Envelopes!

For question #1:

I can't see any way to maximize your expectation, unless you can make some kind of reasonable assumption about the range/distribution of possible values.

For me personally, I'd probably choose some value Y which I'd be satisfied with. If the envelope I pick has more than Y, I stick, otherwise I swap -- no regrets either way.

Unless, of course, he's kind enough to tell you how much money's in the unopened enveloped. That would make your choice much easier.

PP

[Inthacup]

If I was dividing the envelopes this is what I would consider. First any number than ends in a 1,3,5,7,9 MUST be the low, so you can always chose the other envelope. Any number over 333 must be the high. So if I were to be dividing these up I would make sure that if I divided the Big by 2 I wouldn't get an odd #. For instance 100/200, not 98/49.

If I was chosing I would take those things into consideration as well as a few miscellaneous thoughts. If I get certain even numbers I know they're most likely low as well. For instance if I get an envelope with $26 then a logical guess would be that the $26 is low. This is because I know that the dealer knows that odds must be low, therefore he wouldn't chose $26 to be the high w/ $13 being the low. So I can make reasonable guesses on any odd number as well as about half of the even ones.

The only real numbers that are problematic the ones that are divided by 4 and still give an even number. For instance, 48/4 is 12, therefore you have no hints on your 48 options. This makes since to me, but if I'm unclear or just wrong feel free to correct me. Great question.

[Mike Haven]

whilst you have done a huge amount of work and your answer makes extremely interesting reading i think that your opponent's tactics might work only for the "first" game

i assume there are an infinite number of envelopes being proffered and although i may be wrong it seems to me that after the first few thousand or so with such a set strategy your opponent may not have to think too much to improve his chances of winning as he will know what you are going to
do in any given situation, and therefore will he not just do the "opposite" for a time?

I was reading it as more of a one-time thing...make your best choice in this game, because you won't get another shot at it.

Thinking about this problem reminded me of that scene in The Princess Bride ("And you'd have KNOWN that I knew iocaine powder comes from Australia, so I can clearly not choose the glass in front of me...").

PP

OK, my solution to #2:

pick a random number from the set {1,2,4,8,16,32,64,128,256} and put it in the low envelope.

Your opponent has a 1 in 9 chance of winning for certain after picking and 8/9 of being 50/50. That gives you a EV of $44.44.

OK.... thought about it for a few minutes... in my answer the other player should always switch, since that would greater EV. It may be possible to exploit this. I could give greater weighting to the lower numbers, so opponent still favors switching (possibly just barely favors it), but now is more likely to switch to a lower number. The down side of that is that I now have greatly increased the chances of randomly picking 1 from the set, which gives me an automatic loss. I still like my set of numbers, but I don't know what the magic balancing point is for their probablility distribution. Anyone more mathematically inclined than me (and there should be a bunch of you on this forum) want to take a shot at it?

Why not just pare your set down to {8, 16, 32}? It seems to me that including possibilities that might give the picker a sure win can only be bad. Just so long as your opponent doesn't know these three are your only possibilities for the low number, it's a 50/50 proposition.

PP

[Inthacup]

I would not put 256 in the low envelope. The large would have to be 512, and if he picked that one, then he has an automatic win. But if he happened to pick 256, he would most likely think that 256 was the high, because no one would put money in a envelope where they gave the choser a guarenteed winner. I would advise against putting either 256 or 128 in the envelopes as the low for the specified reasons above.

[irchans]

>pick a random number from the set {1, 2, 4, 8, 16, 32, 64, 128, 256} and put it in the low envelope.

Many people had very interesting stuffer and guesser strategies. I don't have time right now to comment on all of them, so I will only comment on Manhattan Jac's strategy.

Let's assume that Manhattan Jac publishes his strategy and then plays the stuffer role. What is his expectation?

The selector will always switch unless he chooses the 512 envelope. (For this stuffer strategy, if the selector picks an envelope containing x dollars and x is neither 1 nor 512, then his expectation for switching is 0.5*0.5*x + 0.5*2*x = 1.25 x, so he must switch.)

So if the selector picks an envelope with

$1 in it, the suffers expectation is 0
$2 in it, the suffers expectation is 50
$4 in it, the suffers expectation is 50
$8 in it, the suffers expectation is 50
$16 in it, the suffers expectation is 50
$32 in it, the suffers expectation is 50
$64 in it, the suffers expectation is 50
$128 in it, the suffers expectation is 50
$256 in it, the suffers expectation is 50.
$512 in it, the suffers expectation is 0.

So Jac's stuffing expectation is 50*8/9 = $44.44 against a smart opponent with that strategy if the selector knows the stuffers strategy.

Could someone check my reasoning here?

My solution is different but it has exactly the same stuffer expectation.

The picker gets to know the stuffer's strategy? That hardly seems fair [img]/forums/images/icons/smile.gif[/img]

PP