game with a not too obvious solution

[well]

[ QUOTE ]
switch with probability f(x) where f(x) is strictly decreasing, 1>f(x)>0, and x is the amount written. For example f(x) = 1/(1+exp(x)).

[/ QUOTE ]

That's a correct solution!
Let me prove it for the ones who don't believe it.

How would you describe a strategy for I?
The only dicision to be made is wether to switch or not, and the only information you have is the number z, the number on your note.
Since this problem is not solvable for pure strategies, I must have a probability for switching based on the number z.
Hence my strategie is a function f(x) from (-Inf,+Inf) to [0,1].

Denote the two numbers by x and y, assuming x<y.
My first pick will be either one of them, both with probability .5.
I win in these two situations:
1. I pick x and I choose not to switch, probability: .5*(1-f(x).
2. I pick y and I choose to switch, probability: .5*f(y).
Addition of these probabilities results in

probability I win equals (1+f(x)-f(y))/2.

For a strictly positive EV I need my winning probability to be over .5.

(1+f(x)-f(y))/2 > 1/2 <=>
f(x)-f(y)>0 <=>
f(x)>f(y)

So any function with f(x)>f(y) for all x>y and 0<=f(x)<=1 for all x in (-Inf,Inf).

f(x) = 1/(1+exp(x)) indeed is such a function.
The function I came up with was f(x)=.5-arctan(x)/pi.

Note that on most opponents you could inprove the effect of your function if you scaled it a little.
For instance, in these functions, replace x by x/m for say m=10, 20... Have a look at it.

Next Time.

[Copernicus]

I believe this is a dead even strategy, not +EV (using the exp version). Ive tested it for 50 number combinations, and it is always .5. Give me a scenario where it is +EV.

[PokerNoob]

How is this different than two signed random numbers of a given precision being generated? Always switch if your number is negative. Reason: there exist more numbers on the the positive side of the number line plus those between the chosen negative number and zero than negative number to negative bound, even if the bound is infinte.

Or did you forget to mention that we must play this game an infinite number of times?

[Copernicus]

There are not more numbers on the positive side...infinite is infinite, there is no "more infinite"

[PokerNoob]

Okay, how about this. You can write down any real number between 0 and +infinity and any rational number between 0 and -infinity.

Seriously, how does "switch with probability" have any meaning? Its a binary choice. You can't .7th switch.

[well]

[ QUOTE ]
Okay, how about this. You can write down any real number between 0 and +infinity and any rational number between 0 and -infinity.

Seriously, how does "switch with probability" have any meaning? Its a binary choice. You can't .7th switch.

[/ QUOTE ]

Switch with probablitiy p means that there is a probability p for the event that you switch, and a probablity 1-p for the event that you don't.
A realization of this could be like: take p=f(x).
Then use a random number generator to generate a random number between 0 and 1 (use Unif(0,1)), denoted by w.
If w<=p then switch, otherwise don't!

I hope I have been clear.

Next Time.

[squiffy]

Well, can you explain your solution in layman's terms. I don't understand the functions or numbers. But Copernicus's observations seem intuitively obvious to me. You have no information about how your Opponent is selecting the two numbers. So they may as well be two random numbers. You are only seeing one of them. Which one you choose seems completely random. And you have no basis for determining what relation the number you select bears to the one you have not selected. If the numbers were finite perhaps it would be a different story. But didn't you specify positive and negative infinity.

[Copernicus]

much clearer, and it works. I think you can use the same technique to contact the dead.

[PokerNoob]

[ QUOTE ]
much clearer, and it works. I think you can use the same technique to contact the dead.

[/ QUOTE ]

Or at least Schrödinger's Cat.

[Copernicus]

The reason it works is that the technique for choosing is biased...it produces a decision to muck with a higher frequency when the card is low, and a decision to keep with a higher frequency when the card is high. (though there is a technical problem in evaluating numbers in all but a fairly narrow range).

Heres an example:

<font class="small">Code:</font><hr /><pre> x f(x)

8 0.00033535
1 0.268941421

pick
8 1
1-f(x) keep 0.99966465 0.731058579
f(x) discard 0.00033535 0.268941421

winning prob (top left bot right): 0.634303036
losing prob (bot left top right): 0.365696964

</pre><hr />

(The sums of the diagonals have to be divided by 2 to keep the total probability at 1).