c4countdown

Sorry Jon, I only really skimmed your second point earlier and just assumed it depended on your probability calculation being right. I'll have a think about it but might not be able to post for a while, as I've got a busy day ahead.

Paul Howe wrote:I'll try a layman's explanation anyway. A continuous probability distribution is represented by a graph of a function that is always positive, and the area underneath the graph is 1.....

Sure.

David Williams wrote: My problem is with the notion
of random numbers up to infinity. Suppose you pick 100 random whole
numbers, and take the largest of them. There are a finite number of
integers less than it, and an infinite number greater than it. So the
numbers are actually all concentrated in an infinitesimally small part
of the range available, and aren't random at all.

David W

I agree that I pick some numbers with higher probability than others, but that doesn't mean they're not random.Before you've seen my first number, you know there are some regions of numbers that are more likely to occur than others, but you have no idea where these regions are. Say I give you 100 numbers from the same distribution, then you'd be able to learn some things about the distribution, like where the regions are, how big they are and so on, and you'd be able to make reasonable guesses about where future numbers from the distribution would go. Obviously, the more samples we have, the better the guess we can make. Now, suppose we have just one sample from our distribution, which is of course the situation that occurs in our game. All the existence of a strategy for our game is saying is that given 1 sample of a distribution, we can make a slightly better guess about the next number we take from that distribution than if we had no samples at all. Put in these terms, I think it seems a lot less mindboggling.

Jon, I think your second objection is wrong for quite a subtle reason. First of all, I don't agree that you can just write infinity or -infinity on a piece of paper, as these are not numbers in the same sense that things like 12, -76234.34234 are. By the rules of the game, a finite number will always appear on my piece of paper, and I don't see how you can replace something finite with something infinite, which almost always results in some weird shit, and claim you still have the same game. What if I modified your objection though, and said that instead of replacing b with infinity if b is higher, I just replace b with something bigger than b. I think this captures the spirit of your objection, as you will have increased x by doing this. The problem, as I see it, is that b has to come from a probability distribution, and, as I said to David, b takes on some values with greater likelihood than others. By replacing b with something else willy nilly, you're effectively taking b from a different probability distribution, but it's implicit in the rules of the game that a and b should come from the same distribution. Sorry for using such technical language, but I really couldn't think of a simpler way of saying things, so I hope its clear

Anyway, this has taken up quite a lot of my time, so unless anyone posts something really interesting I think I'll just let it be. I will put up a much simpler, non-probability, non-counterintuitive puzzle later tonight though.

Corby wrote:

Gevin-Gavin wrote:It's amusing that in your explanation here, for the first two possibilities you blithely state that it's a 50/50 chance whether you're seeing the higher or lower of the two numbers, when that's the very thing you're trying to disprove...

Is it? But I've been converted now though. If I pick a number, and the numbers on both cards are higher than it, then there's no reason why the the number on the first card is going to be higher than the number on the second card more than half the time and vice versa. We are just talking about the case where the numbers on both cards are higher than the number on mine, and the situation is symmetrical for the two numbers. Likewise for when they are both lower. But when my number is in the middle, I will always win.

I think people may be confused by the notion of infinity and how this only works when you consider it, but that's not the case. If we are just looking at continuous distributions from -infinity to +infinity where the probability density is always positive, this will work for any distribution and for any arbitrary number I pick.

For example, if the "dealer" picks the normal distribution with mean 8 and standard deviation 72 and uses this one distribution for many games, I could still pick any number (say 63.7) and continue using that number and ignore any inferences that come into my head about the distribution, and I will win long term. I will do better if I happen to pick a number nearer 8 in this case, but will still win overall whatever number I pick.

Corby wrote:I started to think about Benford's Law and wondered if I could use this at all.

Completely irrelevant to the puzzle, but a quote from the Wikipedia article:

Benford's law, also called the first-digit law, states that in lists of numbers from many real-life sources of data, the leading digit is 1 almost one third of the time, and larger numbers occur as the leading digit with less and less frequency as they grow in magnitude, to the point that 9 is the first digit less than one time in twenty. This is based on the observation that real-world measurements are generally distributed logarithmically, thus the logarithm of a set of real-world measurements is generally distributed uniformly.

I haven't worked out any of the proportions, but it would still be the case that 1 occurred more frequently than the rest, then 2 and so on even if the data was uniformly distributed, as long as it started at zero, and ended at some point.

For example, if the last available number was 5000, then all the digits from 5 up would be less frequent than 1-4 as the first digit. Obviously if it was 2000, then 1 would have an advantage over all the rest. 1 can never lose out to 2, 2 can never lose out to 3 etc.

Speaking of inifinity, I think it's quite funny that in a distribution such as the normal distribution where you always get a finite positive value for the density at that point, the average value (which must be infinitessimally small) must be lower than it is for any particular value!

Gevin-Gavin wrote:Speaking of inifinity, I think it's quite funny that in a distribution
such as the normal distribution where you always get a finite positive
value for the density at that point, the average value (which must be
infinitessimally small) must be lower than it is for any particular
value!

Easily the most bizarre thing is that there are events that can actually happen, and yet have probability zero. For example, if you take a sample from the normal distribution, the probability of it being rational is zero. I hope Jon doesn't read that, as I fear it will incline him to think the whole thread is bollocks

Anyway, I really should leave it now. I've even started posting about probability in one of the other forums, so it's time to get it out of my head.

Paul Howe wrote:Jon, I think your second objection is wrong for quite a subtle reason. First of all, I don't agree that you can just write infinity or -infinity on a piece of paper, as these are not numbers in the same sense that things like 12, -76234.34234 are. By the rules of the game, a finite number will always appear on my piece of paper, and I don't see how you can replace something finite with something infinite, which almost always results in some weird shit, and claim you still have the same game. What if I modified your objection though, and said that instead of replacing b with infinity if b is higher, I just replace b with something bigger than b. I think this captures the spirit of your objection, as you will have increased x by doing this. The problem, as I see it, is that b has to come from a probability distribution, and, as I said to David, b takes on some values with greater likelihood than others. By replacing b with something else willy nilly, you're effectively taking b from a different probability distribution, but it's implicit in the rules of the game that a and b should come from the same distribution. Sorry for using such technical language, but I really couldn't think of a simpler way of saying things, so I hope its clear

Hmmm, I think I have a slight issue with this, I took it to mean that you were as equally likely to pick 1 as you were to pick -312 4/7, or some number with 18 trillion decimal places - ie, it can be anything, as long as it's a number, with some kind of value, such that one is higher than the other. The original question didn't mention anything of probability distributions.

Therefore I don't accept that it is breaking the game, once the number you are going to see has been decided by whatever means, to replace the other number with either "higher" or "lower" - that is all the game is about. Once you're gonna see a number, it matters not a jot what the other number is, just that it is higher or lower. We can therefore remove the notion of having some point between a & b, and instead just a or b extending off indefinitely in the appropriate direction. And we haven't actually changed the game at all (I don't think), yet altered the odds of winning using your strategy. Which has to be wrong.

Paul Howe wrote:Easily the most bizarre thing is that there are events that can actually happen, and yet have probability zero. For example, if you take a sample from the normal distribution, the probability of it being rational is zero. I hope Jon doesn't read that, as I fear it will incline him to think the whole thread is bollocks

Now this I don't understand at all, I did stats at A-Level, and admittedly I missed possibly a 1/3 of the lessons by opting to get pissed and/or play football instead, but I don't think I was ever exposed to this. My understanding was always zero = no chance, however small. If I was brave, I'd assert that this is the case here, that "x" (our probability of intersecting), while it represents something that is possible, is actually zero. But I'm now wondering, given our differing interpretations of the initial question, whether I'm actually answering a different question altogether...

I'll answer this as Paul is trying to get his life back from the clutches of this thread.

Hmmm, I think I have a slight issue with this, I took it to mean that you were as equally likely to pick 1 as you were to pick -312 4/7, or some number with 18 trillion decimal places - ie, it can be anything, as long as it's a number, with some kind of value, such that one is higher than the other. The original question didn't mention anything of probability distributions.

It's not possible to have a distribution that makes it equally likely to pick any number. The original question didn't mention probability distributions but the theory is fundamental. It is possible to have a distribution which makes it possible to pick any number, but they won't be equiprobable. (Paul suggests a normal distribution, which is probably the best-known example.)

Easily the most bizarre thing is that there are events that can actually happen, and yet have probability zero. For example, if you take a sample from the normal distribution, the probability of it being rational is zero. I hope Jon doesn't read that, as I fear it will incline him to think the whole thread is bollocks.

Now this I don't understand at all, I did stats at A-Level, and admittedly I missed possibly a 1/3 of the lessons by opting to get pissed and/or play football instead, but I don't think I was ever exposed to this. My understanding was always zero = no chance, however small. If I was brave, I'd assert that this is the case here, that "x" (our probability of intersecting), while it represents something that is possible, is actually zero. But I'm now wondering, given our differing interpretations of the initial question, whether I'm actually answering a different question altogether...

They probably protect you from infinitesimals at A-level. Imagine throwing a (perfectly sharp) dart at a dart board. The probability of hitting the dart board is 1, yet the probability of hitting any given point on it is infinitely small. This isn't a contradiction at all, it's just that an infinite large number of infinitely small things can combine to give a finitely large thing. The Greeks were unable to appreciate this (see Xeno's Paradox) so it's not simple but it is sound.

Charlie Reams wrote:They probably protect you from infinitesimals at A-level. Imagine throwing a (perfectly sharp) dart at a dart board. The probability of hitting the dart board is 1, yet the probability of hitting any given point on it is infinitely small. This isn't a contradiction at all, it's just that an infinite large number of infinitely small things can combine to give a finitely large thing. The Greeks were unable to appreciate this (see Xeno's Paradox) so it's not simple but it is sound.

That's fine, as far I can see there you are saying that the probability of hitting any given point is really really small. You aren't saying it's zero.

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.

If we know the parameters of the range, it's easy. Picking a random number from the range as suggested would work, but not as well as just noting whether the exposed number was above or below the 50th percentile, and picking accordingly. If we don't know the parameters, picking a random number from some other probability distribution won't help at all. The only time the suggested solution helps is if we don't know the distribution, but we can select a random number from it.

And if it's a real world, there-obviously-is-a distribution-but-I-don't-know-what-it-was,-but-probably-I'll-be-using-something-similar, sort of situation, then picking a random number will work, but again not as well as estimating what the 50th percentile number might be (otherwise known as guessing).

Suppose the chances of any number being positive or negative are equal. Then half the selections will have one of each. It's not hard to see a strategy that will get all of these and half the rest right - 75% success! And if the chances of positive and negative aren't equal, that was certainly implied at the beginning of all this.

David

David Williams wrote: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.

Quite (I think). There was no mention of probability distributions in the first post, just that there were two numbers written down on bits of paper. Dragging all this kind of stuff into it somewhat goes against the earlier assertions that "no advanced maths is needed here, and anyone could come up with the answer".

I particularly still don't see "any guarantee of being ahead in the long run" as promised. As far as I can tell, it's akin to posing the game as "I write down two numbers, which can be anything, positive, negative etc etc. I show you one. You guess what the other one is. You win £1 if you guess it right" and giving the answer as "you always guess 1.23498723948235623" because on average your gain is positive.

I'm not happy.

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.

No, sorry, that's meaningless. There has to be a probability distribution. You may not be thinking of the distribution, any more than if I ask you to pick a random set of lottery numbers, but nevertheless there is one. Newtonian mechanics might not be mentioned in a question about ballistics but the laws still apply.

To answer Mr Corby, I think Paul was roughly right in saying that no advanced knowledge is needed to devise the solution. The intuition is that the probability of the number falling within some range is a normal finite number between 0 and 1, even though the chance of it being any particular value is infinitesimal. If you don't accept the intuition and instead want to be completely formal then that's entirely reasonable, but clearly you need some knowledge of the formalism to follow it through. The problem seems to be that you want to have your cake and eat it, in the sense that you want to be convinced intuitively by an area of maths that humans on the whole find entirely unintuitive.

Charlie Reams wrote:To answer Mr Corby, I think Paul was roughly right in saying that no advanced knowledge is needed to devise the solution. The intuition is that the probability of the number falling within some range is a normal finite number between 0 and 1, even though the chance of it being any particular value is infinitesimal. If you don't accept the intuition and instead want to be completely formal then that's entirely reasonable, but clearly you need some knowledge of the formalism to follow it through. The problem seems to be that you want to have your cake and eat it, in the sense that you want to be convinced intuitively by an area of maths that humans on the whole find entirely unintuitive.

I dunno. I am well aware that my maths is (very) limited, but I like to think I have a pretty good grasp of certain things. If Paul's original post had begun talking of pdf's and calculus, I may well have sat back and watched. When it says "two numbers are written down. You are shown one. Is the other higher or lower?" then I'd like to feel this is within my area of understanding. I'm prepared to accept I'm wrong though.

If you agree that the answer is actually pretty much akin to my modified (and absurd) question (and subsequent) solution of guessing the precise number on the other sheet of paper, then I don't feel there's much else to be said here, and I'll just file this away under what I think is an appropriate classification.... otherwise, I am willing to learn

Aw guys, when I was asked earlier in the thread to clarify what random meant, I very clearly said that the numbers came from a probability distribution (this is a tautology) and that it wasn't a uniform distribution, as it's impossible to have a uniform distribution over the reals. Maybe it just got lost in the torrent of discussion. While I don't think you need any mathematical knowledge to find the strategy, I was perhaps a bit naive to think I wouldn't need any maths to counter the various (and very clever, in some cases) objections that were raised.

Here's a link to the legendary (to mathematical puzzle solvers at least) John Scholes's discussion of this problem, which I think you'll be able to follow with A-level stats, and will hopefully convince you that I'm not making it all up, even if it's only because you're hearing the same thing from a more authoritative voice than mine

http://www.math.niu.edu/~rusin/known-math/94/weirdpro

Your patience is much appreciated, Paul

Paul Howe wrote:Here's a link to the legendary (to mathematical puzzle solvers at least) John Scholes's discussion of this problem, which I think you'll be able to follow with A-level stats, and will hopefully convince you that I'm not making it all up, even if it's only because you're hearing the same thing from a more authoritative voice than mine

Lol, never heard of him, ergo I have more respect for your word I did say earlier on that I was fairly sure you were right (and would have the necessary backup from the world of maths) but nonetheless I like to try and follow these things through and try and satisfy myself...

I think it's fairly apparent that if there is any kind of distribution of the random numbers, then it's obvious you can make a "best guess" as to whether the other number is higher or lower. I suppose my error must be thinking that there wasn't one, when you & Soo claim there has to be one...?

I also thank you for your patience Paul

Good thread, it has been very interesting, and I've probably learnt something... I think...

Yes, very interesting!

All the stuff about distributions and functions and random numbers is surely over the top. If the number you come up with is higher than both the written numbers, or lower than both of them your chance is 50/50. If the number is between them you'll be correct. So your chance is greater than 50%. If you can maximise the chance of picking a number between the written numbers you increase your overall percentage. So just guess!

David

That's a right-on summary, yes.

Anyone got any other puzzles? I like these

Just as a postscript I showed this to my son, who pointed out that there is actually a finite number of numbers that can be selected from. It's limited by the age of the writer, the size of pieces of paper that a human being can turn over, the size and strength of the table the paper rests on, the number of atoms of ink, etc. So there.

I don't know.

I think it's just blind luck, like the show that follows "Countdown" every day.

Paul Howe wrote:
All I can suggest is to read my above post again, I think it was pretty clear. I agree it's fairly mindboggling stuff, most things that involve infinity in some way are, but it all follows logically. I don't really have time to spend all night deconstructing your counterexamples, but the argument behind saying you can win more than half the time is sound, and if you concentrated on proving it wrong by trying to find a false step in the reasoning, rather than trying to show me a situation where its wrong, I think you'd struggle.

I'll dig up some discussion of this problem on the internet later if you still don't believe me.

OK, I know it's 9 months on, but I was re-reading and saw Paul was online and couldn't resist pointing out the logical flaw which I believe I have FINALLY spotted.

One of the assumptions in the solution, which has been taken for granted and never tested, is that if your number c is lower than both a (the turned up number) and b (the hidden number), you have a 50-50 chance of being right. This isn't true. The probability of b being less than a, in the absence of further information, is 0.5. The probability of b being less than a, given that b is greater than c, is less than 0.5 - because all the numbers below c are excluded from the new possible subset that b can be in.

David, I'm glad you still find this entertaining so long after its creation, alas I'm beginning to wonder whether it will ever truly die.

David Roe wrote:
One of the assumptions in the solution, which has been taken for granted and never tested, is that if your number c is lower than both a (the turned up number) and b (the hidden number), you have a 50-50 chance of being right. This isn't true. The probability of b being less than a, in the absence of further information, is 0.5. The probability of b being less than a, given that b is greater than c, is less than 0.5 - because all the numbers below c are excluded from the new possible subset that b can be in.

Lets turn this round a bit. Is there anything to stop me, using your words, saying the probability of a being less than b, given that a is less than c, is less than 0.5 - because all the numbers below c are excluded from the new possible subset that a can be in. I don't think so, as after all a and b are chosen independently of c and in the case by case analysis we don't even know what a and b are yet- remember when I was explaining the solution I said it was crucial to choose c before either a or b were known! So now we have a situation where, according to your logic, Prob(a<b)<0.5, and Prob(b>a)<0.5, implying Prob(b=a) != 0. But, as those versed in the art of continuous probability know, this probability must be 0, so your logic leads to a contradiction.

Remember in the case by case analysis we don't even know what a and b are yet, so you seem to be arguing that its impossible to win this game with probability >0.5 by claiming that we can guess which of two random numbers is the largest with probability >0.5!

Paul Howe wrote:
All I can suggest is to read my above post again, I think it was pretty clear. I agree it's fairly mindboggling stuff, most things that involve infinity in some way are, but it all follows logically. I don't really have time to spend all night deconstructing your counterexamples, but the argument behind saying you can win more than half the time is sound, and if you concentrated on proving it wrong by trying to find a false step in the reasoning, rather than trying to show me a situation where its wrong, I think you'd struggle.

I'll dig up some discussion of this problem on the internet later if you still don't believe me.

That sounds really arrogant reading it back. I must've been getting exasperated

[EDIT - my next two posts contradict this one to some extent, so don't waste too much time trying to understand what I'm on about! ]

The intrinsic probability of a>b or a<b is not affected by c. It's 0.5, whether c is chosen before a is turned up or after. What changes in your computation of the probability is our knowledge of what b is. (We don't actually know anything about b until the card is turned up, but for the purposes of your calculation you have assumed the knowledge. Which is the correct way to calculate it, but it does change the sum.)

As an example of how some knowledge can change the probability of an event, even though the event has already happened and its intrinsic probability cannot change, imagine three coins have been tossed and you don't know the result. What's the probability that the third one is heads? 1/2.
Then, you are told that at least one coin is a head. What now is the probability that the third coin is a head? 4/7, because of the eight possible answers, 1 (TTT) is impossible. There are only 7 possible results, 4 of which end in H.
Then, you are told the first coin is a tail. Now the odds are 2/3 that coin 3 is a head, because all that's left is HH, HT, TH. And then if you are told coin 2 is a tail as well, coin 3 must be a head with certainty. Nothing has changed except our knowledge of the partial result.

In the higher/lower case, there is certainly something to stop you turning it round to say the probability of a<b, given that a<c, is less than 0.5. That's a non-starter because my example was based on a>c. Remember that we know for certain what a is, because a is turned up before you say whether you're going higher or lower. c is fixed and known, because you picked it forst; a is fixed and known, because that card has been turned over; b is the only unknown.

I'm going to post this bit (in case the computer loses it) and then I think I've worked out some sort of proof - but I won't know if I have until I write it down.

Right. Let's assume we pick c as 0 - this means we can use positive and negative as dividing markers for >c and <c which might make the example easier to read. The example works exactly the same for any value of c, "positive" and "negative" are just naming conventions.

So c = 0. Then a is turned up - it's either positive or negative, let's assume positive [because the argument is identical but reversed for negative, so it makes no odds]. b is unknown, but you guess lower, because c < a.

b has three ranges it might fit in.
(i) negative, less than c - you win.
(ii) positive but less than a - you win.
(iii) positive but greater than a - you lose.

If it's negative, that's the case where c is between a and b, which you called x earlier on.
If it's positive, that's the case where you assumed 50-50 whether you win or lose - ie. 50-50 whether it's case (ii) or case (iii). I don't agree that assumption, so let's call the probabilty (b<a given that b is positive) = y.

This means the total probabilty of victory is now

x + (1 - x) y

which converts to

x + y - xy

(which of course is identical to your formula iff y = 0.5).

Now, a is a totally random positive number. b, given that b is positive, is a totally random positive number. None of the above has made any stipulation about the size of a or b, just that they are positive. So the probability that a > b is 0.5. Which means I've just proved you're right, and if I want to prove I'm right, I've got to find the flaw in my own argument.

This needs more thinking about. I'll be back tomorrow.

Am I allowed to keep replying to my own posts? Here goes, anyway.

It's easy to work out the probabilities on a very limited number distribution - eg. if a and b are both picked from the 11 whole numbers between 0 and 10. (a=b is not allowed.) If c is the optimum value of 5.5, then this system gives a 77% chance of winning; at c=2.5, 72% chance of winning; even at c=0.5, there's a 59% chance of winning. Only when c goes outside the distribution altogether, eg. c=12, does the chance of winning = 50%. It never goes below 50%. I believe that this applies equally to all real number systems.

Intuitively, this is how I now see it. Where a is somewhere near the middle of the number distribution, it's pretty close to 50-50 whether b is higher or lower, and no system will beat the odds. But where a is relatively extreme, such as in the top or bottom quartile of the number distribution, then you want a system which will lead you towards the middle of the distribution because that's where b is likely to be. If a is extreme, the chances are your number c is nearer the middle of the distribution and so it will lead you in the right direction.

I'm happy now. Goodnight.

I'm not sure I want to respond to all that unless you can summarise your thoughts more concisely. For the record, here is the explanation in slightly more technical terms that you might appreciate. We can regard the numbers on the paper a,b as being drawn from some unidentified continuous probability distribution function that is non-zero everwhere (we say it has "infinite support"), and the guess c as coming from a different continous pdf with infinite support. Here are the two cases:

1. c doesn't lie between a and b
2. it does

These two cases exist before the values of either a or b are known, and we don't eliminate any of the events that comprise these cases simply by knowing the value of one of the numbers. I haven't used the value of b in my "calculation" as you seem to insist, in fact the only "calculation" I've done is to say that, before we know ANYTHING about a and b, a is larger than b with 50% probability. (it occurs to me your confusion could result from a slight abuse of notation, since the numbers are both face down on the paper at this point and either could be picked to be a).In the first case c doesn't split a and b so our win probabilty is exactly this 50% probability. In the second case we always win by the strategy described, how often we win depends on the exact shape of the distributions a,b and c are drawn from, and it is always a winning strategy as long as the distributions are non-zero everywhere.

Now I'm off to have a well deserved drink.

Impressively patronising. Actually, the reason I didn't write " P(a>b|a>c) " wasn't because I didn't know how, it was that I thought other people who might (just possibly!?) be still reading wouldn't understand it if I did.

If you'd read my posts, you'd have seen I was agreeing with you. Still liked the puzzle though - got any more?

David Roe wrote:Impressively patronising. Actually, the reason I didn't write " P(a>b|a>c) " wasn't because I didn't know how, it was that I thought other people who might (just possibly!?) be still reading wouldn't understand it if I did.

Sorry if it came across that way, it wasn't my intention. I was just trying to explain my solution and answer your objections as clearly and quickly as possible, and seeing as you expressed your own views very stridently I didn't think you'd mind the bluntness! And I don't think I ever implied a lack of knowledge on your part, so not sure where that last sentence has come from.

If you'd read my posts, you'd have seen I was agreeing with you. Still liked the puzzle though - got any more?

I did read them, and sorry if it sounds harsh but it was like reading a stream of consciousness style essay where you changed tack half way through and your last paragraph wasn't something I completely agreed with, so it was difficult for me to think you were agreeing with me! Is that patronising enough for you? Anyway, glad you enjoyed the puzzle, I think I've been very patient answering everyone's questions so the patronising comment wasn't really appreciated, as you might have guessed.

Paul Howe wrote: I think I've been very patient answering everyone's questions so the patronising comment wasn't really appreciated, as you might have guessed.

The moral of the story is don't be patient, because no one ever appreciates it.

David Roe wrote:If a is extreme, the chances are your number c is nearer the middle of the distribution and so it will lead you in the right direction.

Is that right though? Not that I disagree with the correctness of the puzzle (it took me a while to get there and I'm not going back now) but since you have no idea what the distribution is, you number c, which doesn't come from that disribution, is not necessarily likely to be in the middle of the distribution.

Sorry, it was uncalled for.

(It was the line about "slightly more technical terms that you might appreciate" that twitched my nerve.)

I did in fact change opinion half way through - when I started the proof I thought it was going to prove my theorem, it was only when I finished I realised you were right. I was trying to set it out step by easy step.

With reference to Gavin, even if c is right outside the distribution, it your odds still don't go below 50-50. (No, I'm NOT going to prove it!!! ) And if c is somewhere on the edge of the distribution, right away from the mean, you're still gaining on those cases where a is even further away.

David Roe wrote:(It was the line about "slightly more technical terms that you might appreciate" that twitched my nerve.)

I'm guessing you misread it as "slightly more technical terms than you might appreciate", since what Paul wrote is a compliment if anything.

David Roe wrote:
(It was the line about "slightly more technical terms that you might appreciate" that twitched my nerve.)

Ha, that was actually meant to be complimentary! I could tell from your posts you had a decent knowledge of probability and would be comfortable with me blathering on about probability density functions and whatnot, which that line was meant to acknowledge. Change that to than and it becomes a bit snarky, but that's not what I wrote. A classic case of meaning getting scrambled in the low bandwidth medium of text communication, methinks.

Charlie Reams wrote:
The moral of the story is don't be patient, because no one ever appreciates it.

.

I think people do appreciate it (several people have said so in this thread), and I tend to find the appreciation of those who don't (talking about life in general here, not this thread) is seldom worth very much anyway.

You're quite right, Charlie, that's exactly how I did read it. All I can say, Paul, is sorry again. Where's the smiley for "hitting myself over the head with a mallet"?

Charlie Reams wrote: It's not possible to have a distribution that makes it equally likely to pick any number. The original question didn't mention probability distributions but the theory is fundamental. It is possible to have a distribution which makes it possible to pick any number, but they won't be equiprobable. (Paul suggests a normal distribution, which is probably the best-known example.)

Can you explain this? Why does a uniform distribution not have equal probability?

Does this explain why I'm a Government statistician and not a real one?

Ian Volante wrote:Can you explain this? Why does a uniform distribution not have equal probability?

Does this explain why I'm a Government statistician and not a real one?

A uniform distribution has equal probability across some range and is zero everywhere else; it's not (and couldn't be) uniform across all real numbers.

Charlie Reams wrote:

Ian Volante wrote:Can you explain this? Why does a uniform distribution not have equal probability?

Does this explain why I'm a Government statistician and not a real one?

A uniform distribution has equal probability across some range and is zero everywhere else; it's not (and couldn't be) uniform across all real numbers.

A fair point. It is indeed true that whenever I use a uniform distribution as a Bayesian prior, it's always just within a range.

Ian Volante wrote:

Charlie Reams wrote:

Ian Volante wrote:Can you explain this? Why does a uniform distribution not have equal probability?

Does this explain why I'm a Government statistician and not a real one?

A uniform distribution has equal probability across some range and is zero everywhere else; it's not (and couldn't be) uniform across all real numbers.

A fair point. It is indeed true that whenever I use a uniform distribution as a Bayesian prior, it's always just within a range.

Ian caused the credit crunch.

Jon O'Neill wrote:Ian caused the credit crunch.

By calculating the total Scottish income from farming? I'm not convinced...

Just re-read this. Great thread. Still not convinced, myself.

Paul Howe wrote: ↑Thu Jan 31, 2008 10:30 pm Say I give you 100 numbers from the same distribution, then you'd be able to learn some things about the distribution, like where the regions are, how big they are and so on, and you'd be able to make reasonable guesses about where future numbers from the distribution would go. Obviously, the more samples we have, the better the guess we can make. Now, suppose we have just one sample from our distribution, which is of course the situation that occurs in our game. All the existence of a strategy for our game is saying is that given 1 sample of a distribution, we can make a slightly better guess about the next number we take from that distribution than if we had no samples at all.

This is buried in a long thread but is a pretty brilliant piece of intuition I think -- it's a different way of thinking about the same game that's a lot easier to grasp, and "feels right" to me in a way that the original formulation doesn't.

It's still weird because you're not just using your sample, you're using your sample plus a previous guess. And if you'd guessed differently, you could potentially conclude differently.

One extra thing I would add to this is that the less you "know" about the distribution and the more "alien" it becomes, the less effective this strategy becomes, and your chances would actually become 50/50 in the limit.

If it's a normal distribution with a mean (and therefore always median) of 0 and variance of 1, then you might choose 0 as your number (the median of the distribution is always the "best" number you could pick, although you might not know what the median is), and you would definitely win in the long run by saying lower if the first number is positive and higher if it's negative, because you're always effectively saying "closer to the median". But this is about as unalien as it gets. If we stick with normal distributions (because it's easier for now), someone might pick a normal distribution based on some random distribution. They might pick from a uniform distribution between -1 and 1 for their mean, and pick from the same distribution for the variance. The your trusty "use zero" strategy becomes slightly less reliable. You win overall, but less so. But the bigger their picking-from distribution gets, the worse you do. And it doesn't matter if you always use zero, or have some other random distribution of your own. You can incrementally make it more and more random or alien, and whatever strategy you use, the further away from the median your number is likely to be, and therefore the less good you become at this game.

So in some true random from the infinite version of this game, you would only have a 50/50 chance of winning. You only win at this game if it's set by your mate basically, not if it's set by God.

(This isn't a rigorous proof obviously, but it's what would happen.)

Edit - The strategy works because your number is picked as a proxy median, and the nearer you get to the median, the better it works. But the further away you are, the less effective it becomes, and in the limit it's useless. And you only ever get near the median with your guess because another human has set up the problem.

Edit 2 - Because there's no reason really why guessing a number should give some insight into a distribution you know nothing about. It's only because you do really know something about it - your guess is likely to be near the median - that it does work.

I should probably read all this again, because it all seems too simple now.

The strategy is to pick your own number first, then go lower if it is lower than the number you see, higher if it's higher. You win every time if your number is between the other two, it's 50/50 if it isn't. But the difference between the two numbers is finite, so the probability of the number you pick being between them is zero. Intuitively you would think it has to be more than zero, but that is the direct and unavoidable consequence of the way the question is framed.

Another thing is that while it works for any distribution and any picked number, you still can't really say that you've got a better than 50/50 chance on any given one-off occasion. The problem is that when you're not playing with God, you're playing with small numbers and because the distribution is not properly random, you can make some inferences about it and you'd have to do some sort Bayesian calculation to determine the likelihood of the next number being higher or lower.

David Williams wrote: ↑Sat May 16, 2020 11:28 pm But the difference between the two numbers is finite, so the probability of the number you pick being between them is zero.

Not necessarily. In particular, if you choose your own random number according to, e.g. a normal distribution, then it has a non-zero probability of landing in any given finite interval.

Sam Cappleman-Lynes wrote: ↑Mon May 18, 2020 9:12 am

David Williams wrote: ↑Sat May 16, 2020 11:28 pm But the difference between the two numbers is finite, so the probability of the number you pick being between them is zero.

Not necessarily. In particular, if you choose your own random number according to, e.g. a normal distribution, then it has a non-zero probability of landing in any given finite interval.

Don't follow. Your number has a 100% chance of being the number you pick. If you decide to pick a random number in some known distribution you can calculate the probability of it being in any particular range. But that's irrelevant. It's the chance that whatever your number is it will fall inside a finite interval within an infinite range with no defined distribution, and that is zero.

David Williams wrote: ↑Mon May 18, 2020 3:44 pm

Sam Cappleman-Lynes wrote: ↑Mon May 18, 2020 9:12 am

David Williams wrote: ↑Sat May 16, 2020 11:28 pm But the difference between the two numbers is finite, so the probability of the number you pick being between them is zero.

Not necessarily. In particular, if you choose your own random number according to, e.g. a normal distribution, then it has a non-zero probability of landing in any given finite interval.

Don't follow. Your number has a 100% chance of being the number you pick. If you decide to pick a random number in some known distribution you can calculate the probability of it being in any particular range. But that's irrelevant. It's the chance that whatever your number is it will fall inside a finite interval within an infinite range with no defined distribution, and that is zero.

I'm afraid I don't follow you. At the point I generate my number, according to (say) a normal distribution, the two numbers written on the paper have already been fixed (I just don't know what they are yet). And the chance that my number lies between them is non-zero. Whatever distribution or method the other player used to come up with the numbers on the paper is what is irrelevant.

Suppose, for the sake of argument, that the two numbers are picked at random, but from a large but finite range, and that they differ by 1,000,000,000,000,000,000,000,000,000,000. Let's call that number X.

Let's also have a very small number, say 0.0000000000000000000000001, which we call Y. If the large but finite range is greater than X/Y, then there are more than 1/Y intervals of size X, and the chance that our number will fall in any one of them is less than Y. No matter what numbers X and Y are there is a finite range that means the chance of success is less than Y. And the range is actually infinite.

Looks like you don't actually know what a normal distribution is, then. It's not one in which intervals of the same size have equal probabilities, which you seem to imply by your post. In fact, your argument correctly shows that it's impossible to have such a distribution (called a uniform distribution) over the reals.

The normal distribution is the distribution represented by the famous bell curve, in which any (and I mean any) interval has a non-zero probability ascribed to it. It's just that those probabilities get small very quickly as they get further away from the mean.

I know what a normal distribution is. It was you that brought it up and it's got nothing to do with the question.

Gavin Chipper wrote: ↑Sun May 17, 2020 7:53 pm Another thing is that while it works for any distribution and any picked number, you still can't really say that you've got a better than 50/50 chance on any given one-off occasion. The problem is that when you're not playing with God, you're playing with small numbers and because the distribution is not properly random, you can make some inferences about it and you'd have to do some sort Bayesian calculation to determine the likelihood of the next number being higher or lower.

Just to follow up on this post - on a one-off occasion, you pick your number - say, 8 - and the first number out is 4. You're about to say "higher" when this other guy comes along and says "I'm playing too! My number was zero, so I'm going to say lower."

Does this change things? You can't both have a better than 50/50 chance of winning, and it seems that neither is in a better position than the other. Of course, in a normal probability problem, if you're confused by the logic, you can work through all the combinations and count them up, but you can't do that here.

If you use this same probability distribution multiple times and rigidly stick to 8, you will win on average (as will the other guy using 0), but the more you play, the more you learn about the probability distribution, so your best strategy becomes something else anyway. But on the one-off occasion, you can't necessarily say that you'll probably win precisely because of what you learn about the distribution.

There's a parallel with this induction-based problem - There are ten closed boxes each with one ball in. The balls were randomly assigned to the boxes. They might all be the same colour, they might all be different, some might be the same as some others - you don't know. Before any of the boxes are opened, you can make the (arguably strange) statement "The ball in box 10 has no more than a 1 in 10 chance of being less white than the others." (a is less white than b when b is white and a isn't.) If all the balls were guaranteed to have a different value of some property, then you'd just say the ball in box 10 had a 1 in 10 chance of having the lowest value. But because there might be ties here, it's at most 1 in 10.

Anyway, you open boxes 1 to 9, and there's a white ball in each of them. What is the probability of the ball in box 10 now being white? As soon as you find out stuff about the distribution, you don't have a clean answer any more. 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.

And the same applies in the higher/lower problem. When 4 comes out, everything becomes messy. Someone picked this distribution that spewed out a 4 - what was their motivation?

Edit - And the only way to clean it up is to play God in the limit where the strategy becomes useless anyway.

Edited again later tidying a few bits up.

David Williams wrote: ↑Mon May 18, 2020 5:59 pm I know what a normal distribution is. It was you that brought it up and it's got nothing to do with the question.

I suggest you read my first post replying to you more carefully then. It's not part of the question but it's definitely part of the answer.

There are two numbers on pieces of paper. I pick a random number. If I land between them, then I win with probability > 50%. If I don't I win with probability = 50%. So I just need a way of making sure I have non-zero probability of landing between the two numbers. A normal distribution is how to do that. If I pick my random number according to a normal distribution, then I have a non-zero probability of landing between the two numbers on the pieces of paper.

I'm not convinced that a valid answer to picking a number is to say that your number is a normal distribution with mean of x and standard deviation of y, but even if it is, the best you can say is that your method produces a chance higher than zero of winning, and I have shown that the chance is lower than all numbers greater than zero. (But I do understand your argument now, so that's something!)

Depending on exactly which numbers are written on the pieces of paper, the chance that my number falls between them certainly can be made as close to 0 as you desire. In any specific case though, it's still strictly greater than 0. I think that's what you're saying, and if so, we agree.

And still it lives! I'm delighted that many are still invested in this, someone actually emailed me to tell me the thread had been resurrected!

I've tried this on a few different people over the years and many take views similar to some of those expressed above, i.e. that the formulation of the question is a bit sneaky and leaves out assumptions which, if stated explicitly, would change the nature of the problem. To a certain extent I agree with this, and I still struggle to pose the question in a way that is both accessible and faithful to the underlying mathematics.

A more mathematical, although still not very rigorous, formulation might go something like this:

Two real numbers a,b are drawn from a random variable X with a probability density function that is non-zero everywhere. You are told the value of a and invited to guess if the unseen value of b is higher or lower than a, receiving a payoff of 1 for guessing correctly, and -1 if incorrect. Can you devise a strategy with a positive expected payoff?

You can probably see why I didn't put the question like this - it's terse, jargon heavy, and obscures a result that anyone can appreciate behind a mathematical facade. The solution given in this thread (choose c independently from a, guess b>a if c>a, guess b<a if c<a) is unambiguously the correct solution to the problem stated as above. Your probability of success in this game is 1/2 + p/2, where p is the probability that c falls between a and b. 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.

I think there's value in trying to present it more accessibly, as it's a powerfully counterintuitive result that can be reached without any special knowledge (although I'm sure many on this forum are far more knowledgable mathematicians than me!). That said, there were a few issues with how the question was presented in this thread, and I tend to present it a bit differently now, when I dare to ask it:

⦁ I now state the problem as playing the game once and finding a strategy to win with probability > 1 / 2. This sidesteps questions around multiple iterations of the game (e.g. shouldn't I be able to learn from previous iterations, does the person picking the numbers play in an adversarial manner, if the distribution X changes with each iteration shouldn't there be some random process that determines how X changes and then it's turtles all the way down etc ). These are natural to ask given how the problem was posed, but are a distraction from the essence of the thing
⦁ I now say a and b are generated by a random number generator, rather that a human being writing them down, as it's easier to think of the numbers as random and not having a psychological component. I still prefer the RNG formulation as more engaging than "two numbers a and b are drawn from a random variable X"
⦁ Are there certain things that should have been stated more explictly? E.g. on this thread many assumed the numbers a,b could or even must occur with uniform probability. Here I disagree, and I think a nice aspect of the problem is that it tests your assumptions about how probability works in more advanced settings, but without that knowledge actually being necessary to solve the problem. E.g. there was some good discussion on how it was impossible for an infinite set of outcomes to occur with uniform probability, and further that individual outcomes in an "uncountably" infinite space must occur with zero probability. It's nice that this problem helps you engage with those concepts (which are fundamental laws of probability, rather than assumptions) in an accessible setting - they're elementary for those who know, hopefully educational for those who don't, and in any case not necessary to solve the problem.

That aside, I'm glad this is helping you stay sane during lockdown.

Paul Howe wrote: ↑Tue May 19, 2020 4:21 pm ⦁ I now state the problem as playing the game once and finding a strategy to win with probability > 1 / 2. This sidesteps questions around multiple iterations of the game (e.g. shouldn't I be able to learn from previous iterations, does the person picking the numbers play in an adversarial manner, if the distribution X changes with each iteration shouldn't there be some random process that determines how X changes and then it's turtles all the way down etc ). These are natural to ask given how the problem was posed, but are a distraction from the essence of the thing

I also do this, and another reason it's a good idea is that it removes the tendency that people have to misapply limits. If you're only playing the game once, there's nothing to take a limit of, and that means there's less cruft obscuring the true nature of the problem.