Naked before God: HELP ME SKLANSKY!

[RubbleRobble]

I'm not sure how quickly Mr. Sklansky and Mr. Malmuth find/respond to post directed mostly at them, but I have a question in the realm of probability theory that I'm afraid is a little too tricky for us mere mortals...

This question was proposed to me by my instructor in probability theory at the University of Maryland, Professor Michael Brin.

You die and arrive at the gates of heaven. St. Peter informs you that, due to budget cuts, people no longer are admitted to heaven based on the merit of good deeds. The new system goes like this:

God has a distribution of random numbers in mind. (No psychology here folks... we cannot deduce the type of distribution, i.e. uniform, gamma, normal, etc; nor can we deduce the nature of the parameters.) From this distribution of numbers, God takes a number, X, and gives it to you to look at, touch, hug, etc. You KNOW the value of X. God then takes another number from the same distribution, called Y, and puts it in his pocket (I forgot to mention... God has a pocket).

In order to enter heaven, you must 'guess' whether the number He handed to you is greater than or less than Y, God's number, which resides in his pocket.

Stressing again that the only things you have to tackle this problem are your wits/knowledge of probability and statistics plus the knowledge of the value of X. GIVE A STRATEGY THAT WILL GET YOU INTO HEAVEN WITH PROBABILITY STRICTLY GREATER THAN 50%.

I would greatly appreciate the help of Mr. Sklansky or Mr. Malmuth (or anyone that can competently and succinctly explain the solution to this 'Naked before God' problem.) Class is at 1 pm EST on Weds, so a solution by then would be awesome, but I'll take what I can get... Thanks in advance.

[chopchoi]

Since we cannot deduce the nature of the parameters, we must assume that they are infinite. Thus, if God gives you a positive number, then you should guess that his number is less than yours, since there are (in theory) more numbers less than yours than there are numbers greater than yours. Likewise, if God gives you a negative number, then you should guess that God's number is greater than yours.

Or, if all of the numbers are positive, then you should always guesss that god's number is greater than yours, since the distance between your number and zero will always be shorter than the other way.

[jason1990]

[ QUOTE ]
From this distribution of numbers, God takes a number, X...

You KNOW the value of X. God then takes another number from the same distribution, called Y...

In order to enter heaven, you must 'guess' whether the number He handed to you is greater than or less than Y

[/ QUOTE ]
You must be assuming that God's distribution is continuous. If it were allowed to be discrete, then God could choose the distribution satisfying P(X=0)=P(X=1)=1/2. If He did that, then 50% of the time, X and Y would be equal. Since you are only apparently allowed to guess higher or lower, you would automatically lose in those cases. Thus, you could never get into heaven more than 50% of the time, no matter what you did.

If we can assume that X and Y are never equal, then you can find your solution in this post, which summarizes someone else's solution that you can read about in that long, long thread.

[RubbleRobble]

Thanks to chop and jason for responses. One edit that I should make is, yes the distribution is continuous, I neglected to mention that fact... sowwy.

However, I looked at the post and replies on the thread indicated by jason, and my first reaction (though I could always be wrong here) is to say that this in insufficient to answer the 'naked before god' problem.

First, the question posed by RMJ in his thread is trying deciding to stay on/switch from a number "A" taken from the list of all real numbers. That you are taking two numbers from the distribution of all real numbers implies knowledge of the distribution. Tomcollins shows that a solution can be found with P(success)>0.5 (and I was able to understand his proof there), but I can't understand what his actual strategy is. What if the distribution is f(x)=x when 0<=x<=1, f(x)=(-x)+2 when 1<=x<=2 and 0 otherwise (a ligitimate pdf); and the number you are given by God is 0.5? 0.5 is a positive number, so tomcollin's solution would have you not switching (in the cotext of this problem, this means telling God that your number is bigger). Again, I could be wrong, but if you follow this strategy, chances are you won't be winning.

Abstract problem I got here, maybe an explanation tailored to the setting of this problem would help me understand things better... then again maybe I've already been given the correct answer, but im too dumb to know it...

[gasgod]

No fancy math here.

The problem is that an integer cannot be drawn at random from an infinite array. To show this, let's consider how such a choice would be made. In order for the target number to be random, the Nth digit must be random (for all possible values of N). So, we choose a random digit for the Units digit, another random digit for the tens digit, and continue the process for all possible digits. Now, clearly, this process cannot terminate. Any time we assert that we have now chosen a particular integer, it must necessarily consist of a finite number of digits. Let's call that number D. So, the D+1th digit must be zero, and this violates the condition that the number must be random.

Therefore, the range must be finite. Since we are not told what the range is, we have only a 50/50 guess. I hope you brought a coin.


GG

[jason1990]

Okay, let me try to get my head on straight before I reply. The idea of TomCollins is this: take an appropriate f (like the one I linked you to). Look at X and compute f(X). Then, with probability f(X), tell God that you think X is bigger than Y. One way to facilitate this is to choose a random number U uniformly on [0,1]. Then, if U < f(X), you tell God you think X is bigger than Y. Otherwise, of course, you guess that Y is bigger than X. The probability you will be right is

P(U < f(X), Y < X) + P(U > f(X), Y > X).

Conditioned on X, this becomes

P(U < f(X))P(Y < X) + P(U > f(X))P(Y > X)
= f(X)F(X) + (1 - f(X))(1 - F(X)),

where F is the cumulative distribution function of God's chosen distribution. Since we conditioned on X, we need to take expectations, so the probability you will be right is

E[f(X)F(X) + (1 - f(X))(1 - F(X))].

What you have to show is that this will always be strictly greater than 1/2, regardless of F.

[jason1990]

God is not picking integers. His random number has a continuous distribution. In particular, for any integer n, P(X=n)=0.

[gasgod]

[ QUOTE ]
God is not picking integers. His random number has a continuous distribution. In particular, for any integer n, P(X=n)=0.

[/ QUOTE ]

Still, cannot the same reasoning be extended to all real numbers? Am I missing something?


GG

[Guruman]

this may be too simplistic, but if we're assuming a finite group positive integers here, I have a theory:

always pick x<y

the reasoning is that if the number that you possess occupies the space between zero and the next number above yours and the range is an odd number, then there will be more room on the other side. If the range is an even number, you have a coin flip.

ex: God's range is 1-9. here's the table

x=1 x<y
x=2 x<y
x=3 x<y
x=4 x<y
x=5 x<y
x=6 x>y
x=7 x>y
x=8 x>y
x=9 x>y

five times we have a greater probability of being correct, where as four time we do not.

It's a guess [img]/images/graemlins/confused.gif[/img]

[pzhon]

[ QUOTE ]

The problem is that an integer cannot be drawn at random from an infinite array.

[/ QUOTE ]
Sorry, but you completely missed the point.

Random does not mean uniformly distributed. Your arguments assume that it does.

There is no problem with choosing a random integer, or a random real number. There is no way to choose a random integer so that each is equally likely, but it was never assumed that the distribution is like that.