Envelopes!

[lorinda]

Please note a couple of things about my solution.

1. These numbers are ratios that you would use a random number generator to provide if you could

2. In a medium term game you could "cheat" your opponent by not putting enough $2 envelopes in, however, when his envelope tracker/envelope stat program found this out, he would stop switching when he opened $4 envelopes and you would lose out in the long term, it is to be considered as a pure solution for millions of trials.

[heihojin]

The main problem is that I find it nonsensical to discuss unbounded probability distributions. It may just be because I haven't yet encountered such in my math education, but I've always considered expectation to be incalculable for such problems.

While I can understand how both of your strategies marginally increase your expectation by an infinitesimal amount, I'm not sure that you could calculate that amount given an unbounded probability distribution.

Furthermore, there doesn't seem to be anything inherently magical about those two strategies; you obtain the same results by switching with probability f(x), where f is any function such that the following are true for all x in [1, infinity):

1. 0 <= f(x) <= 1
2. f(L) >= f(H) when L < H
3. there is some number a such that f(L) > f(H) when L < a < H.


heihojin

[irchans]

Heihojin,

Great observations. Any monotonically increasing switching function beats random. (i.e. 0 <= f(L) < f(H) < 1 when L < H).

[irchans]

Lorinda,

I believe that your solution is the best possible stuffer strategy! [img]/forums/images/icons/ooo.gif[/img] (i.e. That is the highest expectation that the stuffer can get if she tells the selector her strategy first.) Your solution has interesting properties:

1) If you open the randomly selected first envelope and it contains more than $2, then there is a 66.6664% chance that it is the larger envelope. (It seems very weird to me that when you open this "randomly" selected envelope, it is more likely to be the larger of the two envelopes! )

2) Even though there is a 66.6664% chance that it is the larger envelope, the guesser is forced to switch because he gets a 2 to 1 pay off on the amount risked.

It is neat that you force the switcher to switch even though is it more likely that he has the larger envelope.


Now, what is the best switching strategy?

(PS: I think you can get just a tiny tiny bit more expectation by going down to $1.)

Just a nitpicking note:
You lose nothing by adding a $1 low envelope, I think, since you right now have the same affect with your $2 envelope. Now you can get even closer to 50%

[lorinda]

Irchans wrote:
(PS: I think you can get just a tiny tiny bit more expectation by going down to $1.)

Damn! Damn! Damn! I missed that.
[img]/forums/images/icons/laugh.gif[/img]

[lorinda]

Ive thought about it again and I'm not certain you should include the $1.
Even though the strategy works if you tell the picker, it also works (assuming a reasonably intelligent picker)if you don't.
Including the $1 takes away some of the edge made by picker error....That's my lame defence Im afraid.

(Okay Okay, you're right, and I'm real annoyed at myself)