Higher or Lower
Moderator: Michael Wallace
Higher or Lower
There seems to be something of an appetite for interesting puzzles on here, so I'll occassionally post some of my favourites. Sadly, most of my absolute favourites are rather mathematical for c4countdown, but I do have a few suitable for a general audience, starting with this one:
Suppose I write down two numbers on separate pieces of paper, and then put them face down on a table so you can't see them. The numbers can be anything: positive or negative, integers or decimals, arbitrarily large or small, and are chosen totally at random. I then let you pick one of the pieces of paper and turn it over so you can see the number on it. After this, you have to guess whether the number on the other paper is larger or smaller than the number you've just seen. If you guess right, I give you £1, if you guess wrong, you give me £1.
Now suppose we play this game over and over again. What strategy would you adopt so that, in the long run, you are guaranteed to win money from me?
Suppose I write down two numbers on separate pieces of paper, and then put them face down on a table so you can't see them. The numbers can be anything: positive or negative, integers or decimals, arbitrarily large or small, and are chosen totally at random. I then let you pick one of the pieces of paper and turn it over so you can see the number on it. After this, you have to guess whether the number on the other paper is larger or smaller than the number you've just seen. If you guess right, I give you £1, if you guess wrong, you give me £1.
Now suppose we play this game over and over again. What strategy would you adopt so that, in the long run, you are guaranteed to win money from me?
Re: Higher or Lower
Hmm... interesting.
Can I ask a really retarded question though  when you say "larger or smaller" do you just mean higher/lower  is 6 larger than 1, or larger than 1...?
(That looks really dumb but hopefully you'll get what I mean)
Can I ask a really retarded question though  when you say "larger or smaller" do you just mean higher/lower  is 6 larger than 1, or larger than 1...?
(That looks really dumb but hopefully you'll get what I mean)
Re: Higher or Lower
Not retarded at all, I should have been clearer. For the numbers you gave, we have
6 < 1 < 0 < 1 < 6
i.e. positive numbers are always bigger than negative numbers and the "bigger" the negative number, the smaller it is.
6 < 1 < 0 < 1 < 6
i.e. positive numbers are always bigger than negative numbers and the "bigger" the negative number, the smaller it is.
Re: Higher or Lower
Thanks Paul. I'm going to have a really good think about this before I look back in this thread again, because I don't want to see anyone else's answer.
At the moment I'm completely stuck that it will always be a complete 50/50 guess..... can't get my brain to escape this at the moment.
Nice puzzle!
At the moment I'm completely stuck that it will always be a complete 50/50 guess..... can't get my brain to escape this at the moment.
Nice puzzle!

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Re: Higher or Lower
I'm going to use my usual thinkasItype method rather than thinking first.
I've heard of something vaguely similar where there are two amounts of money and one is double the other and you "should" always swap because you'd average more by doing so. But I won't get into why that's silly here.
With your problem, then surely the numbers are probably going to be infinite? Or even if it has to be a finite number (but can be any), then surely the next number will be nearer infinity with infinite likelihood!
I'm not sure that this is a sensible problem to be honest. But in any case, it should be symmetrical around zero. So I would say higher for a negative number and lower for a positive number. But that's quite a boring answer.
Edit  Having said that, it should be symmetrical around any finite number, so as long as the first number is finite, it doesnt matter what you do!
I've heard of something vaguely similar where there are two amounts of money and one is double the other and you "should" always swap because you'd average more by doing so. But I won't get into why that's silly here.
With your problem, then surely the numbers are probably going to be infinite? Or even if it has to be a finite number (but can be any), then surely the next number will be nearer infinity with infinite likelihood!
I'm not sure that this is a sensible problem to be honest. But in any case, it should be symmetrical around zero. So I would say higher for a negative number and lower for a positive number. But that's quite a boring answer.
Edit  Having said that, it should be symmetrical around any finite number, so as long as the first number is finite, it doesnt matter what you do!
Last edited by Gavin Chipper on Tue Jan 29, 2008 11:48 pm, edited 2 times in total.
Re: Higher or Lower
I'm going to assume the numbers are independent of each other, but I'll say always guess higher if it's negative and always guess lower if it's positive. And if it's zero, then it doesn't matter.
Re: Higher or Lower
I assure you it's a sensible problem!GevinGavin wrote: I'm not sure that this is a sensible problem to be honest.
There is no requirement for symmetry about zero. The answer you and Conor gave supposes that 0 is the "middle" of the numbers, which seems sensible at first glance but is actually false; for any number we can think of there are an infinity of numbers above it and an infinity below, so there is no middle.GevinGavin wrote: But in any case, it should be symmetrical around zero. So I would say higher for a negative number and lower for a positive number. But that's quite a boring answer.
Re: Higher or Lower
Can you invent a middle point to help you?Paul Howe wrote:I assure you it's a sensible problem!GevinGavin wrote: I'm not sure that this is a sensible problem to be honest.
There is no requirement for symmetry about zero. The answer you and Conor gave supposes that 0 is the "middle" of the numbers, which seems sensible at first glance but is actually false; for any number we can think of there are an infinity of numbers above it and an infinity below, so there is no middle.GevinGavin wrote: But in any case, it should be symmetrical around zero. So I would say higher for a negative number and lower for a positive number. But that's quite a boring answer.

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Re: Higher or Lower
I did realise that just after I posted (see my edit  it does predate your post, honest!)Paul Howe wrote:GevinGavin wrote: There is no requirement for symmetry about zero. The answer you and Conor gave supposes that 0 is the "middle" of the numbers, which seems sensible at first glance but is actually false; for any number we can think of there are an infinity of numbers above it and an infinity below, so there is no middle.
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Re: Higher or Lower
frivolous law of large numbers, anyone?
Re: Higher or Lower
Nope!GevinGavin wrote:
Edit  Having said that, it should be symmetrical around any finite number
Michael, no advanced mathematics is required for this problem. I think anyone on here could come up with the solution, given a suitable flash of insight.

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Re: Higher or Lower
Well, I'm going to bed now, but I'll leave you with  do these numbers have to be finite?Paul Howe wrote:Nope!GevinGavin wrote:
Edit  Having said that, it should be symmetrical around any finite number
Michael, no advanced mathematics is required for this problem. I think anyone on here could come up with the solution, given a suitable flash of insight.
Re: Higher or Lower
Yes, they have to be finite. They are however, allowed to be arbitrarily large or small, i.e. there is no upper or lower limit on the numbers I can choose.
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Re: Higher or Lower
Are you allowed to use the 2nd number in more than 1 game? if so, then my strategy then would be to ask you each time to use that ORIGINAL 2nd number, so you only write 1 new number.
Say you have 2 numbers 'A' and 'B' and you turn over 'A', you will have a 50/50 shot at working out whether 'B' is higher or lower. Then because I never saw 'B', I ask you to use it again and you write out a new number 'C', which I will then turn over.
Going by what 'C' is and how it relates to 'A's answer, I could then start to work out if 'C' is higher or lower than 'B'. I would carry this on and it shouldn't take long to find out what 'B' is.
Is this sorta right?
EDIT: I assume that the 2nd one doesn't get turned over, right?
Say you have 2 numbers 'A' and 'B' and you turn over 'A', you will have a 50/50 shot at working out whether 'B' is higher or lower. Then because I never saw 'B', I ask you to use it again and you write out a new number 'C', which I will then turn over.
Going by what 'C' is and how it relates to 'A's answer, I could then start to work out if 'C' is higher or lower than 'B'. I would carry this on and it shouldn't take long to find out what 'B' is.
Is this sorta right?
EDIT: I assume that the 2nd one doesn't get turned over, right?
Re: Higher or Lower
Joseph, I'm afraid I choose two new numbers each time the game is played.
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Re: Higher or Lower
the frivolous law of large numbers states that most numbers are very very large  surely that doesn't count as advanced mathematics?
Re: Higher or Lower
Right, I've slept on this and given it more thought and pretty much drawn a blank...
I started to think about Benford's Law and wondered if I could use this at all. I don't believe I can, but this then led me to something else  you state that I can adopt a strategy which guarantees, in the long run, that I will win money from you. Does this mean that even if you are aware of my strategy, there is still nothing that you can do to prevent my advantage?
(Let's for example say that I choose lower if the first significant digit is 1, 2 or 3, and higher otherwise  while this may work for random numbers from an arbitrary set (nb I'm not saying it does, this is just an example), you are choosing the numbers  so once you were aware of the strategy, you could deilberately choose pairings which will make me lose...
Edit: I've just reread, and I'm now unclear about the numbers on the paper  are you "choosing" random numbers, or are they "genuine" random numbers, or doesn't it really matter? This is probably all irrelevant, but I'm pretty stumped by this, so when I think of something to explore I have to do so, because I have bogall else. Great puzzle!
I started to think about Benford's Law and wondered if I could use this at all. I don't believe I can, but this then led me to something else  you state that I can adopt a strategy which guarantees, in the long run, that I will win money from you. Does this mean that even if you are aware of my strategy, there is still nothing that you can do to prevent my advantage?
(Let's for example say that I choose lower if the first significant digit is 1, 2 or 3, and higher otherwise  while this may work for random numbers from an arbitrary set (nb I'm not saying it does, this is just an example), you are choosing the numbers  so once you were aware of the strategy, you could deilberately choose pairings which will make me lose...
Edit: I've just reread, and I'm now unclear about the numbers on the paper  are you "choosing" random numbers, or are they "genuine" random numbers, or doesn't it really matter? This is probably all irrelevant, but I'm pretty stumped by this, so when I think of something to explore I have to do so, because I have bogall else. Great puzzle!
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Re: Higher or Lower
I wonder if there is a significance to the fact that if one guesses wrong, you get £1  does that mean if they're the same then you get £1 regardless? That would desymmetricise the problem I suppose
edit: indeed, now I reread the problem, it could be a trick question  you could always choose two numbers the same, and we are left with no option but to lose (but that seems like a rather unlikely solution!)
edit: indeed, now I reread the problem, it could be a trick question  you could always choose two numbers the same, and we are left with no option but to lose (but that seems like a rather unlikely solution!)
Re: Higher or Lower
Nah, it's us  the guesser  who can adopt the strategy to guarantee profit, not the person writing the numbers down.Raccoon wrote:edit: indeed, now I reread the problem, it could be a trick question  you could always choose two numbers the same, and we are left with no option but to lose (but that seems like a rather unlikely solution!)
I need Paul to clarify whether the numbers he writes down are truly random, or whether he is choosing the numbers (such that they may as well be random, but he could counter certain strategies by intentionally picking pairs with certain characteristics).
However, once he has done so, I have absolutely no idea how I am going to use this information.
Re: Higher or Lower
Haha, I've never heard that expression before. I have to admit that after you said that I spent 5 minutes feeling embarrassed because I couldn't figure out how to apply LLN to solve this problemRaccoon wrote:the frivolous law of large numbers states that most numbers are very
very large  surely that doesn't count as advanced mathematics?
If the two numbers are equal, I'll be generous and give you the money. However, as this happens with probability zero, it doesn't affect the solution.
Anyway, the only caveat I had with posing this problem is that it's difficult to say what is meant by picking the numbers totally at random without being overly mathematical. Nonmathematically, each time we play I pick the numbers independently, i.e. you can't just look at the numbers I used in previous games and infer some sort of pattern. Furthermore, I am not choosing the numbers to counter my opponent's strategy, they're just random picks. Mathematically, in any given round of the game I pick both numbers from some probability distribution defined over the open interval ( infinity, infinity). However, each round I use a different, randomly chosen probability distribution to select the numbers.
If that mathematical bit didn't make sense to anyone, don't worry, as its totally unnecessary to solve the problem.
Re: Higher or Lower
Could you counter the strategy then, by not picking truly randomly?Paul Howe wrote:Furthermore, I am not choosing the numbers to counter my opponent's strategy, they're just random picks.
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Re: Higher or Lower
In fact, by the frivolous law of large numbers you'd go bankrupt buying enormous pieces of paper (to fit the numbers on) and so I'm doomed to never get my money off you!

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Re: Higher or Lower
I assume mugging you isn't a valid strategy. Going to have a think about this ... Nice problem in any event.
EDIT: Is this definitely just not the same as the envelope problem Gevin mentioned which shows that there is no strategy but a form of paradoxical logic that leads you to the erroneous conclusion that a strategy does exist?
EDIT: Is this definitely just not the same as the envelope problem Gevin mentioned which shows that there is no strategy but a form of paradoxical logic that leads you to the erroneous conclusion that a strategy does exist?

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Re: Higher or Lower
Are you sure it is totally unnecessary to solve the problem? You now state that you are just using a particular probability distribution. Previously I'd assumed that the probability distribution was uniform, and I'm sure things are different now!Paul Howe wrote:Haha, I've never heard that expression before. I have to admit that after you said that I spent 5 minutes feeling embarrassed because I couldn't figure out how to apply LLN to solve this problemRaccoon wrote:the frivolous law of large numbers states that most numbers are very
very large  surely that doesn't count as advanced mathematics?
If the two numbers are equal, I'll be generous and give you the money. However, as this happens with probability zero, it doesn't affect the solution.
Anyway, the only caveat I had with posing this problem is that it's difficult to say what is meant by picking the numbers totally at random without being overly mathematical. Nonmathematically, each time we play I pick the numbers independently, i.e. you can't just look at the numbers I used in previous games and infer some sort of pattern. Furthermore, I am not choosing the numbers to counter my opponent's strategy, they're just random picks. Mathematically, in any given round of the game I pick both numbers from some probability distribution defined over the open interval ( infinity, infinity). However, each round I use a different, randomly chosen probability distribution to select the numbers.
If that mathematical bit didn't make sense to anyone, don't worry, as its totally unnecessary to solve the problem.
So how are you randomly choosing probability distributions? Is there a probability distribution for that? If so, then surely we would be able to make inferences from previous games. Also, I'm not sure it can be assumed that draws take place with zero probability because the probability distribution of probability distributions could well include discrete distributions.
Aside from that, for you to have nonzero values on your probability density graph, the values would have to eventually tail off as you reach plus or minus infinity. So in each distribution, there would always be two points where it is incredibly unlikely for a higher or lower number to be picked. But these could be anywhere!
Basically, for this to make any sense, I cannot see a way of doing this where we could not infer from previous games. There would have to be some way of you picking your distributions. And these would have to follow some pattern. If I observed enough games, I would be able to see for which firstcard value ranges the next card was normally higher and for which it was lower. Otherwise if it's always 50/50, I might as well pick at random anyway!
So I would be able to pick based on these observations. Obviously the observations could have been freak chance, but freak chance probably won't happen and there's nothing I can do if it does!
Re: Higher or Lower
You took the words right out of my mouth.GevinGavin wrote:Are you sure it is totally unnecessary to solve the problem? You now state that you are just using a particular probability distribution. Previously I'd assumed that the probability distribution was uniform, and I'm sure things are different now!Paul Howe wrote:Haha, I've never heard that expression before. I have to admit that after you said that I spent 5 minutes feeling embarrassed because I couldn't figure out how to apply LLN to solve this problemRaccoon wrote:the frivolous law of large numbers states that most numbers are very
very large  surely that doesn't count as advanced mathematics?
If the two numbers are equal, I'll be generous and give you the money. However, as this happens with probability zero, it doesn't affect the solution.
Anyway, the only caveat I had with posing this problem is that it's difficult to say what is meant by picking the numbers totally at random without being overly mathematical. Nonmathematically, each time we play I pick the numbers independently, i.e. you can't just look at the numbers I used in previous games and infer some sort of pattern. Furthermore, I am not choosing the numbers to counter my opponent's strategy, they're just random picks. Mathematically, in any given round of the game I pick both numbers from some probability distribution defined over the open interval ( infinity, infinity). However, each round I use a different, randomly chosen probability distribution to select the numbers.
If that mathematical bit didn't make sense to anyone, don't worry, as its totally unnecessary to solve the problem.
So how are you randomly choosing probability distributions? Is there a probability distribution for that? If so, then surely we would be able to make inferences from previous games. Also, I'm not sure it can be assumed that draws take place with zero probability because the probability distribution of probability distributions could well include discrete distributions.
Aside from that, for you to have nonzero values on your probability density graph, the values would have to eventually tail off as you reach plus or minus infinity. So in each distribution, there would always be two points where it is incredibly unlikely for a higher or lower number to be picked. But these could be anywhere!
Basically, for this to make any sense, I cannot see a way of doing this where we could not infer from previous games. There would have to be some way of you picking your distributions. And these would have to follow some pattern. If I observed enough games, I would be able to see for which firstcard value ranges the next card was normally higher and for which it was lower. Otherwise if it's always 50/50, I might as well pick at random anyway!
So I would be able to pick based on these observations. Obviously the observations could have been freak chance, but freak chance probably won't happen and there's nothing I can do if it does!
Re: Higher or Lower
I can't escape that it's 50/50
I'm dumb.
My strategy is therefore to "think positive", much like the audience do on Deal Or No Deal. That always seems to work for them. Except when it doesn't.
I'm dumb.
My strategy is therefore to "think positive", much like the audience do on Deal Or No Deal. That always seems to work for them. Except when it doesn't.
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Re: Higher or Lower
my post got eaten, hmm...
but yes, I also would presume that the distribution of the random numbers is uniform on (infinity,+infinity), but the only thoughts I've had on this problem have involved me thinking about my first year course on number theory, and countable and uncountable infinites (I'd ask if the numbers need to be rational, but given that this is meant to be a simple answer, I prsume it doesn't matter), but I can't really seem that being the way to go...
but yes, I also would presume that the distribution of the random numbers is uniform on (infinity,+infinity), but the only thoughts I've had on this problem have involved me thinking about my first year course on number theory, and countable and uncountable infinites (I'd ask if the numbers need to be rational, but given that this is meant to be a simple answer, I prsume it doesn't matter), but I can't really seem that being the way to go...
Re: Higher or Lower
I wouldn't stake my life on it, but I'm fairly sure that I couldn'tCorby wrote: Could you counter the strategy then, by not picking truly randomly?
Yes, I'm sure. They're random numbers, so implicitly they have to come from a probability distribution, even if I didn't explicitly state this to avoid scaring people away. However, they emphatically DON'T come from a uniform distribution, because the integeral of such a distribution over the real numbers is infinity, and not one, as required. Once again though, you don't need this information to solve the problem.GevinGavin wrote:
Are
you sure it is totally unnecessary to solve the problem? You now state
that you are just using a particular probability distribution.
Previously I'd assumed that the probability distribution was uniform,
and I'm sure things are different now!
Fair enough, the problem can be restated as: "the game is played once, find a strategy that allows you to win with probability greater than 50%". A stronger, but less compelling formulation I think.GevinGavin wrote:
So how are you randomly
choosing probability distributions? Is there a probability distribution
for that? If so, then surely we would be able to make inferences from
previous games. Also, I'm not sure it can be assumed that draws take
place with zero probability because the probability distribution of
probability distributions could well include discrete distributions.
Basically, for this to
make any sense, I cannot see a way of doing this where we could not
infer from previous games. There would have to be some way of you
picking your distributions. And these would have to follow some
pattern. If I observed enough games, I would be able to see for which
firstcard value ranges the next card was normally higher and for which
it was lower.
No! There is a winning strategy that doesn't depend on being able to infer things from previous games. You are severely overcomplicating things, the actual answer is very simple.GevinGavin wrote:
Otherwise if it's always 50/50, I might as well pick at
random anyway!
Last edited by Paul Howe on Thu Jun 24, 2010 11:34 pm, edited 1 time in total.
Re: Higher or Lower
No, it's more like the opposite actually, because it really looks like a strategy shouldn't exist, when in fact it does. That's the beauty of the problem.David O'Donnell wrote: EDIT:
Is this definitely just not the same as the envelope problem Gevin
mentioned which shows that there is no strategy but a form of
paradoxical logic that leads you to the erroneous conclusion that a
strategy does exist?

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Re: Higher or Lower
Bugger!Paul Howe wrote:No, it's more like the opposite actually, because it really looks like a strategy shouldn't exist, when in fact it does. That's the beauty of the problem.David O'Donnell wrote: EDIT:
Is this definitely just not the same as the envelope problem Gevin
mentioned which shows that there is no strategy but a form of
paradoxical logic that leads you to the erroneous conclusion that a
strategy does exist?
Is there anything of similarity in that problem though? Here is a wiki link: http://en.wikipedia.org/wiki/Two_envelopes_problem
Re: Higher or Lower
I didn't read through the whole page, but it seems like a completely different problem to mine. Looks interesting though!
Re: Higher or Lower
Your strategy should clearly work for that "two envelopes" problem as well though (it matters not a jot that one number happens to be the double of the other presumably)  yet that wiki page appears to conclude that your intuition that there is no strategy is correct.Paul Howe wrote:I didn't read through the whole page, but it seems like a completely different problem to mine. Looks interesting though!
Can you explain that? I'm getting very confused here
Re: Higher or Lower
Pick a number and then if the first number revealed is bigger than that number, say the second one is smaller and vice versa. If the two numbers are either both bigger or both smaller than the number you've picked, then there's a 50/50 you've won, if it's in the middle, then you've definitely won. So you should make money. Hopefully.
Re: Higher or Lower
Interesting... so if we say x = [probability of our number being between the two], our probability of winning = x + 0.5 * (1x). Which is always higher than 0.5.Ben Pugh wrote:Pick a number and then if the first number revealed is bigger than that number, say the second one is smaller and vice versa. If the two numbers are either both bigger or both smaller than the number you've picked, then there's a 50/50 you've won, if it's in the middle, then you've definitely won. So you should make money. Hopefully.
That seems to work, but surely it's nonsense? There must be a flaw here somewhere, still sounds like balls to me.... I need to think about this some more.
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Re: Higher or Lower
I can't see the logic myself. If you pick a number and then the number turned over is higher, the 2nd number aint necessariy going to be lower, so its still a 50/50 decision isn't it?Corby wrote:Interesting... so if we say x = [probability of our number being between the two], our probability of winning = x + 0.5 * (1x). Which is always higher than 0.5.Ben Pugh wrote:Pick a number and then if the first number revealed is bigger than that number, say the second one is smaller and vice versa. If the two numbers are either both bigger or both smaller than the number you've picked, then there's a 50/50 you've won, if it's in the middle, then you've definitely won. So you should make money. Hopefully.
That seems to work, but surely it's nonsense? There must be a flaw here somewhere, still sounds like balls to me.... I need to think about this some more.
Re: Higher or Lower
OK, from the Wikipedia article, we're working in the second case of the two envelopes problem, where you can open the first envelope before deciding to swap. Let N be the amount of money in the first envelope. All the different different versions of the two envelopes problem are based on a fallacious argument that concludes that the expected value of the second envelope is > N, and therefore you should always switch, when in fact if you do the calculations correctly, you get the expected value to be N, so there is no advantage in switching. Notice however, that once we see the first value, we know the distribution of the second envelope, it can either be N/2 or 2N, and from this you can calculate the expectation of the second envelope.Corby wrote:
Your strategy should clearly work for that "two
envelopes" problem as well though (it matters not a jot that one number
happens to be the double of the other presumably)  yet that wiki page
appears to conclude that your intuition that there is no strategy is
correct.
Can you explain that? I'm getting very confused here
In my problem, we have no such distributional information and cannot calculate the expectation of the value. Even if we could calculate the expected value, we couldn't draw any conclusions, because, unlike the two envelopes problem, your goal in my problem is merely to guess higher or lower, not how much higher or lower. For example, suppose the first envelope contained 2 and we've somehow discovered that the distribution of the 2nd envelope is 1,1,1,1,1,1,1,1,10,10. The expected value of this is 2.8, which is higher than 2, but guessing higher would on average lose you 8 games from 10, so even having knowledge of the expected value would not guarantee you success in my game without further distributional information, which is not available.
I hope that's clear! It's always much easier for me to explain things face to face, with a pencil and paper, I find it much harder to obtain clarity over the internet.
Yes! Well done mate. The only way this could be wrong is if your number never lies between my numbers, but since this happens with finite (although admittedly very small) probability, this is not the case.Ben Pugh wrote:Pick a number and then if the first number revealed is bigger than that
number, say the second one is smaller and vice versa. If the two
numbers are either both bigger or both smaller than the number you've
picked, then there's a 50/50 you've won, if it's in the middle, then
you've definitely won. So you should make money. Hopefully.

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Re: Higher or Lower
Well, you could tell us we have just one go at it. Or that our memories are wiped after each game. Or that we only play once, but so do many other people and we are all trying to make as much as we can for "the team" as it will be split evenly afterwards.Paul Howe wrote:Actually, there is a way to construct this so that you can't make inferences from previous games, unfortunately it's less easy to tell you why this is so, as the proof is rather nonelementary and technical.

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Re: Higher or Lower
This sounds a bit shaky to me. I could pick my number to be zero. And then I would always go lower for a positive number and higher for a negative number. So I would win on average?Paul Howe wrote:Yes! Well done mate. The only way this could be wrong is if your number never lies between my numbers, but since this happens with finite (although admittedly very small) probability, this is not the case.
What if I simply ignored this strategy and simply went lower for a positive number and higher for a negative one? Would the results be different because I don't have the number zero in my mind ans a "third number"? This is what I originally went for before I'd given it that much thought and it was wrong!
What if two separate people who can't communicate with each other are being shown the same cards and one is adopting each strategy  i.e making exactly the same choices but throught different reasoning?
Re: Higher or Lower
Actually it is correct, just picking 0 every time is a special case, which I was a bit too dismissive to realise as your justification was wrong. The correct solution is also significantly more general in that you could pick ANY number and it would still work, or pick a different number each time.GevinGavin wrote:What if I simply ignored this strategy and simply wentPaul Howe wrote:Yes! Well done mate. The only way this could be
wrong is if your number never lies between my numbers, but since this
happens with finite (although admittedly very small) probability, this
is not the case.
lower for a positive number and higher for a negative one? Would the
results be different because I don't have the number zero in my mind
ans a "third number"? This is what I originally went for before I'd
given it that much thought and it was wrong!

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Re: Higher or Lower
Also, I don't see why it would make any difference if I picked a number before or after seeing the first card. So I could make it better to pick higher or lower depending on what I felt like.
But then I could have an opponent who is shown the same cards and always does the opposite of me. We can't both be right!
But then I could have an opponent who is shown the same cards and always does the opposite of me. We can't both be right!

 Postapocalypse
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Re: Higher or Lower
Also I think I sneaked in an edit while you must have been writing that, as there's more to my post now (as well as a new one).Paul Howe wrote:Actually it is correct, just picking 0 every time is a special case, which I was a bit too dismissive to realise as your justification was wrong. The correct solution is also significantly more general in that you could pick ANY number and it would still work, or pick a different number each time.
Re: Higher or Lower
It DOES make a difference and it ONLY works if you pick your number before you see my first bit of paper. Lets call my first number a, my second b, and your guess c. Now, suppose you choose c after a is known, that you choose c > a (the analysis is symmetrical for c < a ) and then follow Ben's strategy. You have two cases:GevinGavin wrote:Also, I don't see why it would make any difference if I picked a number
before or after seeing the first card. So I could make it better to
pick higher or lower depending on what I felt like.
But then I could have an opponent who is shown the same cards and always does the opposite of me. We can't both be right!
a<c<b in which you guess higher and win 100% of the time.
a,b<c in which you guess higher but how often do you win? When c is picked a priori, its half the time, but what about when c is picked afterwards? Well, suppose you picked c larger, but very close to a. To win you need a<b<c but the closer c is to a, the more likely we have b<a and you lose. Of course, picking c close to a also increases the chances of falling into the 100% win case, but you'll never fall into this case more than 50% of the time, and when you do fall into it exactly 50% of the time (on average), you always lose in the other case, so your overall odds of winning never go beyond 50%.
So, choosing your number after you've seen mine destroys the guarantee that on average you win half the time when your number doesn't split mine. Again, counterintuitive, but true.

 Postapocalypse
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Re: Higher or Lower
To be honest, I'm not 100% sure I followed all of that properly, but this seems to rest on the assumption that there is a finite probability that the second number will be between the first number and the picked number. For a uniform distribution, this wouldn't be the case (there is an infinitessimally small chance (zero, basically) of the number being in any finite range), but we did abolish the uniform distribution. But what about other distributions?Paul Howe wrote:It DOES make a difference and it ONLY works if you pick your number before you see my first bit of paper. Lets call my first number a, my second b, and your guess c. Now, suppose you choose c after a is known, that you choose c > a (the analysis is symmetrical for c < a ) and then follow Ben's strategy. You have two cases:GevinGavin wrote:Also, I don't see why it would make any difference if I picked a number
before or after seeing the first card. So I could make it better to
pick higher or lower depending on what I felt like.
But then I could have an opponent who is shown the same cards and always does the opposite of me. We can't both be right!
a<c<b in which you guess higher and win 100% of the time.
a,b<c in which you guess higher but how often do you win? When c is picked a priori, its half the time, but what about when c is picked afterwards? Well, suppose you picked c larger, but very close to a. To win you need a<b<c but the closer c is to a, the more likely we have b<a and you lose. Of course, picking c close to a also increases the chances of falling into the 100% win case, but you'll never fall into this case more than 50% of the time, and when you do fall into it exactly 50% of the time (on average), you always lose in the other case, so your overall odds of winning never go beyond 50%.
So, choosing your number after you've seen mine destroys the guarantee that on average you win half the time when your number doesn't split mine. Again, counterintuitive, but true.
When the first card is revealed (let's say my number is higher than it), something is already being stated about the distribution, and I don't think we can assume any more that there is a 50/50 chance that the second number will be higher than both. This is because we cannot assume any more that any finite number is equally likely. We have defined this to be the case by making the distribution nonuniform.
But going back to your post. I still don't see how picking the number before or after makes a difference. Let's say I pick a number first. 9. the first number comes out as an 8. Or maybe 8 comes up and I pick a 9. All we really know is that an 8 has come up using some predetermined random distribution. Nothing else has happened that has given us any information. What I pick has no effect on the distribution, which was chosen beforehand.
Modifying my earlier example, let's say me and my opponent both pick a number before either card comes up and do not confer. I pick 9 and he picks 7. Then the first card is 8. I say higher and he says lower. Then we discuss the situation before the second card comes up. What now?
And you could not do this in practice. So does that not make you suspicious? I think it's all smoke and mirrors.
Re: Higher or Lower
Any properly defined (ie its integral over the reals is one) distribution will work. Randomly drawn numbers implicitly come from such a distribution. Saying "what about other distributions" is kind of like saying "well what if 1=2?". Just doesn't make sense. All that's required for the splitting case to have nonzero probability is that the distribution from which I draw my numbers and the distribution from which you draw yours overlap somewhere. If you draw your numbers from a distribution that is nonzero everywhere (e.g. the normal distribution) that will guarantee it.GevinGavin wrote:
To be honest, I'm not
100% sure I followed all of that properly, but this seems to rest on
the assumption that there is a finite probability that the second
number will be between the first number and the picked number. For a
uniform distribution, this wouldn't be the case (there is an
infinitessimally small chance (zero, basically) of the number being in
any finite range), but we did abolish the uniform distribution. But
what about other distributions?
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.GevinGavin wrote:
But going back to your post. I still
don't see how picking the number before or after makes a difference.
Let's say I pick a number first. 9. the first number comes out as an 8.
Or maybe 8 comes up and I pick a 9. All we really know is that an 8 has
come up using some predetermined random distribution. Nothing else has
happened that has given us any information. What I pick has no effect
on the distribution, which was chosen beforehand.
I'll dig up some discussion of this problem on the internet later if you still don't believe me.

 Postapocalypse
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 Joined: Mon Jan 21, 2008 10:37 pm
Re: Higher or Lower
When I said "What about other distributions?" I was just referring to distributions other than the uniform one that we rejected. My followon point being that by having a distribution that is not uniform and then specifying a data point, the symmetry is ruined. But I will now be more specific in a minute.Paul Howe wrote:Any properly defined (ie its integral over the reals is one) distribution will work. Randomly drawn numbers implicitly come from such a distribution. Saying "what about other distributions" is kind of like saying "well what if 1=2?". Just doesn't make sense. All that's required for the splitting case to have nonzero probability is that the distribution from which I draw my numbers and the distribution from which you draw yours overlap somewhere. If you draw your numbers from a distribution that is nonzero everywhere (e.g. the normal distribution) that will guarantee it.
OK.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.
 Joseph Bolas
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 Location: Liverpool, UK
Re: Higher or Lower
On to the next mathematical conundrum .
Re: Higher or Lower
Hmm... okay, I had a few hours to think about this while driving to/from my son's football training, and obviously then standing outside watching him. (I definitely didn't sit in the bar and have a beer, in case my bird reads this)Corby wrote:Interesting... so if we say x = [probability of our number being between the two], our probability of winning = x + 0.5 * (1x). Which is always higher than 0.5.
Anyway, I am unhappy on a couple of points. The first of which Gevin has already touched upon, that being that more than one person could play this game at the same time and it wouldn't make any difference, as the rules wouldn't change. Two of us could each think of our own numbers, see the number on the card, and then assert higher or lower individually. In fact, we could have an inifinite number of people doing this, such that we're covering the entire set of real numbers*, and yet we'd still all win more than 50% of the time? That smells bad to me.
The second problem I have is kinda marked with the * above. It's all about infinity. I have a layman's view of infinity. To me, it means "limitless". I don't know if this therefore actually means that an infinite number of people would cover the entire set of real numbers. I don't know if this means that an infinite number of monkeys would eventually type the complete works of Jordan. And I certainly don't know how you use it in mathematical formulae/equations. And yet there I have above. My x, while it is a probability of something that might happen, and therefore I know it is greater than 0 and less than/equal to 1, is actually something divided by infinity. In actual fact, it's infinity divided by infinity. And that messes me up. There are an infinite number of numbers between your two numbers, and an infinite number of numbers to choose from. Where the f**k does that leave my x? 0? 1? I don't know.

 Postapocalypse
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Re: Higher or Lower
Right, I've thought about this and it is right. Sorry Paul for doubting you.
To explain in my own words:
C1 os the number on the first card and C2 the number on the second.
We have four possibilities:
C1 and C2 are both higher than x  50/50 chance of winning
C1 and C2 are both lower than x  50/50 chance of winning
C1 is higher than x and C2 lower  I win
C2 is higher than x and C1 lower  I win
This does not require any counterintuitive notions of infinity to work. I think it works for any distribution (so could be demonstrated). If someone tells you they are going to use the normal distribution, for example, but not necessarily with mean 0, it would work.
If I pick 1 and the mean happens to be 4, this is what happens:
There is a small chance that both C1 and C2 are lower than x. It is 50/50 which is lower than which, so we have a 50/50 chance of winning when we say lower.
There is a high chance that both C1 and C2 are higher than x. Again it is 50/50 which is higher than which. I would say higher and have a 50/50 chance. The small and high chances in these cases do not matter though.
I would always win when I split the two.
Sorry again.
To explain in my own words:
C1 os the number on the first card and C2 the number on the second.
We have four possibilities:
C1 and C2 are both higher than x  50/50 chance of winning
C1 and C2 are both lower than x  50/50 chance of winning
C1 is higher than x and C2 lower  I win
C2 is higher than x and C1 lower  I win
This does not require any counterintuitive notions of infinity to work. I think it works for any distribution (so could be demonstrated). If someone tells you they are going to use the normal distribution, for example, but not necessarily with mean 0, it would work.
If I pick 1 and the mean happens to be 4, this is what happens:
There is a small chance that both C1 and C2 are lower than x. It is 50/50 which is lower than which, so we have a 50/50 chance of winning when we say lower.
There is a high chance that both C1 and C2 are higher than x. Again it is 50/50 which is higher than which. I would say higher and have a 50/50 chance. The small and high chances in these cases do not matter though.
I would always win when I split the two.
Sorry again.
Re: Higher or Lower
OK guys, I'm not sure I'm ever going to be able to completely convince you, but I'll do one last post where I outline my strategy, say why I think it works, and address some of your more persistent objections.
My strategy would be:
1. Pick a number c randomly, BEFORE any of the numbers on the paper are revealed.
2. If a, the number on the first piece of paper, is smaller than c, guess that b, the second number, is higher. Otherwise guess lower
I claim that this works because there are two cases:
a,b < c or a,b > c (the guess doesn't split the numbers)
a < c < b or b < c < a (the guess splits the numbers)
In the first case, since there is no way to distinguish between a and b, you win and lose with probability 1/2. In the second case, you always win. As long as the second case actually occurs occasionally, then our overall probability of success is greater than 1/2. The simplest way to guarantee this is for you to pick your numbers from a continuous distribution that is nonzero everywhere, the normal distribution will do
So, why does it make a difference if you choose c after a is revealed? Surely you have exactly the same cases as before? The difference is that knowing a, c has to be chosen with reference to a. See my post above for the reasoning, but here is a numerical example. Suppose a = 15 has been revealed. Now suppose you choose c to be 15.0000000001. Will you win half the time in the first case, with this choice of c? No, you will almost certainly lose unless a,b have been drawn from an extremely strange looking distribution. You might try and get away from this by choosing c to be much different to a, but no matter how you choose it it is still subject to the same effect.
What about two (or more) people playing the game simultaneously? Well, as long as they're playing independently, then yes in the long run they'll both win more than half the time. In the case Gevin gives one wins and one loses, but it's completely possible for them to both win at the same time. If there is a strategy to beat me in the long run, then it stands to reason that you can play for long enough to take me to the cleaners, stop, and then start a new game and take me to the cleaners again. This I think seems very sensible. But, it is equivalent to the situation where two (or more) people play separately, because the player's strategies don't depend on each other and they may as well not be aware that the other players exist. You have to be a bit careful with infinities, but if a countably infinite number (that is, you can label each person with an integer) of players adopted the strategy, I think they would all win in the long run, although I might find that a very expensive game!
Again, I'll try and dig up some links where people more authoratitive than me discuss this problem, but everything I've found so far treats it in a more mathematical setting, I'm still trying to find something that everyone will be able to follow.
Oh and Gevin don't be sorry, you should never accept anything unless you're totally convinced of it. I think you'd make a very good mathematician!
My strategy would be:
1. Pick a number c randomly, BEFORE any of the numbers on the paper are revealed.
2. If a, the number on the first piece of paper, is smaller than c, guess that b, the second number, is higher. Otherwise guess lower
I claim that this works because there are two cases:
a,b < c or a,b > c (the guess doesn't split the numbers)
a < c < b or b < c < a (the guess splits the numbers)
In the first case, since there is no way to distinguish between a and b, you win and lose with probability 1/2. In the second case, you always win. As long as the second case actually occurs occasionally, then our overall probability of success is greater than 1/2. The simplest way to guarantee this is for you to pick your numbers from a continuous distribution that is nonzero everywhere, the normal distribution will do
So, why does it make a difference if you choose c after a is revealed? Surely you have exactly the same cases as before? The difference is that knowing a, c has to be chosen with reference to a. See my post above for the reasoning, but here is a numerical example. Suppose a = 15 has been revealed. Now suppose you choose c to be 15.0000000001. Will you win half the time in the first case, with this choice of c? No, you will almost certainly lose unless a,b have been drawn from an extremely strange looking distribution. You might try and get away from this by choosing c to be much different to a, but no matter how you choose it it is still subject to the same effect.
What about two (or more) people playing the game simultaneously? Well, as long as they're playing independently, then yes in the long run they'll both win more than half the time. In the case Gevin gives one wins and one loses, but it's completely possible for them to both win at the same time. If there is a strategy to beat me in the long run, then it stands to reason that you can play for long enough to take me to the cleaners, stop, and then start a new game and take me to the cleaners again. This I think seems very sensible. But, it is equivalent to the situation where two (or more) people play separately, because the player's strategies don't depend on each other and they may as well not be aware that the other players exist. You have to be a bit careful with infinities, but if a countably infinite number (that is, you can label each person with an integer) of players adopted the strategy, I think they would all win in the long run, although I might find that a very expensive game!
Again, I'll try and dig up some links where people more authoratitive than me discuss this problem, but everything I've found so far treats it in a more mathematical setting, I'm still trying to find something that everyone will be able to follow.
Oh and Gevin don't be sorry, you should never accept anything unless you're totally convinced of it. I think you'd make a very good mathematician!
Re: Higher or Lower
Cheers Paul, I don't think you've really covered anything there that I wasn't already thinking, but thanks for taking the time
I understand the formula  I mean, it's ludicrously simple really  and I certainly can't see any flaws in it at the moment, but my intuition is still nagging at me. I completely accept that you're right about it (particularly as you have supermathematical backup resources at hand!), but I still need to think about it a bit more to completely okay myself with it. It's really annoying when you set people puzzles like this, and they don't get it but are happy just to leave it saying "nah, I don't agree". Makes my blood boil, that.
I'm still a little unhappy about the case where we have infinite players, all picking their own random numbers before seeing yours. I don't see how we can all always win more than 50% of the time if I have infinity people guessing lower and infinity people guessing higher... but then as I already said, I don't really understand how to use infinity in "normal" equations and that. I'm making the assumption here that my "infinity people guessing higher" is always equal in my number to my "infinity people guessing lower", which almost certainly isn't true I suppose.
(Is it?)
I understand the formula  I mean, it's ludicrously simple really  and I certainly can't see any flaws in it at the moment, but my intuition is still nagging at me. I completely accept that you're right about it (particularly as you have supermathematical backup resources at hand!), but I still need to think about it a bit more to completely okay myself with it. It's really annoying when you set people puzzles like this, and they don't get it but are happy just to leave it saying "nah, I don't agree". Makes my blood boil, that.
I'm still a little unhappy about the case where we have infinite players, all picking their own random numbers before seeing yours. I don't see how we can all always win more than 50% of the time if I have infinity people guessing lower and infinity people guessing higher... but then as I already said, I don't really understand how to use infinity in "normal" equations and that. I'm making the assumption here that my "infinity people guessing higher" is always equal in my number to my "infinity people guessing lower", which almost certainly isn't true I suppose.
(Is it?)
Re: Higher or Lower
No worries Jon, I went through the exact same "this has to be bollocks" line of thinking the first time I saw the solution and it took me a while to completely accept it. I'm not too sure about the infinities either, so I won't try and convince you one way or the other.
I think I'll lay off the proabibility the next time I do a puzzle, as keeping up with this one was tiring.
I think I'll lay off the proabibility the next time I do a puzzle, as keeping up with this one was tiring.
Re: Higher or Lower
Nah, go for it if you have any more! (Though maybe give me a day or so to rest up...) Probability puzzles are ace, I love stuff like this! I can't believe I haven't come across this one before actually, it's a beaut. I put it to some of the guys at work this morning, and Ben's answer was posted just before I left tonight  I had enough time to relay it to them and the general consensus was "bollocks". Should be some interesting discussions tomorrow... unfortunately the ultimate proof is usually to play out games (often using a simulator) which is going to be a bit tricky with this one though
Re: Higher or Lower
Yeah, I'd like to see some more puzzles, I find them very enjoyable.
Re: Higher or Lower
Probably my final word on this is particular question is to query the assertion that this strategy guarantees me to be ahead in the long run, playing the game as described in Paul's original post.
Our advantage is based entirely on the probability of our number being between the two numbers written down. Provided this is not zero (and I certainly can't sensibly claim it to be zero), we do indeed have an advantage.
However, this probability is calculated by basically dividing some kind of infinity by some larger kind of infinity. As I have said before, I don't know how to do maths with these kinds of concepts, but this must be so ludicrously close to zero that I can't accept it would ever provide any guarantee of profit, other than in an infinite series of games.
The original puzzle is given in a very "real world" kind scenario, with Paul writing down numbers, and £1 exchanging hands each game. I don't accept that this strategy provides any kind of guarantee of profit in such a game. Even if we played for 20 hours a day for a hundred years at a rate of one game every 30 seconds, that would still only make a little over 100 million games  I don't reckon this is enough to guarantee anything in a game where the chances are so close to 50/50. Realistically we would never even play anywhere near that many games  we could actually spend our entire lifetimes writing down just one of the numbers!
Sorry if this seems overly pedantic, but I think it's a valid point. When people disbelieve the "Monty Hall" puzzle, I'll happily set the game up and play it for real money (offering them a greater than 50% return, with certain provisos regarding the minimum number of games to give me enough confidence of being ahead). I'd never do that in this instance.
I can understand Paul's reasons for wording the question like this, as to state the problem in purely mathematical terms would make it less appealing, but I do think it's an issue...
Our advantage is based entirely on the probability of our number being between the two numbers written down. Provided this is not zero (and I certainly can't sensibly claim it to be zero), we do indeed have an advantage.
However, this probability is calculated by basically dividing some kind of infinity by some larger kind of infinity. As I have said before, I don't know how to do maths with these kinds of concepts, but this must be so ludicrously close to zero that I can't accept it would ever provide any guarantee of profit, other than in an infinite series of games.
The original puzzle is given in a very "real world" kind scenario, with Paul writing down numbers, and £1 exchanging hands each game. I don't accept that this strategy provides any kind of guarantee of profit in such a game. Even if we played for 20 hours a day for a hundred years at a rate of one game every 30 seconds, that would still only make a little over 100 million games  I don't reckon this is enough to guarantee anything in a game where the chances are so close to 50/50. Realistically we would never even play anywhere near that many games  we could actually spend our entire lifetimes writing down just one of the numbers!
Sorry if this seems overly pedantic, but I think it's a valid point. When people disbelieve the "Monty Hall" puzzle, I'll happily set the game up and play it for real money (offering them a greater than 50% return, with certain provisos regarding the minimum number of games to give me enough confidence of being ahead). I'd never do that in this instance.
I can understand Paul's reasons for wording the question like this, as to state the problem in purely mathematical terms would make it less appealing, but I do think it's an issue...
Re: Higher or Lower
It probably isn't possible to convince you of this without using calculus, but 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. This is called the proabability density function (pdf). Is it even possible for such a graph to have a finite area, given that it extends infinitely in all directions? Yes, its similar to when you have a sum like a 1/2 + 1/4 + 1/8 + 1/16 +.... etc, which has an infinite number of terms, but whose value never exceeds 1. Then, to calculate the probability that a continuous quantity lies between a and b, we calculate the area underneath the graph and between a and b. This gives a sensible probability because it always lies between 0 and 1 (and satisfies some other axioms), but avoids issues about dividing infinity by infinity.Corby wrote:Probably my final word on this is particular question is to query the
assertion that this strategy guarantees me to be ahead in the
long run, playing the game as described in Paul's original post.
However, this probability is calculated by basically
dividing some kind of infinity by some larger kind of infinity. As I
have said before, I don't know how to do maths with these kinds of
concepts, but this must be so ludicrously close to zero that I can't
accept it would ever provide any guarantee of profit, other than in an
infinite series of games.
I quite agree, your advantage over me in this puzzle is ludicrously (but not infinitely!) small, and we might have to play for a very long time indeed for you to emerge a clear winner.Corby wrote: Realistically we would never
even play anywhere near that many games  we could actually spend our
entire lifetimes writing down just one of the numbers!
Re: Higher or Lower
Lol, I lied Paul, that wasn't my final post at all. I'm still not happy here!
I've just thought of another way of approaching this, which could well be flawed, as it includes some kind of interference. I don't think it messes it up, but I know how careful you have to be when you do stuff like this.
Okay  here we go. As already stated, our advantage comes from the probability that our number (in our heads) is between the two numbers written down. If we call this probability x, our chance of correctly guessing higher or lower is x + 0.5(1x), which simplifies to 0.5 + 0.5x. x is some kind of infinity (the range of numbers between the two numbers written down  I'll call this a), divided by some larger kind of infinity (the total range of numbers  I'll call this b). I know nothing about doing calculations with infinity, but I accept this to be greater than 0 because a is a subset of b.
Right. Now  once we've turned over one piece of paper to see a number, all that is now important is whether the other number is higher or lower. We never even need to know what this other number is, somebody could come along, look at both pieces of paper, and then change the one we haven't seen to simply say either "higher" or "lower" as appropriate. This doesn't mess up the game, doesn't change our guess, doesn't change the outcome. However, instead of saying "higher", let's say the scribble the number out and replace it with "infinity", and instead of "lower" replace it with "negative infinity". Our range (a) now extends from either + or  infinity to the number we can see. Hasn't our value of x just massively increased? If we accepted that a/b was greater than zero because a was a subset of b, hasn't a just increased by the same logic? (ie that the "old" range is a subset of the "new" range)
That would be lunacy  nothing has changed, our chances of being correct can't possibly have increased via this amendment. I think it kinda suggests that x is zero all along regardless... but I'm not sure.
NB  this almost certainly isn't my last post on the subject
I've just thought of another way of approaching this, which could well be flawed, as it includes some kind of interference. I don't think it messes it up, but I know how careful you have to be when you do stuff like this.
Okay  here we go. As already stated, our advantage comes from the probability that our number (in our heads) is between the two numbers written down. If we call this probability x, our chance of correctly guessing higher or lower is x + 0.5(1x), which simplifies to 0.5 + 0.5x. x is some kind of infinity (the range of numbers between the two numbers written down  I'll call this a), divided by some larger kind of infinity (the total range of numbers  I'll call this b). I know nothing about doing calculations with infinity, but I accept this to be greater than 0 because a is a subset of b.
Right. Now  once we've turned over one piece of paper to see a number, all that is now important is whether the other number is higher or lower. We never even need to know what this other number is, somebody could come along, look at both pieces of paper, and then change the one we haven't seen to simply say either "higher" or "lower" as appropriate. This doesn't mess up the game, doesn't change our guess, doesn't change the outcome. However, instead of saying "higher", let's say the scribble the number out and replace it with "infinity", and instead of "lower" replace it with "negative infinity". Our range (a) now extends from either + or  infinity to the number we can see. Hasn't our value of x just massively increased? If we accepted that a/b was greater than zero because a was a subset of b, hasn't a just increased by the same logic? (ie that the "old" range is a subset of the "new" range)
That would be lunacy  nothing has changed, our chances of being correct can't possibly have increased via this amendment. I think it kinda suggests that x is zero all along regardless... but I'm not sure.
NB  this almost certainly isn't my last post on the subject
Re: Higher or Lower
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...GevinGavin wrote:Right, I've thought about this and it is right. Sorry Paul for doubting you.
To explain in my own words:
C1 os the number on the first card and C2 the number on the second.
We have four possibilities:
C1 and C2 are both higher than x  50/50 chance of winning
C1 and C2 are both lower than x  50/50 chance of winning
C1 is higher than x and C2 lower  I win
C2 is higher than x and C1 lower  I win
Re: Higher or Lower
Except that this really isn't how you calculate x. If we let a and b be my two numbers (sorry to conflict with your notation, but it is simpler to write like this), then as per my previous post the chance of your number splitting a,b is the area under the pdf between a and b (which is between 0 and 1) divided by the total area under the pdf (which is 1), a finite number divided by a finite number!Corby wrote:
Okay  here we go. As already stated, our advantage
comes from the probability that our number (in our heads) is between
the two numbers written down. If we call this probability x, our chance
of correctly guessing higher or lower is x + 0.5(1x), which simplifies
to 0.5 + 0.5x. x is some kind of infinity (the range of numbers between
the two numbers written down  I'll call this a), divided by some
larger kind of infinity (the total range of numbers  I'll call this
b). I know nothing about doing calculations with infinity, but I accept
this to be greater than 0 because a is a subset of b.

 Kiloposter
 Posts: 1060
 Joined: Wed Jan 30, 2008 9:57 pm
Re: Higher or Lower
I thought I posted something yesterday, but it hasn't appeared, and in any case I hadn't noticed there was more than one page of postings, so here's something more uptodate!
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.
What if your two numbers had to be whole numbers between 1 and 360, representing degrees, and the question was whether the shortest route between them was clockwise or anticlockwise? Would your method work? If not, why not?
David W
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.
What if your two numbers had to be whole numbers between 1 and 360, representing degrees, and the question was whether the shortest route between them was clockwise or anticlockwise? Would your method work? If not, why not?
David W
Re: Higher or Lower
Fair enough  I think this may actually help me on my second point though (with regards to altering one of the numbers to either  or + infinity, as appropriate). Note that this can all be done entirely independently of you choosing your number and seeing the number. The card you will be shown can be selected entirely at random, and then the other card amended to +/ infinity, and then you can choose your number in your head. This should all make absolutely no difference to the outcome of the game, yet now your portion of the graph between a and b is much larger than before, ergo your chance of success should increase accordingly. It doesn't, it just can't.Paul Howe wrote:Except that this really isn't how you calculate x.
Essentially, we end up calculating the probability of whether your number is either higher than the lower of the two numbers, or lower than the higher of the two. The whole notion of intersecting the two is a bumsteer.
(I know I'm going to be wrong but I can't let this go until I'm happy!)
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