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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1237629&page=&view=&s b=5&o=&vc=1), 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 /images/graemlins/confused.gif

[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.

[Mr Mojo Risin]

[ QUOTE ]
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.

[/ QUOTE ]

This theory is very flawed in my opinion because there are an infinate amount of negitive numbers and an infinate amunt of positive numbers. Due to the fact that the number line goes on forever in either direction, the number being negitive or positive should have no effect on the outcome.

[RubbleRobble]

Jason,

Thank you VERY much for the help with this problem. I was able to follow the logic in your more recent posts and the story checks out.

Just a point of clarification though... When tomcollins has that we "must find an increasing function that maps from R-> [0,1]," what does it mean to map R-> [0,1]? He says later that his chosen function "has a range within [0,1]." What does that mean?

I want to have a commanding grasp on all points of the proof before I speak up on this tomorrow. Thanks.

[jason1990]

It just means that for any real number x, 0 <= f(x) <= 1. That, and needing f to be continuous and strictly increasing are all you should need (I think) for this to work. You may need even less than that.

By the way, the requirement that f take values in [0,1] is obvious, because f(X) is supposed to be the probability that you guess X>Y.

[M127]

Usually I just lurk here, but I figured I'd humbly offer what seems like one possible solution. Here goes:

Naively, we'd like to just guess X > Y when X is positive, X < Y when X is negative. In fact, it turns out that this strategy almost works by itself - to see why, notice there are four possibilities:

1. X,Y>0 : when this happens, our strategy works 50% of the time, since X and Y are identically distributed.

2. X>0, Y<0: here, we we always get in, since in this case we guess X>Y correctly

3. X<0, Y>0: again, our strategy always works here, since it causes us to guess X<Y

4. X, Y<0: in this case, our strategy works 50% of the time, for the same reason as case 1.



Its not hard to see intuitively (and pretty easy to prove numerically) that unless cases 2 and 3 never happen, our probability of getting the right answer will be strictly > .5. Also, when cases 2 and 3 are impossible (ie when all God's numbers are positive or all are negative), our strategy still works 50% of the time. So our answer is "almost" right.

Now lets call the above strategy S(0). Its pretty easy to see that the number 0 is irrelevant to the above discussion, and if we let r be any real number, and let S(r) be the strategy "say X>Y when X>r and say X<Y if X<r", then S(r) will work strictly > 50% of the time, unless all of God's numbers are greater than r or all of them are less than r. In that case, S(r) would work exactly 50% of the time.

Now for every real number r, we have a strategy S(r) that works except in a few situtations, and even then it comes close. So we take a countable dense set of reals - say the rationals. List them in some infinite sequence, say r1, r2, r3....


Now flip a coin repeatedly until it comes up tails (or use some other method) to pick a random positive integer, in such a way that every positive integer has a positive probability of being picked. Then use the corresponding strategy. (so if you picked 7, use the strategy S(r7)). This new Meta strategy is guaranteed to give you strictly greater than 50% probability of being right.

The reason this works is that any distribution God picks (we are assuming it is continuous) will have to have an interval [m1, m2] containing the middle 50% of the values taken on by X. (ie X lies in the interval 50% of the time, below the intercal 25%, above 25% - i think m1 and m3 are called the 1st and third quartile or something - been a long time since stat) Since the set (r1, r2,r3...) is dense, some rn lies in [m1,m2], meaning X can be smaller or larger than rn, and so S(rn) works strictly more than 50% of the time.

Since at least 1 strategy will work strictly greater than 50% of the time, and the rest of the strategies work > or = to 50%, since you choose every strategy with positive probability, you will be in good shape. Sorry for the long, rambling nature of this post, hope the answer makes sense.

[jason1990]

This is nice and I think I've seen it before, if not something very similar. What you've done during this process is to construct a random variable which is rational almost surely and the probability of each rational is positive.

I think that if we take the f in TomCollins's strategy to be the CDF of your random variable, then we get a strategy which is equivalent to yours.

[XChamp]

gasgod,

You are forgetting that any number never terminates.

The number 2 is really 2.00000000000000...


Likewise 1.5708 is really 1.5708000000000....

[Kristian]

[ QUOTE ]
1. X,Y>0 : when this happens, our strategy works 50% of the time, since X and Y are identically distributed.

2. X>0, Y<0: here, we we always get in, since in this case we guess X>Y correctly

3. X<0, Y>0: again, our strategy always works here, since it causes us to guess X<Y

4. X, Y<0: in this case, our strategy works 50% of the time, for the same reason as case 1.

Its not hard to see intuitively (and pretty easy to prove numerically) that unless cases 2 and 3 never happen, our probability of getting the right answer will be strictly > .5.

[/ QUOTE ]

This argument is decievingly nice, but ultimately wrong.

It is not easy to see however, since it has to do with the very nature of probability. The problem is that the argument presumes that there is a positive probability that X>0 and y<0 or opposite. While this SEEMS very reasonable it is actually impossible to consider this probability. The reason for this is that the distribution of X and Y is unknown, and you would need this to be able to use P(X>0, Y<0) in a logical argument.
Thus, also the argument in the linked post by Tom Collins is wrong, sorry Tom.

The only correct answer to the riddle/paradox is that it is in fact impossible to devise a good strategy, since you have no information at all.

[Kristian]

The only correct answer to the riddle/paradox is that it is in fact impossible to devise a good strategy, since you have no information at all.

The reason why otherwise smart people THINK they have a solution is that they don't take into account whether certain probabilities are well defined, which they are not in this case, since the distribution God uses is completely unknown.

[jason1990]

[ QUOTE ]
[ QUOTE ]
1. X,Y>0 : when this happens, our strategy works 50% of the time, since X and Y are identically distributed.

2. X>0, Y<0: here, we we always get in, since in this case we guess X>Y correctly

3. X<0, Y>0: again, our strategy always works here, since it causes us to guess X<Y

4. X, Y<0: in this case, our strategy works 50% of the time, for the same reason as case 1.

Its not hard to see intuitively (and pretty easy to prove numerically) that unless cases 2 and 3 never happen, our probability of getting the right answer will be strictly > .5.

[/ QUOTE ]

This argument is decievingly nice, but ultimately wrong.

It is not easy to see however, since it has to do with the very nature of probability. The problem is that the argument presumes that there is a positive probability that X>0 and y<0 or opposite. While this SEEMS very reasonable it is actually impossible to consider this probability. The reason for this is that the distribution of X and Y is unknown, and you would need this to be able to use P(X>0, Y<0) in a logical argument.
Thus, also the argument in the linked post by Tom Collins is wrong, sorry Tom.

The only correct answer to the riddle/paradox is that it is in fact impossible to devise a good strategy, since you have no information at all.

[/ QUOTE ]
/images/graemlins/smile.gif I guess this is why that other thread was so long.

It is not impossible to consider P(X>0,Y<0). You can consider it all day. It's a number. It exists. It is between 0 and 1. M127 already said that if that number is 0, then the strategy S(0) doesn't work, but if that number is positive, then it does work. You don't need to know the specific value of that number to use it in a logical argument and prove these statements.

And I guess you just joined the "I don't believe TomCollins" club, which as you can see from the other thread, is quite large indeed. But nonetheless, his solution makes your probability of winning greater than 50%.

However, given a number bigger than 50%, say 60%, is there a strategy that lets you win 60% of the time? No. How about 50.0001%? Still no. If you fix any p>.5, then you will not be able to find a strategy that guarantees you a p probability of winning. This is because if God knows the f you chose, then for any positive epsilon, God can choose a distribution which makes your probability of winning less than .5 + epsilon. This is the effect of having "no information."

But God can never choose a distribution which makes your probability of winning less than or equal to .5. This is why your probability of winning is always bigger than 50%.

(Looking at what I just wrote, I now think the proper place to debate this is in one of Sklansky's psychology posts.)

[reubenf]

[ QUOTE ]
If we can assume that X and Y are never equal

[/ QUOTE ]

Is it sufficient to assume that the probability that x=y is zero?

[jason1990]

[ QUOTE ]
[ QUOTE ]
If we can assume that X and Y are never equal

[/ QUOTE ]

Is it sufficient to assume that the probability that x=y is zero?

[/ QUOTE ]
I believe it is necessary and sufficient to have the CDF of X be continuous, which happens if and only if f(y):=P(X=y)=0 for all y. Do a little logical song and dance to show that f(y)=0 for all y if and only if f(Y)=0 almost surely. Then note that since f(Y) is nonnegative, f(Y)=0 almost surely if and only if E[f(Y)]=0. Finally, observe that

P(X=Y) = E[P(X=Y|Y)] = E[f(Y)].

So the CDF of X is continuous if and only if P(X=Y)=0.

[pzhon]

[ QUOTE ]
The only correct answer to the riddle/paradox is that it is in fact impossible to devise a good strategy, since you have no information at all.

[/ QUOTE ]
False. You can't give a uniform bound greater than 50%, but strategies have been presented that win more than 50% of the time against any continuous distribution.

[ QUOTE ]
The reason why otherwise smart people THINK they have a solution is that they don't take into account whether certain probabilities are well defined...

[/ QUOTE ]
You are disagreeing with multiple professional mathematicians who are telling you that the probabilities are well-defined, and that this problem is solved. Perhaps you should think about it a bit more.

[Kristian]

[ QUOTE ]
...strategies have been presented that win more than 50% of the time against any continuous distribution.


[/ QUOTE ]

It is very difficult for me to believe this. The space of continous distributions is so unbelievably large that it is impossible to make such a statement. I am also quite sure that the professional mathematicians you mention can confirm that they make some additional assumptions before presenting their solution.

[PairTheBoard]

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

If you mean by "at random" that all integers would be equally likely to be chosen then of course you are right. You don't need such a complicated argument though. Just notice that the sum of the equal probabilities for all the integers exceeds 1. This is the kind of hidden bogus assumption people make in the Two Envelope Problem.

Rubble is talking about a probability distribution though. Certainly, a probability distribution exists for the integers. Take 1/2^n for example. For Rubble's problem he restricts it to continuous probabilty distribtions though.

PairTheBoard

[PairTheBoard]

Jason, My intuition tells me there might need to be some kind of tightness restrictions on God's distribution. Suppose X has no finite Expectation. Are you sure in that case that the Expectation you take in your calculation is actually finite?

PairTheBoard

[jason1990]

Both f and F are continuous (hence measurable), nonnegative functions bounded by 1, so the expectation certainly exists, regardless of the distribution of X. If you want to think of X as having a density p(x), then having no finite expectation means int{xp(x)} does not exist. But int{p(x)}=1, so E[G(X)]=int{G(x)p(x)} exists whenever G is a bounded function.

[pzhon]

[ QUOTE ]
[ QUOTE ]
...strategies have been presented that win more than 50% of the time against any continuous distribution.


[/ QUOTE ]
It is very difficult for me to believe this.

[/ QUOTE ]
It is not something you were asked to take on faith. Logical arguments were given.

[ QUOTE ]
The space of continous distributions is so unbelievably large that it is impossible to make such a statement.

[/ QUOTE ]
No, it isn't impossible to make or prove such a statement. The space is not unbelievably large, either.

[ QUOTE ]
I am also quite sure that the professional mathematicians you mention can confirm that they make some additional assumptions before presenting their solution.

[/ QUOTE ]
I suggest you be less sure of things that are not true. The only assumption necessary is that the distribution is continuous (the CDF is continuous), which I mentioned above.

There many real numbers. It is still easy to prove that for any real number x, x^2+x+1>0. No additional assumptions are necessary there, either.

[PairTheBoard]

OK!

[PairTheBoard]

I accept the result. Now can you explain in intuitive terms what's going on in the solution? The proof convinces me. But I'm never really satisfied until I have a feel for some principle at play in making it work. Thanks.

PairTheBoard

[pzhon]

[ QUOTE ]
I accept the result. Now can you explain in intuitive terms what's going on in the solution?

[/ QUOTE ]
There are two approaches given/referenced in this thread. They are equivalent, but I think the way described by M127 is more intuitive. I'll quote the start:

<ul type="square">
Naively, we'd just like to guess X &gt; Y when X is positive, X &lt; Y when X is negative. In fact, it turns out that this strategy almost works by itself - to see why, notice there are four possibilities:

1. X,Y&gt;0 : when this happens, our strategy works 50% of the time, since X and Y are identically distributed.

2. X&gt;0, Y&lt;0: here, we we always get in, since in this case we guess X&gt;Y correctly

3. X&lt;0, Y&gt;0: again, our strategy always works here, since it causes us to guess X&lt;Y

4. X, Y&lt;0: in this case, our strategy works 50% of the time, for the same reason as case 1.



Its not hard to see intuitively (and pretty easy to prove numerically) that unless cases 2 and 3 never happen, our probability of getting the right answer will be strictly &gt; .5. Also, when cases 2 and 3 are impossible (ie when all God's numbers are positive or all are negative), our strategy still works 50% of the time. So our answer is "almost" right.
[/list]
More explicitly, let p=P(X&gt;0). This strategy works with probability 1/2 + p(1-p). That is only 1/2 when p=0 or 1. When 0&lt;p&lt;1, this strategy works with probability greater than 1/2.

It is worth thinking about this until it is clear that this strategy works between 50% and 75% of the time (inclusive).

<ul type="square">
Now lets call the above strategy S(0). Its pretty easy to see that the number 0 is irrelevant to the above discussion, and if we let r be any real number, and let S(r) be the strategy "say X&gt;Y when X&gt;r and say X&lt;Y if X&lt;r", then S(r) will work strictly &gt; 50% of the time, unless all of God's numbers are greater than r or all of them are less than r. In that case, S(r) would work exactly 50% of the time.
[/list]
S(r) will always work at least 50% of the time. We want to ensure that our strategy works strictly more than 50% of the time, so we would like to vary r so that there is a chance it is in the middle of God's distribution (neither above 100% of God's numbers nor below 100%).

Let R be a random variable from a distribution that has positive probability on any interval, e.g., choose r from a normal distribution, or a Cauchy distribution (density dx/pi 1/(x^2+1)). Then R has a positive probability of being in the middle of God's distribution, so S(R) works strictly greater than 50% of the time.

Instead of generating a random number from that distribution, we can just skip to the consequence: If you are shown x, you will say X is greater than Y any time R is less than x. That means you switch with probability P(R&lt;x). If you are using the Cauchy distribution, P(R&lt;x) = 1/2 + arctan(x)/pi. So, instead of generating R, you can declare X is greater than Y 1/2 + arctan(x)/pi of the time, and declare X is less than Y 1/2 - artcan(x)/pi of the time. This strategy guesses correctly more than 50% of the time against any continuous distribution.

[PairTheBoard]

Thanks. That's what I was looking for.

PairTheBoard

[JoshuaD]

Wow, I thought that when I first looked at the thread, and it seemed way too simple to be right.

[Kristian]

Sorry about the late answer, I have been on a short vacation. In the mean time, I have given the matter some more thought, and also discussed it with an equally geeky friend of mine.
My statement that the strategies posted used undefined probabilities was wrong, and I am sorry about posting stuff that I had not thought through.

[ QUOTE ]
[ QUOTE ]
I am also quite sure that the professional mathematicians you mention can confirm that they make some additional assumptions before presenting their solution.

[/ QUOTE ]
I suggest you be less sure of things that are not true. The only assumption necessary is that the distribution is continuous (the CDF is continuous), which I mentioned above.

[/ QUOTE ]

...but in this matter you are wrong. The problem certainly needs a further assumption in order to devise an above 50% strategy, but the assumption needed is much weaker than I first thought.
What you need to know is the interval of possible outcomes, since every strategy posted requires that you are able to choose a seperation point of the distribution with a positive probability of outcomes on either side of this point. If you don't know this interval, then it is impossible to construct a &gt;50% strategy unless you start considering some sort of 'distribution of possible distributions', and this would very quickly make the problem very boring.
Note however, that it is enough to know that the distribution has positive probabilities on the whole real axis. This would, for instance, be the case with the normal distribution with any parameters. Of course, if there are no bounds on the parameters God chooses for his normal distribution, he can put the probability of success as close to 50% as he likes.

[Kristian]

[ QUOTE ]

It is not impossible to consider P(X&gt;0,Y&lt;0). You can consider it all day. It's a number. It exists. It is between 0 and 1.


[/ QUOTE ]

You are right, sorry /images/graemlins/smile.gif

[ QUOTE ]

But God can never choose a distribution which makes your probability of winning less than or equal to .5. This is why your probability of winning is always bigger than 50%.


[/ QUOTE ]

You are wrong. He can choose 'any' distribution, and you will have no chance at all of finding that crucial seperation point in this distribution if you know nothing further about it.

[pzhon]

[ QUOTE ]
What you need to know is the interval of possible outcomes, since every strategy posted requires that you are able to choose a seperation point of the distribution with a positive probability of outcomes on either side of this point. If you don't know this interval, then it is impossible to construct a &gt;50% strategy unless you start considering some sort of 'distribution of possible distributions', and this would very quickly make the problem very boring.

[/ QUOTE ]
That is still wrong. Let me quote a paragraph from elsewhere in this thread:

<ul type="square">
Let R be a random variable from a distribution that has positive probability on any interval... Then R has a positive probability of being in the middle of God's distribution, so S(R) works strictly greater than 50% of the time. [/list]
Once again, there are strategies that work strictly more than 50% of the time against any continuous distribution. If you don't believe that, try to find a continuous distribution against which guessing X is greater with probability 1/2 + arctan(x)/pi wins at most 50%. Please post such a distribution or retract your objection to the proof.

[Kristian]

[ QUOTE ]

That is still wrong. Let me quote a paragraph from elsewhere in this thread:

<ul type="square">
Let R be a random variable from a distribution that has positive probability on any interval... Then R has a positive probability of being in the middle of God's distribution, so S(R) works strictly greater than 50% of the time. [/list]


[/ QUOTE ]

But here the further assumption has been made that any interval has a possible probability. I agree that that is enough to make a &gt;50% strategy.

[ QUOTE ]

If you don't believe that, try to find a continuous distribution against which guessing X is greater with probability 1/2 + arctan(x)/pi wins at most 50%.


[/ QUOTE ]

I don't understand this, please clarify.

[jason1990]

[ QUOTE ]
[ QUOTE ]

But God can never choose a distribution which makes your probability of winning less than or equal to .5. This is why your probability of winning is always bigger than 50%.


[/ QUOTE ]

You are wrong. He can choose 'any' distribution, and you will have no chance at all of finding that crucial seperation point in this distribution if you know nothing further about it.

[/ QUOTE ]
As pzhon has suggested, you either need to submit to our mathematical prowess /images/graemlins/shocked.gif and admit you are wrong, or "play God" and defeat one of these strategies.

So you are God. I will tell you beforehand what I'm going to do. (As God, you know it anyway.) I am going to take the number you give me, X, and compute 1/2 + arctan(X)/pi. Whatever this number is, call it p, I'm then going to flip a biased coin that has probability p of being heads. After I flip it, if it's heads, I'm going to guess that X is the larger number. If it's tails, I'm going to guess that Y is the larger number.

So now, as God, what distribution will you choose? You can choose any continuous distribution you like. So please choose it. And then please prove that my probability of winning is not more than 50%.

This is the challenge that pzhon has laid down for you. If you cannot complete it, then you have no ground to dispute the claims in this thread other than uninformed speculation. (And, yes, I am claiming the challenge cannot be completed. Hence, I'm claiming that you are uninformed and speculating.)

[Kristian]

Please disregard that last post. I misunderstood your first point, and had the second one explained in the last post by Jason1990. /images/graemlins/confused.gif /images/graemlins/blush.gif

[Izverg04]

[ QUOTE ]

University of Maryland, Professor Michael Brin.


[/ QUOTE ]
How did this thread go so long without anyone mentioning Google, as in Michael Brin, grandfather of Google.

[Moonsugar]

some infinities are bigger than others

[Moonsugar]

You dazzle the prof and make the math chicks swoon or what?

[Kristian]

I will think about it some more, but yeah, it looks like I am screwed. /images/graemlins/smile.gif

In my defense I will just say that this strategy is far more complicated than the one I originally commented on.
I will shut up now.