem right. The odds are higher that when you open and find an even sum, then you are in the EE case, but not necessarily with the higher envelope. Let's see the math:

We want to find P(H|Even).

P(H|Even) = P(H|Even & EE)*p(EE)+P(H|Even & EO)*p(EO)

Also:
P(Even & EE) = P(Even|EE)*P(EE) = ...[text shortened]... source of the paradox), so all this math should be seen with the caveats I mentioned before.
If you want to field test it.......send me two envelopes.I promise to send one back!