Higher or Lower

[Gavin Chipper]

There is no way to know the size of your advantage p, and it may indeed be incredibly small, but it is definitively positive, there is no limit style argument that makes it vanish to zero.

Are you sure about this? I would say that the more "Godlike" (basically less limited to using "small numbers") your adversary becomes, the further away from your pick the median of the distribution is likely to be, and therefore the less effective your pick becomes.

But then I suppose the standard deviation could also increase with this, which perhaps cancels this out (to some extent? Completely?) I suppose it depends on the subjective nature of increasingly advanced adversaries and how they choose to play the game. But certainly they could play it so that it vanishes to zero in the limit - by increasing the range of the possible median point but also keeping the standard deviation low.

For example, all beings use a normal distribution for the main distribution. But they pick the mean of the distribution randomly first using a separate sub-distribution, and always use a standard deviation of 1 for the main distribution. The sub-distribution is just a uniform distribution between -x and x with x corresponding to the being number. So:

Being 1 uses a sub-distribution that is a uniform distribution between -1 and 1 to pick the mean for their main distribution (which is a normal distribution with standard deviation 1).

Being 2 uses a sub-distribution that is a uniform distribution between -2 and 2 to pick the mean for their main distribution (which is a normal distribution with standard deviation 1, not 2).

Being 3 uses a sub-distribution that is a uniform distribution between -3 and 3 to pick the mean for their main distribution (which is a normal distribution with standard deviation 1).

And so on. The higher up the beings you go, the less effective the picking strategy becomes, with it being useless in the limit.

Obviously you can't have this case where your probability becomes exactly 0.5, but you can get arbitrarily close to it.

[David Williams]

Surely the whole essence of the original problem was that the numbers were selected randomly from a boundless selection. If there is some sort of probability distribution it's a different matter altogether.

I knew I should have read at least some of this before I started again.

[Gavin Chipper]

I would say that the more "Godlike" (basically less limited to using "small numbers") your adversary becomes

Etc. Well, to add to this, I would say that the more Godlike your adversary becomes, the more the distribution should become like a uniform distribution across the real numbers (God wouldn't arbitrarily favour some numbers over others as in my previous example). Achieving this exactly seems to be impossible, but you can have beings that iteratively get closer to it - using something similar to what I did with the normal distribution in my last post. This way we wouldn't have to worry about keeping the standard deviation small and the mean big as I did with the normal distribution, because we no longer have these variables. If we do this, then as we work our way up to more Godlike beings, your probability of winning should approach 0.5.

[Paul Howe]

Let's fix some terminology - the "dealer" in the game picks two numbers a,b from distribution X. The "player" picks c from distribution Y and plays according to the strategy described above (i.e guess b>a if c>a, guess b<a if c<a)

There is no way to know the size of your advantage p, and it may indeed be incredibly small, but it is definitively positive, there is no limit style argument that makes it vanish to zero.

Are you sure about this? I would say that the more "Godlike" (basically less limited to using "small numbers") your adversary becomes, the further away from your pick the median of the distribution is likely to be, and therefore the less effective your pick becomes.

Yes. The player's advantage is strictly positive under very trivial conditions - the only thing needed to guarantee the result is a non-zero probability of c falling between a and b, i.e. at least one of X or Y must have non-zero probability density everywhere. The player can guarantee it through their choice of Y alone. Certainly the advantage could be very small indeed, but not infinitesimally so. If the game is played perpetually with any choice of X and Y satisfying these basic conditions, the player will always eventually come out ahead, regardless of the size of his advantage (and even if X is of divine provenance!)

So let's say the dealer fixes some probability, e.g. 50.000001%, and asks "is there a distribution X that is guaranteed to reduce the win probability below this level, regardless of the player's choice of Y". In one sense the answer is no, e.g. if the player happened to choose Y=X, then they would do very nicely indeed, regardless of how localised or otherwise complicated X is. Of course that's obviously a cheat as the player has no knowledge of X and is cosmically unlikely to pick Y=X. But this guards against the idea that you can find a sequence of distributions X_1, X_2, ..., X_n,... for which the win probability is guaranteed to approach 1/2. Such a sequence cannot exist independently of the player's choice of Y. Your question probably lives in a setting where X and Y themselves are in some sense "random", and I have a sense that's not an easy space in which to answer even basic questions (e.g. if you think about how you might actually go about choosing Y "at random" from, say, the set of all continuous pdfs that are non-zero everywhere, you'll see it's not an easy thing to formalise).

Ultimately, most of the complication here just seems less interesting than the very straightforward underlying result, i.e. that the player has a guaranteed advantage in this game merely from making a single observation from X. Doesn't feel like it should be true, but it is.

I would say that the more "Godlike" (basically less limited to using "small numbers") your adversary becomes

Etc. Well, to add to this, I would say that the more Godlike your adversary becomes, the more the distribution should become like a uniform distribution across the real numbers (God wouldn't arbitrarily favour some numbers over others as in my previous example). Achieving this exactly seems to be impossible, but you can have beings that iteratively get closer to it - using something similar to what I did with the normal distribution in my last post. This way we wouldn't have to worry about keeping the standard deviation small and the mean big as I did with the normal distribution, because we no longer have these variables. If we do this, then as we work our way up to more Godlike beings, your probability of winning should approach 0.5.

So you define a uniform distribution with pdf 1/2x on [-x,x) and 0 outside of this, and claim the limting distribution is what you get as x goes to infinity. But in the limit this assigns zero probability to everything. You're being told that the set of real numbers is so vast that you can't assign the same probability to everything, you simply have to make some events more likely than others. You might as well say 2+2=5, as god surely wouldn't arbitrarily favour 2+2 to equal 4, and go with whatever nonsense follows from that. You can prove everything and nothing in such an environment!

[Gavin Chipper]

Fair enough. It definitely works for any specific distribution, but I wondered if that itself might not be enough and whether it would always be down to some sort of bias in the distribution, caused by it being set by a human. But if I'm defining bias as any distribution that isn't a uniform distribution across the reals, and such a distribution is impossible, then that would make an unbiased distribution impossible, so it would negate my argument. But then if God would play the game in an unbiased way, then this proves the non-existence of God.

[Charlie Reams]

I suggest using the probability distribution of human ages, which of course has zero density above 120 years. Finally something we can all agree on!

[Gavin Chipper]

I thought I might give this a revisit, for a laugh.

Basically we have two competing things that both seem correct, but are also seemingly in conflict with each other.

1. We just have a single random number from a distribution we don't really know anything about and no other relevant information about it, so can have no way of knowing if another number taken from that distribution is likely to be higher or lower, so it must be 50/50.

2. It is true that for any distribution (one that's relevant for this anyway), you can pick any number (c) and if you pick a number (a) from the distribution, then more often than not, another number (b) from the distribution will be c-side of a.

What those setting the puzzle do is double down on (2). They say it's correct, so (1) must be incorrect. What those being set the puzzle do is double down on (1). It's correct so (2) must be incorrect. But there's little interaction between (1) and (2) in the discussions, so both sides go away thinking that they must still be right. (It reminds me of the aeroplane on a treadmill in that respect, though I can't remember any of the details about that.)

So what's the resolution? Well, I would suggest it might be a bit more philosophical than mathematical. And it's largely what I said in this post, but I'll make it more consise here.

Basically, in the abstract, you can pick any distribution and show that the reasoning in (2) follows. Say you pick a normal distribution with mean 4, standard deviation 5, and arbitrarily choose 6 for your "starting number". If the first number out is 4.3, that's lower than 6, so you go higher. If you played this lots of times, you'd win overall. It works for any distribution and any "starting number". So it will work for the distribution used in the actual game. Great.

But then the first number out is 4. Someone has picked out a distribution from the set of distributions that spewed out a 4. What was the thinking behind this distribution? Is this a likely thing to have happened a priori?

Analogously, if someone has 10 balls and they put them randomly into 10 boxes labelled 1 to 10, then you can say confidently that the ball in box 10 has at most a 10% chance of being the "least white". (There might be ties so it could be less than 10%.) And you can make money by betting according to these odds. But when you open the first 9 boxes and see they're all white, you no longer have this confidence. Someone chose the balls. What are the chances that they chose just one different ball versus all the same? You end up with unanswerable Bayesian stuff, and you can no longer give a clear answer. That happens as soon as you know anything about the distribution. It breaks your reasoning power.

And that happens when the 4 comes out. That solves the conflict. (1) and (2) are both right, but as soon as you see the first number, it breaks (2). (1) remains correct and you cannot win more than 50% of the time.

Fight me.

[Gavin Chipper]

Yeah, but what if the guy doesn't look at the number that comes out and sets up a bot to deal with it?

Also given that it works for all distributions and all starting numbers, it will work for the one used in the game. So are we saying it would work on over 50% of occasions but just not the first one?

[Gavin Chipper]

I've been thinking a bit more about this, and I don't think what I've quoted below has been fully resolved. And I wonder if this is some sort of philosophical paradox rather than a purely mathematical question. Like e.g. "This statement is false" but with more baggage around it so that it's difficult to distil it to its purest form to isolate and understand it.

But to revisit a couple of things: You can pick any distribution that fits the parameters of the question and show that the solution given seems to basically work. I.e. pick the distribution, and then pick your starting number, and you can work out mathematically that you will win over 50% of the time if you play multiple times.

E.g. you could pick a normal distribution with a mean of 2 and standard deviation of 2 and pick 5 as your starting number. Play that out hundreds of times and you'll win more often than not (that's been discussed elsewhere but I can go through it again if needed).

But it works for any number that you pick. And despite what was said previously in this thread, it has nothing to do with picking the number beforehand, because the above paragraph is true regardless of when a number is picked, and that's where it starts to get paradoxical.

For example, I have 2 as my number. The first card is revealed as a 4. I'm about to say "lower" when I think to myself that whatever underlying distribution is being used here, it also works for 6, so I could just as easily go for higher.

Alternatively, if you don't like that, someone could walk into the room and say that he's playing too and that 6 is his number. Paul brushed over this by saying that playing long term, both numbers would win in the long run, but that's not the point. We have a one-off event. One strategy is saying higher and the other is saying lower and there's no way to distinguish between them. Do I have a greater than 50/50 chance of winning? Does the other guy? We can't both have.

Also if you ask someone who wasn't playing the game what they know about the distribution if they see a 4 come out as the first number, they know exactly the same amount about the distribution as someone who is playing and picked 2 beforehand. Picking your number and seeing a card turned over gives you no more knowledge about the underlying distribution than simply seeing a card turned over.

But as I said previously, it would seem very strange if the strategy given as the solution would work over a long run, but not give you a greater than 50/50 chance on the first attempt. After all, the first go is exactly the same as the rest of them.

But the bottom line is that solving this problem doesn't mean doubling down on either 1 or 2 in the quote below. It means resolving the tension between the two of them. So if someone comes back to this thread now just to double down on 2 and spell it out in minute detail, they would be missing the point.

I thought I might give this a revisit, for a laugh.

Basically we have two competing things that both seem correct, but are also seemingly in conflict with each other.

1. We just have a single random number from a distribution we don't really know anything about and no other relevant information about it, so can have no way of knowing if another number taken from that distribution is likely to be higher or lower, so it must be 50/50.

2. It is true that for any distribution (one that's relevant for this anyway), you can pick any number (c) and if you pick a number (a) from the distribution, then more often than not, another number (b) from the distribution will be c-side of a.

What those setting the puzzle do is double down on (2). They say it's correct, so (1) must be incorrect. What those being set the puzzle do is double down on (1). It's correct so (2) must be incorrect. But there's little interaction between (1) and (2) in the discussions, so both sides go away thinking that they must still be right. (It reminds me of the aeroplane on a treadmill in that respect, though I can't remember any of the details about that.)

Edit - One other thing that could be said is that it's been said many times on here that you can't have a uniform distribution across all real numbers. Because of this, any underlying distribution will be in some way "biased". Specifically, the median of the distribution must be "approximately" zero. That is, I will be able to pick a number n where the median of the distribution is zero to the nearest n.

Also the number that the player picks will also come with exactly the same bias, so the distributions they come from are therefore related in some manner. So the player using their own pick as a proxy for a sample of the dealer's distribution makes more sense from that perspective.

But that still doesn't change the fact that the guy standing behind you is saying the opposite of what you're saying...

[Gavin Chipper]

I suggest using the probability distribution of human ages, which of course has zero density above 120 years. Finally something we can all agree on!

Underrated comment.