[Guest]

My mom told me that I had been a good kid, and for being so she has decided to award me with a special gift. She has prepared 2 envelopes, in two rooms. One contains twice the amount of the other and that it's in the multiples of 2 (my mom knows that I love puzzles ). So the distribution is either [1,2]; [2,4]; [4,8] ... and so on. She says it can be as large as I can imagine.

Now I go to the first room and I peek into my envelope, but just at that moment she catches me ! I feel guilty and ask sorry to her, and she being her motherly self says, "Ok, Aatif. I forgive you. I give you one more chance to choose, even though I know that you have peeked into this envelope!" ...

What should I do?!!!

While I was muling over my problem, I heard mom saying the exact same thing to my brother (we are twins, you see ). Should I still do the same as above? Who has the chance of making more money?

The Probability of choosing:

P( [2ⁿ,2ⁿ⁺¹] )= (2ⁿ/3ⁿ⁺¹)

Elliot Linzer, American Mathematical Monthly, Volume 101, Number 5, May 1994, p. 417

[unreality]

Unless you peeked and saw $1, you dont know if the amount you got is the smaller or the larger. But since $1 is the smallest, it would be advantageous to switch if you saw $1. Since that shifts the probably from infinity over 2*infinity chances (or 1/2) to infinity+1 over infinty*2 chance. A slightly higher chance. So I would say "why not... switch".

What I would actually do is make a pact with my twin brother and agree to split the money evenly no matter what. That way, if you took the smaller amount, or x, you get 1.5x cash. If you took the larger one, 2x, you get 1.5x cash, which is 3/4 of what you would have gotten anyway- so the person with the lesser envelope gets 3/2 as much, the person with the better envelope gets 3/4 as much, which is almost as much... either way it's good enough for me. ;D

[Guest]

Unless you peeked and saw $1, you dont know if the amount you got is the smaller or the larger. But since $1 is the smallest, it would be advantageous to switch if you saw $1. Since that shifts the probably from infinity over 2*infinity chances (or 1/2) to infinity+1 over infinty*2 chance. A slightly higher chance. So I would say "why not... switch".

What I would actually do is make a pact with my twin brother and agree to split the money evenly no matter what. That way, if you took the smaller amount, or x, you get 1.5x cash. If you took the larger one, 2x, you get 1.5x cash, which is 3/4 of what you would have gotten anyway- so the person with the lesser envelope gets 3/2 as much, the person with the better envelope gets 3/4 as much, which is almost as much... either way it's good enough for me. ;D

Ummmm .. Correct though not neat ..

If you look at the hint:

P( [2ⁿ,2ⁿ⁺¹] )= (2ⁿ/3ⁿ⁺¹) for all n>= 1

Now say, you found x = 2^m in the envelope.

The Conditional Probability that it's the smaller of the distribution is:

P( [2^m-1,2^m] | 2^m) = {P([2^m-1,2^m] )/(P([2^m-1,2^m] )+P([2^m,2^m+1] )} = 3/5

Now, you have 60 percent chance that it's [2^m-1,2^m] and 40 percent chance that it is [2^m,2^m+1]

Thus expected value of money (E1) in the Other envelope = 2^m-1 * 0.6 + 2^m+1 * 0.4 = 1.1*2^m

For all values of m more than 0. For 0 its 100 percent probable that the other envelope has 2$.

Thus irrespective of what you see, you should switch!

Now about my twin, your solution would work if regardless of what we see in the envelope we decide to share it equally. Then because the quantity inside is unknown, the expected average value of each envelope would be (E2) = 1/2*2^m+1/2*2* 2^m =1.5*2^m ; irrespective of what we see. Now we can see that E2 > E1 ; so if my brother doesn't agree its better for both of us to switch else it's always better to be in peace and split! .

Ahh .. If only my Boss could agree that dividing his money with me is not so Bad! <_<

[bonanova]

You can argue that even odds of winning more or losing less says you should switch.

But symmetry says your expectation stays the same.

Symmetry wins.

And here's another.

[Guest]

And here's another.

You can argue that even odds of winning more or losing less says you should switch.

But symmetry says your expectation stays the same.

Symmetry wins.

Nope Bona .. I don't agree ..

The problem that you mentioned here doesn't define the range from which you are picking your money. In that case theres 50% probability that the money in opponents pocket is more than you by say dx and 50% probability that it is less by dx.

Hence expected gain is E = 0.5(M+dx)+0.5(M-dx) = M . That is no loss, no gain! Hence its pointless to play such a bet!! Here you can use symmetry argument because equal indeterminate distribution of money.

Whereas, once you know the amount in the envelope, the symetry is broken. Now, you know, that there is 2^m amount in the envelope. So there is finite probability to find out that there can be either 2^m-1 or 2^m+1 in the other envelope. Hence, your choice cannot be driven by symetry.

On the other hand, if you had no idea how much was there in your envelope (or if you closed your eyes at the right moment!), then the statement still maintains its symetry, and as I had showed earlier, the expected gain remains constant at 1.5*2^m, for both envelope. Hence, no switch.

Peace!

[Guest]

I think this old question should be reviewed. If you and your twin chooses accidentally identical envelopes. And say it, you see the smaller number and your twin sees the greater number. Who will switch then? I agree here with Bonanova, symmetry is in charge in this case, and switching is meaningless.

[Guest]

I don't know if this will be any consolation, but a lot of actual mathematicians have been trying to find a resolution to this paradox for about 20 years. There has been a pattern of someone basically finding a technicality to invalidate the problem's reasoning, and then someone else making a small change to the problem to overcome the technicality.

No consensus has yet been reached. The explanation most satisfying to me is that a precise formulation of the problem requires that the envelopes have infinite expected value, and that normal reasoning doesn't apply to an infinite expected value (as in the St. Petersburg Paradox). This explanation isn't really all that satisfying, but it's the best I've got.

For more info, see http://en.wikipedia.org/wiki/Two_envelope_paradox