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#1
09-29-2002, 12:32 PM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Envelopes!

Two envelope questions to think about:

1) A kind man tells you that he has two envelopes each with money in them. He says that you can choose either envelope, open it, and keep the money in that envelope, or trade the contents of the opened envelope for the unopened one. Is there any strategy that has a higher expectation than randomly choosing, opening, and not trading?

2) A millionaire is funding a game between you and another person. The other person is given 1000 one dollar bills and two identical envelopes. He is instructed to put any integral dollar amount up to \$333 in the first envelope and put exactly twice that amount into the second envelope. He shuffles the envelopes and presents them to you. As in 1) you get to choose one envelope, open it, and then decide whether to trade for the unopened envelope. If you end up with the envelope containing less money, he gets \$100. If you end up with the envelope containing more \$, he gets nothing. What is a good strategy for you? What is a good strategy for him? What is his worst strategy?
#2
09-29-2002, 08:15 PM
 Guest Posts: n/a
Re: Envelopes!

I'm confused? Does anybody end up with the 1000 one dollar dollars? It sounds like the 1000 \$1 bills are just the instruments of the game (not unlike a deck of cards in poker). I want more information. The rules (as I see them) state that the other guy gets either \$100 or nothing -- I don't understand. Define the problem better.

What's in it for me? Please explain my possible gain in this game -- not his.

I have no stragety before hand -- I'm not putting any money in the envelopes -- he is....
I think his worse stragety would be to but \$333 in one of the envelopes (also if he is above the age of reason, he should not use any odd integer amounts in the envelopes). I think his best stragery would be to put \$2 &amp; \$4 in the envelopes or maybe \$124 &amp; \$248. (any two even integers where their sum is less than 372).
#3
09-29-2002, 08:32 PM
 lorinda Senior Member Join Date: Sep 2002 Location: England Posts: 2,478
Re: Envelopes!

I'll not post all my conclusions yet (partly because they are still incomplete) but i believe in case 2, that if we get to keep the money in the envelope we end up with (which is what i believe is the case) then we should always switch, except in some very obvious situations.
I also believe the opponent (who has motivation to win \$100) needs a variable strategy.

As to case one, well, Im always going to switch if there's \$1 in the envelope...
..now Im off to consider, and to give others a chance to do this one, I like this puzzle a lot
#4
09-29-2002, 09:19 PM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Clairification of question #2

Carl,

For game 2, the \$1000 is just the maximum amount that can be put into the envelopes. You want to do your best to guess which envelope has the most money because you get to keep the money in the envelope that you choose. The other guy wants you to choose the envelope with less money because he then gets \$100. If you choose the envelope with more money, then he wins nothing.

Example of game # 2: The other guy puts \$3 into envelope A and \$6 into envelope B. You choose randomly between A and B. Let' say you choose envelope A. You open it and find \$3. You say a-ha! The other envelope must contain either \$1.50 or \$6 and \$1.50 is not allowed because game #2 specifies that the envelopes must contain whole dollar amounts. Therefore, the other envelope must contain \$6. You switch and get \$6. The other guy gets nothing because you ended up with the richer envelope.

Another example of game # 2: Other guy puts \$30 into envelope A and \$60 into envelope B. You choose randomly between A and B. Let' say you choose envelope A. You open it and find \$30. You reason that the other envelope must contain either \$15 or \$60. Suppose you decide to keep the \$30 envelope rather than risking the \$15 loss. The other guy now gets \$100 because you ended up with the poorer envelope.

Did that make the game clear?

#5
09-30-2002, 02:58 AM
 Guest Posts: n/a
Re: Clairification of question #2

irchans

#6
09-30-2002, 09:59 AM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Worst Strategy for game 2

Carl wins the ribbon for finding the worst envelope stuffer strategy for game #2. If the stuffer puts \$333 in one envelope and \$666 in the other, then you always win \$666. The stuffer wins nothing.

#7
10-01-2002, 10:39 AM
 Guest Posts: n/a
Re: Envelopes!

The best strategy for the envelope stuffer:

- Tell the picker that you will put \$333 in one envelope and \$666 in the other if he agrees to split the money with you (\$200 / \$466). If he doesn't agree, you'll put \$1 and \$2 in.

- If he agrees to split, hoorah! That's a guaranteed \$200 instead of a 50/50 \$100. The picker also gets more money this way, since it's pretty obvious that if you were playing 'correctly' you would never put more than \$332 in the larger envelope.

- If he doesn't agree to split the money, actually put \$2 and \$4 in the envelopes...that way if he chooses the \$2 envelope, he'll stay (netting you \$100), plus if he chooses the \$4 envelope he might switch just because you double-crossed him and he's confused / out-thinking himself.

PP
#8
10-01-2002, 03:12 PM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Re: Envelopes!

Pseudo,

If the two players in game #2 (stuffer and guesser) are permitted to collaborate, then their combined expectation is \$666 as in your solution.

If no communication is permitted between the stuffer and the guesser, what strategy would you recomend?

#9
10-01-2002, 04:20 PM
 Mike Haven Senior Member Join Date: Sep 2002 Location: Northern Ireland Posts: 2,288
Re: Envelopes!

2 is a good problem, irchans

i'm having difficulty getting beyond random even number stuffing and random choosing unless i am lucky and hit an envelope with \$2 or \$334 or more in it - but that's not my final guess! [img]/forums/images/icons/confused.gif[/img]
#10
10-01-2002, 04:42 PM
 Guest Posts: n/a
Re: Envelopes! (Problem #2)

That's a lot less fun. Here goes:

As the picker:
- If I see an odd number, I switch (it's got to be the low).
- If I see a number over \$333, I stay (it's got to be the high).

- If I see a number that's not a multiple of 4, it's probably the low, so I switch. (If it is the high, that means the stuffer put an odd number of bills in the other envelope, which is an odd strategy.)
- If I see a number that's over \$166, I stay. (If it is the low, that means the stuffer put over \$332 in the high envelope, which is an odd strategy.)

- If I see a multiple of \$4 between \$4 and \$164, it's trickier. I pick X. The envelopes contain either (X and 2X) or (X/2 and X).
-- For (X and 2X) to be a good stuffing choice, 2X should be \$164 or less (and thus 'ambiguous' whether it's the high or low value)...otherwise, if I had picked 2X instead of X, I'd be pretty sure it was the high value and stay. So if 2X is over \$164, then X is probably not the low value.
-- For (X/2 and X) to be a good stuffing choice, X/2 should be a multiple of \$4...otherwise, if I had picked X/2 instead of X, I'd be pretty sure it was the low value and switch. So if X is not a multiple of 8, it's probably not the high value.
-- So, if X is over \$82, I stay. If X is not a multiple of \$8, I swap.

- What if X is a multiple of \$8 between \$8 and \$80? Here we go:
-- Again, the stuffing choices are (X and 2X) and (X/2 and X).
-- If it's (X and 2X), the stuffer would have to be worried about me choosing (2X) and thinking it might be either (X and 2X) or (2X and 4X). Thus 4X should be an ambiguous choice as well...so 4X must be under \$164. So if X is a multiple of 8 over \$41, X is probably not the low value, and I'll stay.
-- Likewise, if it's (X/2 and X), the stuffer would have to worried about me choosing (X/2) and considering the pair (X/4 and X/2). So X/4 should be a multiple of 4 as well. Thus, if X is not a multiple of 16, it's probably the low value, and I'll switch.
-- So, if X is over \$41, I'll switch. If X is not a multiple of 16, I'll stay.

- So, what do I do with 16 and 32 (the multiples of 16 under \$41)? It depends on my read of the stuffer:
-- If the stuffer (a) did not think out the problem or (b) thought out the problem only to the point of figuring out the (16, 32) is the most 'ambiguous' pairing, then I would stay on 32, swap on 16.
-- If the stuffer (a) figured out that (16, 32) is the most 'ambiguous' pair and (b) realizes that I, the picker, also would figure that out, then I would probably stay on 16, swap on 32...since he's probably counting on me to figure him for (16, 32) and make the appropriate move based on that.

- On the other hand, you could think about it this way: since I don't care about what happens to the picker, I just have to choose some value Y so that for all X &gt; Y I'm equally happy and all X &lt; Y I'm equally sad.
-- If X &gt; Y, then I stay -- no regrets if it's high or low.
-- If X &lt; Y but 2X &gt; Y, then I swap -- I'm happy if I get 2X and would have been equally unhappy with either X or X/2.
-- If 2X &lt; Y, it doesn't matter if I swap or not -- I'm not happy no matter what I do.
-- And, of course...this is FREE MONEY! Who cares if it's the most free money you could have gotten or not? It's FREE!

- The way NOT to look at this problem is this: I have X. If I swap, I'll either get 2X or X/2...so I'm risking X/2 to gain X with a 50% chance of winning. That's a 2:1 bet on even odds. Of course I'll swap! (Stupid paradoxes.)

As the stuffer:
- It depends on what I know of the picker.
-- Against a dunce I'd choose (16, 32).
-- Against anyone else, I'd probably go with (8, 16). If I get lucky and he chooses 16, he'll probably swap, since he reasoned that (16, 32) is a likely choice for me. If I get unlucky and he gets 8, well, maybe he'll get confused and outsmart himself by staying on 8. (I think this is better than choosing a combo like (19, 38), where you definitely lose on (19) and only probably win on (38).)

That's my mostly-muddled thinking on the problem.

Of course, depending on the exact interpretation of the rules, the stuffer's best strategy for X and 2X might be X = 0.
- 0 is an interger, after all...so both envelopes would be empty.
- If the rule is that you get paid off if the pick ends up with the least amount of money, then this is an automatic win.
- Even if not, if you really hate the picker, you can screw him over this way.

PP

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