Some funky paradox...

[RocketManJames]

A friend brought this up, and I am pretty sure I know why this doesn't work, but I wanted to throw this out at all of you.

Two distinct real numbers are chosen randomly from the universe of real numbers. So we have number A and number B.

Now, your goal is to pick the bigger number. You are allowed to look at A... then you must decide if you want to keep it or take number B instead.

This is obviously a 50/50 proposition.

Now, here's a strategy that seems to give you better than 50/50. It shouldn't work, but it's sort of clever.

Choose any number... let's call it C.

Now, when you look at A, you compare it to C. If A is bigger than C, take A, otherwise take B.

There is some probability P_bigger that your number C is bigger than A and B. There is some P_smaller that your number C is smaller than both A and B. There is some P_between that C is between A and B.

So, using this strategy, the probability that you pick the bigger number is:

P_smaller * 0.5 + P_bigger * 0.5 + P_between * 1.00

which is

P_between + (1 - P_between) * 0.5

So, only if P_between is 0, do we end up with 50/50. If P_between is anything greater than 0, then we have better than 50% chance at picking the bigger number.

In Edit: In case it wasn't clear... the reason why P_between is 100% probability, is that if your number C is between A and B, then you are guaranteed to choose the bigger number following the above strategy as A relates to C.

This can't be right, because you cannot create information out of non-information.

I am pretty sure why this doesn't work has to do with the fact that the random numbers A and B are chosen from all real numbers. If it were chosen from a bounded set of numbers then the strategy does give you a boost in probability.

Any thoughts?

-RMJ

[TomCollins]

This is not a paradox, but it has been discussed on this forum before.

The problem is you are assuming it is a 50-50 proposition. It is not.

[TomCollins]

I only have a few minutes, but I can put a math behind it showing that it is not 50-50.

Suppose I have some increasing function that maps R -> [0,1]. I can't think of any off the top of my head, but at least grant me one exists.

So for f(x) > f(y) iff x > y.

If I find f(A) and keep my number that probability of the time. I will come out ahead.

I will let others elaborate (ok, I just have to get to lunch). I'll have the rest of the proof later.

[TomCollins]

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

[Jman28]

[ QUOTE ]
Since x is bigger than y, we know that f(x) > f(y).

[/ QUOTE ]

I think this is a mistake.

You're saying that since the random number X is greater than the random number Y, the probability of you picking and staying with X is greater than the probability of you picking and staying with Y?

I don't see how you proved this, but I may be wrong.

-Jman28

[Jman28]

[ QUOTE ]
So, only if P_between is 0, do we end up with 50/50. If P_between is anything greater than 0, then we have better than 50% chance at picking the bigger number.


[/ QUOTE ]

This statement is true. I think here's the solution...

P_between = 0.

Hear me out on this one...

Let the B - A = Z

The probability of C falling between A and B = (Z / infinity)

Z/infinity = 0

So, P_between = 0

I think this is right... Unless we say that Z = infinity, which it sort of does?

-Jman28

[RocketManJames]

[ QUOTE ]

This statement is true. I think here's the solution...

P_between = 0.

-Jman28

[/ QUOTE ]

This is how I figured it also... that P_between is 0, because of the infinity argument. I don't follow Tom Collins' proof. I guess if he were to provide the mapping function for R to [0,1], I'd be more inclined to believe it. As it stands, it seems that you've got a "range" for all Reals, when there really isn't one. You're essentially creating information out of thin air.

But, that said, I'm not in a position to judge its correctness, so I'll let others discuss.

-RMJ

[TomCollins]

You can and are wrong.

But its ok, I wasn't as clear as I could have been.

Suppose you have two numbers A and B. One of these two is larger, we will call it x, and the smaller one y. You have A, which is either x or y. Since A and B are equally likely to be x or y, we have a 50% chance at each. First we must find an increasing function that maps from R-> [0,1]. Since you would not take my word on it, I will find one.

So the function I will choose is f(x)=tan^(-1)(x)/4+.5. This function is increasing, defined for all real numbers, and has a range within [0,1]. This is just one example, but it any function that is increasing and maps from R->[0,1] works.

So back to the real proof. We will have x 50% of the time, and y 50% of the time. (If it were any more or any less, then the proof is done, since we would simply stay with the first number we picked if it were more, or always switch if it were less). So we take f(A) and stay that much of the time. (We could take a random number between 0 and 1, and if it is greater than f(A) we switch, and less we stay). So 50% of the time f(A) = f(x), and 50% of the time, f(A) = f(y). Again, this MUST be true, otherwise the proof is trivial. So when we end up with x, we end up correct f(x) of the time, and wrong 1-f(x) of the time. If we end up with y, we end up switching 1-f(y) and being correct, and are wrong f(y) of the time. So our odds of being correct come down to .5f(x)+.5-.5f(y). Simplified this is .5+ .5(f(x)-f(y)). We know f(x) > f(y) since f is increasing, so P(correct) > .5.

QED

[TomCollins]

This argument doesn't work because there are an infinite amount of real numbers inside of any range. So we are really dividing an infinite number by an infinite number. This is why it is possible to map all real numbers to a range of numbers R-> [0,1].

[Leo99]

You choose A. Now there is an infinite number of real numbers smaller and an infinite number of real numbers bigger. I don't see how probability applies when you're dealing with infinite numbers like this.

By the way, what do you guys do for a living that you remember all this math stuff?