Some funky paradox...

[RocketManJames]

So, Tom... I'm trying to follow this a bit better.

You have provided me with an increasing function that maps Reals to [0,1]... the range of your function is:

(.5 - PI/8) to (.5 + PI/8); or approximately +0.1073 to +0.8927.

So, this function can map every real number to exactly one number in that range. An input of 0 yields +0.500, which is the center point of this range.

So essentially, if you're playing this game with A and B, and you see that A is greater than 0, then you should always keep A. And, if A is less than 0, you should switch. And, if A = 0, then you're just as well off either keeping or switching.

Interesting... so it's really not 50/50. Just seems like you're creating information out of nowhere, but maybe I'm mistaken. Can you try (if possible) to explain this further in layman's terms (using some non-mathematical conceptual reasoning)? Because, intuitively, it sure seems like a 50/50 proposition.

And I have a follow-up question...

If 0 is the cut-off point (using your mapping function), it sort of makes sense, because you'd think this is the midpoint of all Reals. What if the A and B were strictly non-negative Reals? Does this change anything? And what value is the "midpoint" if this is the new game?

-RMJ

[Leo99]

Good follow up RMJ,

I guess in simple terms you don't need any advanced math. If A is greater than 0 you keep it. If A is less than zero you switch. What does happen when you don't have a middle point?

[TomCollins]

This is not what I said. I said you change a probability of the time. If you change always at 0, you are at a 50-50 proposition. You change (in the long run) f(x) number of times. So for 0, you change half of the time. for 1, you change slightly less than 50% of the time, and for 1 million, you change about 20% of the time. Follow the math and it works. For your followup, it still does not matter where I am more likely to switch than not, just that the larger number I have, the less likely it is that I switch.

[RocketManJames]

I got it now (I think). So you map the reals using an increasing function to [0,1]... then this new [0,1] is like a probability map. So, it tells you to keep the number with higher probability the higher it is. Thus, the bigger A is, the more likely you are to keep it.

I think this is making sense now. Neat. I would never have imagined that a better than 50/50 strategy existed.

-RMJ

[benthehen]

[ QUOTE ]
So say you randomly choose one of the two numbers. Suppose you chose x, and x > y.

Then you would stay f(x) of the time, and switch 1-f(x) of the time.

Your probability of being right would then be f(x) of the time.

If you chose y, you would switch (1-f(y)) of the time.

So you will end up being right .5f(x) -.5f(y) + .5.

Since x is bigger than y, we know that f(x) > f(y). So your probability of being correct is .5(f(x)-f(y)) + .5.

Since f(x) - f(y) > 0 for all x > y, p(correct guess) > .5.

QED

[/ QUOTE ]

I followed everything except for the switching part. Suppose I'm given x, how do I switch 1-f(x) of the time? I have to choose one or the other, there is no choice for f(x) of one and 1-f(x) of the other.

[Piz0wn0reD!!!!!!]

This thread always gets long.....

[Leo99]

I don't get it. The middle point concept I understand. I don't understand the concept of the "the bigger A is." How big is big when you're dealing with infinity? If you only include positive numbers so you're dealing with 1 to infinity. Whatever A is, there is a finite number of numbers smaller and an infinite number of numbers bigger.

[TomCollins]

One way to do this is to choose another random number between 0 and 1, and if it is bigger, switch, if not, don't switch. So you are varying your strategy based on a random number. This, surprisingly, makes the strategy work. This also, by the way is the same as picking another random number without bounds and switching if A is smaller than it. Because you can take f(C) and if f(C)> f(A), switch, which is exactly what I said to begin with. My point is, if you run this test a million times, you will get it above 50% correct. You can switch 50% of the time if A is 0. You just have to randomly decide.

This is no different than TOP which has you bluffing x% of the time. You either bluff a particular hand or you don't. But in the long run, you bluff x% of the time, based on some random event (ideally). This provides an optimial strategy, much like in the number switching game.

[TomCollins]

With an increasing function, the larger A is, the larger f(A) becomes. There is no concept of "big", just "bigger". Every pair of real numbers has one that is "bigger".

And no, there are not a finite number of numbers less than A, even if A is bound by 1. We are talking about real numbers here. The set of integers 1 to infinity is smaller than the set of reals. I forget the name of this, but they are different sizes. Anyone with a math background is recommended to give the name of this.

[Leo99]

Density?