More head-busting probability

[kirisutogomen]

rifmeister posts a wonderful little problem here, and it reminded me of an entirely different wacko problem posed on the xkcd balrog bolgia blog "blag". I don't particularly recommend slogging through all the comments, as a reasonable number of them are useless or worse, but the link to the Wikipedia article on the "two envelopes problem" is certainly worth a look.

Comments

[mathhobbit.livejournal.com]

But is it equivalent to the Monte Hall problem?

http://en.wikipedia.org/wiki/Monty_Hall_problem

[rifmeister.livejournal.com]

Classic! That has always been one of my favorites. I actually consider Randall's explanation [in the first comment] to be one of the better explanations I've read.

[treptoplax.livejournal.com]

I think Randall's strategy can be made a little less abstract, and simpler:

I pick any distribution that has non-zero probability over any subset of the reals. Use it to pick a number. If the revealed number is higher, guess lower, and vice-versa. Ta-da! You win 50% + epsilon/2.

I think. There's still just a whiff of paradox there, but maybe that's just me.

[treptoplax.livejournal.com]

(Ok, I actually mean non-zero probability for the interval between any two distinct positive real numebrs, but you get the point...)

[firstfrost.livejournal.com]

I think the question isn't asking about whether there is a strategy that gives you better than 50% in the long term, but one that gives you better than 50% every time.

(For example, if Alice's strategy is to choose two very huge numbers that happen to both be larger than your two fenceposts, then you have a 50% chance.)

[treptoplax.livejournal.com]

No, no matter how large her numbers are, there's a chance I'll pick a number between them.

[firstfrost.livejournal.com]

Oh, I see, I'm misinterpreting "over any subset of the reals" as "over some particular subset of the reals that I choose" and you mean "over all reals"?

(Darned English...)

Or maybe I'm still confused, which is also quite possible.

[fredrickegerman.livejournal.com]

I totally bought the explanation, until I asked myself why the whole business wasn't equivalent to guessing the sign of a single randomly-chosen real number. There's an interesting thing about dense-ness of uncountable sets in there somewhere. It's not like the cardinalities involved are actually different, even...

[fredrickegerman.livejournal.com]

Actually, strike "randomly-chosen" in the above.

[treptoplax.livejournal.com]

Ow, my head. Although... I'm not 100% clear it's equivalent. They did get the first number, which is non-zero information. "Guess the sign or a single real number" is equivalent to "I picked two numbers, is A or B larger", not "I picked [number] and another number, which is larger."

I'm still not entirely satisfied, though.

[fredrickegerman.livejournal.com]

Yeah, I then tried to back-fill by picking two numbers at random from (0,1). But of course that's just like flipping a coin!

Except for one corner case, which also occurs in the original problem: you can pick exactly the number chosen (with probability 0, of course). Which leads me to wonder if we're not in one of these "integrate a function that's everywhere 0 and get 1" kinds of situations.