About there being a finite amount of time, that's right, but remember that in an infinite EV game like the St. Petersburg game, even ONE trial gives us the infinite expected value. You don't need to play the game forever to get the infinite EV, just one chance at it is all you need. And yet, your payoff is always a finite amount... paltry compared to this infinity which is promised to us by the math. They never actually write you a check for infinity dollars.

But I think what you are getting at is 100% true: that if you look at any finite run of trials of the envelope game, and ask yourself "Would I have gained money, or lost money, had I switched my envelope every single time?" you will find that it doesn't really matter if you switch or not. And while this is true, it is ALSO true that if you open an envelope for any specific number of dollars, you really do want to switch it.

The problem really is that I asked a question and then sort of changed my mind for a reason that may not be obviously related, and I don't know how to best express all that. So, just for the record now, let's start with P being p(N=q and not q+1).

You gave an equation for calculation P and I wanted to know how you derived that equation. Later, I realized that whenever P=1/3 (and only then), then the EV of switching equals the EV of not switching. So, in order for the EV calculations over the entire infinite distribution to show no paradox (as I wanted), we would have to have calculated P=1/3 for every distribution.

This is the important part -- I do not believe (and know for a fact, in fact) that P=1/3 for each and every infinite distribution imagineable. Therefore, questioning your equation for P would get me nowhere. Almost regardless of your equation used, it will not be 1/3 all the time, so I shouldn't care what you use for the equation, it will not help us out of the paradox.

Hope that cleared it up, but I don't have that much faith in my wording skills.

<About there being a finite amount of time, that's right, but remember that in an infinite EV game like the St. Petersburg game, even ONE trial gives us the infinite expected value.>

I'd have to question this. In theory, it may be ok. But in practice -- I know that I have a limit on my life. There are only so many coin flips I can make before I die. Is my EV really going to be infinite?>

Far more important than the time limit on your life (which could be extended in effect, by using a computer to flip the coins very rapidly, instead of physically flipping a real coin) is the fact that the amount of money on planet earth is finite. How many times do you have to double a dollar before it exceeds the amount of money in circulation?

R Blau vs A Ammann, 1993

<Sneaky: I defy anybody to find a more patzerish game of chess in the database than this one.>

But in defense of R Blau vs A Ammann, 1993, that game is filled with real stinkers from both sides of the board. In K Kanakari vs M Przezdziecka, 2006 Black actually played solid chess. Black has a 2300+ rating! It's like Crispin Glover vs. Mike Tyson. It happens a lot during the first round of big open tournaments, total mismatches.

There is a pile of n chips. The first player to move may remove as many chips as desired, at least one chip not not the whole pile. Thereafter, the players alternate moving, but the next player may remove at most twice the number of chips his opponent took on the previous move. What is the optimal move for the first player if n=43.

I started out looking at a couple simple cases. The first case I tried was n=5, and discovered it was unwinnable. But n=4 and n=6 were winnable.

That's when it clicked that 5 is a term of the Fibonacci series, and I hypothesized that my strategy should be to stick my opponent with a Fibonacci term, e.g. 1, 2, 3, 5, 8, 13, 21, 34, 55 ...

Sometimes, I may have to play a game within a game to force this condition, as seen in the example below:

For n=43, I should take just enough chips to reduce it to the highest Fibonacci term -- so I take 9 chips, leaving <34>.

Now, suppose you take 1 chip, leaving 33. I know want to get it down to <21> (on your turn), so I must play as if I want to eliminate 12 chips.

I take 1 getting to 32 [11 of 12 remaining],
you take 1 (31)[10/12],

I take 2 (29)[<8>/12], you take 1 (28)[7/12],

I take 2 (26)[<5>/12], you take 1 (25)[4/12],

I take 1 (24)[<3>/12], you take 1 (23)[<2>/12],

I take 2 leaving <21>[0/12]

The hint we were given was Zeckendorf’s Theorem which states that every positive integer can be written uniquely as a sum of distinct non-neighboring Fibonacci numbers.

The proof is long and boring and I still haven't had the time to go over it, but you got it!

This may be one of those problems where a hint like that may cause the problem to seem more difficult that it is without the hint. :-)

Not sure whether you've seen this, but it could be of interest to compare it to your results. It's number 266 here:

Not but really, it has helped me a lot in understanding physics. In fluid mechanics, "pressure" is a paradigm just like the envelops. I have learned to not "twist conclusions" to demystify something incredibly complex. Instead, I just acknowledge all sides of the facts and realize my brain might not rest easy without distorting the problem into something I can understand, but if I were to go about this strategy throughout life, then I'd be no different than everyone else.

I hope your forum never dies, and other people can enjoy the puzzle. My favorite line was something like... "yeah, I solved it the same way but came to a different conclusion"!

"Maybe you expect me to explain quantum mechanics by metaphors, saying it's just like billiard balls, or it's just like little springs, or magnets covered with glue, or something like that. But I won't, because if I did, I'd be lying. Metaphors like that only take you so far before they break down, and they ALL break down, so it's best to just avoid them from the start."

We all try to simplify things to get our heads around them, that's not the sign of a stupid man, that's the sign of wisdom. That's how grandmasters see a 10 move combination in a split second. But then one day you run into something like the Envelopes paradox, or quantum mechanics, or a Sunday puzzle at Chessgames, in which the simplified metaphors don't help at all. That's when you have to roll up your sleeves and brace yourself.

I'm a little confused by Krabbe's simulation though. When I did it, I discovered that everything revolves around the question of what the natural percentage of games are drawn. Krabbe doesn't mention this variable whatsoever, but I suspect that he probably used the 60% figure that was used in the google posting http://groups.google.com/group/rec....

A draw rate of 60% sounds very reasonable, but it has to be stressed that a tiny change in this variable changes the final conclusion.