Heights

[Gavin Chipper]

John has no idea how tall he is relative to other people, so he decides to compare his height with some. After nine random comparisons, he is taller than five and shorter than four. Assuming that no two people are exactly the same height, what is the probability that he will be taller than the next person?

[Dinos Sfyris]

This seems too obvious so is probably wrong but I may as well go ahead and say it. 5/9?

[Joseph Bolas]

John has no idea how tall he is relative to other people, so he decides to compare his height with some. After nine random comparisons, he is taller than five and shorter than four. Assuming that no two people are exactly the same height, what is the probability that he will be taller than the next person?

If you were to take all the possibilits of 10 people regardless of height then it should be 1024? Then if you take all the combinations out of those 1024 where there are 6-small & 4 large (first 9 in any order, the 10th being short) then that number, over 1024, simplified (if possible), should be the answer?

[Gavin Chipper]

This seems too obvious so is probably wrong but I may as well go ahead and say it. 5/9?

Nope.

[Gavin Chipper]

If you were to take all the possibilits of 10 people regardless of height then it should be 1024? Then if you take all the combinations out of those 1024 where there are 6-small & 4 large (first 9 in any order, the 10th being short) then that number, over 1024, simplified (if possible), should be the answer?

That's not how I'd look at it, but feel free to work out what it comes to.

[Joseph Bolas]

If you were to take all the possibilits of 10 people regardless of height then it should be 1024? Then if you take all the combinations out of those 1024 where there are 6-small & 4 large (first 9 in any order, the 10th being short) then that number, over 1024, simplified (if possible), should be the answer?

That's not how I'd look at it, but feel free to work out what it comes to.

Knowing me, I am definitely overcomplicating this, but I think this way is going to work .

[Conor]

Hmm. 6/11?
I'm not too sure, and I can a potential flaw with what I've done, but it's the best I've got.

[Joseph Bolas]

If you were to take all the possibilits of 10 people regardless of height then it should be 1024? Then if you take all the combinations out of those 1024 where there are 6-small & 4 large (first 9 in any order, the 10th being short) then that number, over 1024, simplified (if possible), should be the answer?

That's not how I'd look at it, but feel free to work out what it comes to.

I have worked this out, and the figure does seem wrong, but I will post what I got anyway (on the very very slim chance I may be right ):

Out of the 1024 possible combinations, only 512 can end with a short person.

There are a total of 126 combinations out of the remaining 512, where there are 4 tall people in the combinations. Therefore if you simplify 126/512 you get 63/256.

[Ben Pugh]

1/2?

[Joseph Bolas]

This may sound weird but is there a 100% chance (so 1/1) that the next guy is shorter than he is?

[David Williams]

I was actually thinking of posting the same puzzle!

I'll just say to anyone who thinks it's 5/9 . . . What would your answer be if all nine were taller than him, or when he'd only met one person?

David

[Gavin Chipper]

Hmm. 6/11?
I'm not too sure, and I can a potential flaw with what I've done, but it's the best I've got.

What's your reasoning for this?

[Gavin Chipper]

I have worked this out, and the figure does seem wrong, but I will post what I got anyway (on the very very slim chance I may be right ):

Out of the 1024 possible combinations, only 512 can end with a short person.

There are a total of 126 combinations out of the remaining 512, where there are 4 tall people in the combinations. Therefore if you simplify 126/512 you get 63/256.

That's not what I've got!

[Gavin Chipper]

1/2?

Nope.

This may sound weird but is there a 100% chance (so 1/1) that the next guy is shorter than he is?

Nope!

[Joseph Bolas]

How about if I simplify 63/256 rounding up to the nearest whole number and say is it approx 1/4?

EDIT: I will also say 1/16 or 1/64, but think these are wrong too, but I gave it a good shot .

[Conor]

Hmm. 6/11?
I'm not too sure, and I can a potential flaw with what I've done, but it's the best I've got.

What's your reasoning for this?

Call john J, and anyone else X.
You can write it out as |X|X|X|X|X|J|X|X|X|X|
| is where the next person must go (since there can't be two people with the same height), 5 of them are taller than John and 6 smaller, so it's 6/11.

[Kirk Bevins]

I don't get this problem. Surely if he is say 190cm, then there are infinitely many heights below this and infinitely above this so the probability is 1/2 or 1 - but these are both wrong. I don't really understand your question.

[Charlie Reams]

It would only be 1/2 if all heights (over some range) were equally probable and he had no evidence about his own height.

[Gavin Chipper]

Hmm. 6/11?
I'm not too sure, and I can a potential flaw with what I've done, but it's the best I've got.

What's your reasoning for this?

Call john J, and anyone else X.
You can write it out as |X|X|X|X|X|J|X|X|X|X|
| is where the next person must go (since there can't be two people with the same height), 5 of them are taller than John and 6 smaller, so it's 6/11.

Yes.

[Gavin Chipper]

I don't get this problem. Surely if he is say 190cm, then there are infinitely many heights below this and infinitely above this so the probability is 1/2 or 1 - but these are both wrong. I don't really understand your question.

It's just based purely on the comparisons he's already made.

[Paul Howe]

I feel compelled to point out that 6/11 is merely a decent guess at the probability, and not the actual probability, which we couldn't possibly know from the information given.

I'm not particularly bothered, just wanted to pick holes in someone else's puzzle. And because it would effect the answer only very slightly, I'll be generous and not mention what's wrong with your bus puzzle

[Gavin Chipper]

I feel compelled to point out that 6/11 is merely a decent guess at the probability, and not the actual probability, which we couldn't possibly know from the information given.

I'm not particularly bothered, just wanted to pick holes in someone else's puzzle. And because it would effect the answer only very slightly, I'll be generous and not mention what's wrong with your bus puzzle

Surely it is just the probability that is based on the information given.

What is an "actual" probability anyway? Surely it's all about perspective. If I toss a coin and see that it lands on heads, the probability that it's landed on heads is 1. But you haven't seen what's happened, so the probability from your perspective is 1/2.

Basically, the point being that if an infinite number of random Johns performed the same task, picking the comparison people at random, 6/11 of them (of the ones that were 5-4 up) would find that they were taller than the next person. Or if it doesn't make sense to talk about infinite, the proportion would home in on 6/11 the more times you did it.

Now you have to tell me what you think is wrong with the bus problem!

[Paul Howe]

Basically, the point being that if an infinite number of random Johns performed the same task, picking the comparison people at random, 6/11 of them (of the ones that were 5-4 up) would find that they were taller than the next person. Or if it doesn't make sense to talk about infinite, the proportion would home in on 6/11 the more times you did it.

Really? Let's say that John is taller than exactly half the people in the population (he doesn't know this of course). If infinite Johns did your experiment, some of them would pick unrepresentative samples that, according to your logic, would lead them to conclude that the probability of them being taller than the next person was, I dunno, 8/11. Would 8/11ths of these Johns be taller than the next person picked? No, half of them would, and the proportion would home in on 1/2, not 8/11. This is all I meant.

Now you have to tell me what you think is wrong with the bus problem!

Nothing really, I was being slightly facetious. The logic is perfectly fine, it's just the fact that a finite number of people live at A means the answer should be very slightly greater than 9. Think about what would happen if 3 people lived at A, and each observed an average of 2 people (including themselves) per journey.

[Gavin Chipper]

Basically, the point being that if an infinite number of random Johns performed the same task, picking the comparison people at random, 6/11 of them (of the ones that were 5-4 up) would find that they were taller than the next person. Or if it doesn't make sense to talk about infinite, the proportion would home in on 6/11 the more times you did it.

Really? Let's say that John is taller than exactly half the people in the population (he doesn't know this of course). If infinite Johns did your experiment, some of them would pick unrepresentative samples that, according to your logic, would lead them to conclude that the probability of them being taller than the next person was, I dunno, 8/11. Would 8/11ths of these Johns be taller than the next person picked? No, half of them would, and the proportion would home in on 1/2, not 8/11. This is all I meant.

By infinite Johns, I meant infinite people picked at random rather than infinite people who are the same height as each other relative to the rest of the population.

Now you have to tell me what you think is wrong with the bus problem!

Nothing really, I was being slightly facetious. The logic is perfectly fine, it's just the fact that a finite number of people live at A means the answer should be very slightly greater than 9. Think about what would happen if 3 people lived at A, and each observed an average of 2 people (including themselves) per journey.

Yeah, that's why I specified it was a large number where someone going on the bus didn't significantly reduce the pool of people.

[Paul Howe]

By infinite Johns, I meant infinite people picked at random rather than infinite people who are the same height as each other relative to the rest of the population.

Oh right, sorry for the misunderstanding. I still don't think what you've written is completely right though, as for one thing it implies the probabilities will converge to a multiple of 1/11.

All I was originally getting it was that most people (or maybe its just me being obtuse) would interpret the probability you asked for as the number of people in the population shorter than John / the total population, and what you have is an estimate of that. Just a difference in semantics I guess.

Yeah, that's why I specified it was a large number where someone going on the bus didn't significantly reduce the pool of people.

Yeah I know mate. I only said it because certain individuals were picking some very minor nits in my other puzzle

[Gavin Chipper]

Oh right, sorry for the misunderstanding. I still don't think what you've written is completely right though, as for one thing it implies the probabilities will converge to a multiple of 1/11.

Shouldn't they then?

[David Roe]

I'll just say to anyone who thinks it's 5/9 . . . What would your answer be if all nine were taller than him, or when he'd only met one person?

David

I would say that his best estimate based on the available data would be that there is a 9/9 chance the next person is taller than him. With, of course, the caveat that there is a very small sample size and data is incomplete. Increasing or reducing the sample size doesn't change the method of calculation, it just changes the confidence interval.

The solution of 6/11 can only be right if the probability of someone falling into each gap is the same for all 11 gaps. Which of course it isn't. Surely what you need to do with this gap method is to work out the probability that the next person falls into each gap, presumably by using some statistical approximation to the actual distribution, and then add up the totals. Which obviously we can't do with the data available.

[Gavin Chipper]

I'll just say to anyone who thinks it's 5/9 . . . What would your answer be if all nine were taller than him, or when he'd only met one person?

David

I would say that his best estimate based on the available data would be that there is a 9/9 chance the next person is taller than him. With, of course, the caveat that there is a very small sample size and data is incomplete. Increasing or reducing the sample size doesn't change the method of calculation, it just changes the confidence interval.

The solution of 6/11 can only be right if the probability of someone falling into each gap is the same for all 11 gaps. Which of course it isn't. Surely what you need to do with this gap method is to work out the probability that the next person falls into each gap, presumably by using some statistical approximation to the actual distribution, and then add up the totals. Which obviously we can't do with the data available.

Even if the probability of falling in each of the gaps isn't equal, I don't see how that makes the probability jump from 6/11 to 9/9.

Anyway, the probability of the final person falling in any of the 11 gaps should be equal. We basically have 11 random people, and we take one random person of the 11 and ask, what is the probability of them being the tallest of the 11? Well, 1/11 obviously. And then we ask, what is the probability of them being the 2nd tallest? Obviously 2/11. The question as originally posed works!

[Jon Corby]

And then we ask, what is the probability of them being the 2nd tallest? Obviously 2/11.

1/11?

[David Roe]

I'll just say to anyone who thinks it's 5/9 . . . What would your answer be if all nine were taller than him, or when he'd only met one person?

David

I would say that his best estimate based on the available data would be that there is a 9/9 chance the next person is taller than him. With, of course, the caveat that there is a very small sample size and data is incomplete. Increasing or reducing the sample size doesn't change the method of calculation, it just changes the confidence interval.

The solution of 6/11 can only be right if the probability of someone falling into each gap is the same for all 11 gaps. Which of course it isn't. Surely what you need to do with this gap method is to work out the probability that the next person falls into each gap, presumably by using some statistical approximation to the actual distribution, and then add up the totals. Which obviously we can't do with the data available.

Even if the probability of falling in each of the gaps isn't equal, I don't see how that makes the probability jump from 6/11 to 9/9.

Anyway, the probability of the final person falling in any of the 11 gaps should be equal. We basically have 11 random people, and we take one random person of the 11 and ask, what is the probability of them being the tallest of the 11? Well, 1/11 obviously. And then we ask, what is the probability of them being the 2nd tallest? Obviously 2/11. The question as originally posed works!

The point of my quoting "What would your answer be if all nine were taller than him" was that my next line was in fact answering that question, ie. 9/9 on the available evidence. I wasn't answering the original question with 9/9.

The big problem in your solution is that you are basically deciding that the sample size isn't big enough to produce a sensible answer, so you're adding 1 above and 1 below to avoid these 9/9 situations. Which is intuitive and perhaps sensible in the real world, but is not mathematically sound. One feature of statistical distributions is that the more data you get, the closer you approach to the actual solution. With your approach, the tallest man in the world will still be expecting to meet someone taller than himself, even after he has met every other person.

The probability of the final person falling into each of the 11 gaps will not be equal. Imagine if your man has met Snow White and the seven dwarfs - the probability of falling into gap 2 will not be the same as that of falling into gap 3. You're assuming that the 9 people met so far have produced a perfectly representative sample of the population as a whole - that isn't an assumption you can make and still produce a logical answer.

[Gavin Chipper]

And then we ask, what is the probability of them being the 2nd tallest? Obviously 2/11.

1/11?

You know what I mean!

[Gavin Chipper]

The point of my quoting "What would your answer be if all nine were taller than him" was that my next line was in fact answering that question, ie. 9/9 on the available evidence. I wasn't answering the original question with 9/9.

OK then.

The big problem in your solution is that you are basically deciding that the sample size isn't big enough to produce a sensible answer, so you're adding 1 above and 1 below to avoid these 9/9 situations. Which is intuitive and perhaps sensible in the real world, but is not mathematically sound. One feature of statistical distributions is that the more data you get, the closer you approach to the actual solution. With your approach, the tallest man in the world will still be expecting to meet someone taller than himself, even after he has met every other person.

The probability of the final person falling into each of the 11 gaps will not be equal. Imagine if your man has met Snow White and the seven dwarfs - the probability of falling into gap 2 will not be the same as that of falling into gap 3. You're assuming that the 9 people met so far have produced a perfectly representative sample of the population as a whole - that isn't an assumption you can make and still produce a logical answer.

Obviously it complicates matters if we look at how much he is taller or shorter than the others, but we are basing it on only knowing that he is taller or shorter. Looking at it from our point of view, or John's assuming someone makes the comparison for him without him seeing any of them, it would still work.

The people are picked at random from the population, so while it's possible that he's picked Snow White and the seven dwarves, it's not something we need to worry about. It's just as likely that he's picked Snow White and the seven giants or whatever. We are basing it on not knowing anything other than a simple yes/no to whether John is taller.

You can test out this problem "empirically" if you want. Write a computer programme that picks a random number - the John number - and then picks nine others. If the John number isn't bigger than exactly five and smaller than four, everything is scrapped and the process starts again (or to speed it up you could just pick the number that is 5-4 up over the others as the John number if you don't see that as cheating). On the occasions where the John number is 5-4 up, another random number is picked and compared to the John number. A few thousand cycles later and I bet you will find it converging nicely on 6/11.

[Charlie Reams]

You can test out this problem "empirically" if you want. Write a computer programme that picks a random number - the John number - and then picks nine others. If the John number isn't bigger than exactly five and smaller than four, everything is scrapped and the process starts again (or to speed it up you could just pick the number that is 5-4 up over the others as the John number if you don't see that as cheating). On the occasions where the John number is 5-4 up, another random number is picked and compared to the John number. A few thousand cycles later and I bet you will find it converging nicely on 6/11.

You will if you make the same incorrect assumption about uniformity you've been making all along. David is correct.

[Gavin Chipper]

You will if you make the same incorrect assumption about uniformity you've been making all along. David is correct.

What incorrect assumption about uniformity? Try it then.

Edit - also if you look at it from this perspective - i.e. do it as this computer program, you can forget about all accusations of vagueness. There would be an answer that it would converge to, whether it be 6/11 or not.

[Charlie Reams]

If your program doesn't model the situation correctly then it will converge to something that has no relevance to the problem.

[Gavin Chipper]

If your program doesn't model the situation correctly then it will converge to something that has no relevance to the problem.

If the program is made as I described it, then it is exactly as I intended the problem to be. If there is any disagreement about whether the program as I described it models the problem, then it's purely down to a misunderstanding between us about what the problem is. Obviously if I write such a program and it comes out as I expected, then people may think I've either cheated or accidently programmed it wrongly in such a way that it comes out with my answer, so it's open to anyone to do. I will give it a go myself.

[Gavin Chipper]

I will give it a go myself.

Actually no, my graphical calculator's way too slow to get anywhere.

[Gavin Chipper]

But anyway, evidence from a computer program should count for less than mathematical reasoning (or proof). And I think that it is quite clear that this situation is identical to asking if eleven people are picked at random, what is the probability that a random person from this eleven (this being the person in the last comparison) is ranked anywhere from 6th to 11th in terms of height. John being currently 5th out of 10, so the last person needs to be 6th to 11th for John to be taller. And that probability is 6/11.

The only argument anyone could have with this is the way the problem was worded, but I think that the wording along with the subsequent clarifications made it quite clear. Anyone who still disagrees is simply wrong!

[David Roe]

You've posed two different questions here.

1. What is the probability of a random individual being in the bottom 6 of 11 individuals ranked in order of height?

2. What is the probability of a random individual being in the bottom 6 of 11 individuals ranked in order of height, given that 4 of the top 5 positions are already filled, and 5 of the bottom 6 are already filled, and John and A.N.Other will occupy the other two?

As it's 20 years since I did any stats, I've forgotten just how to formulate the difference between the two questions, but there is a difference.

I think that the wording along with the subsequent clarifications made it quite clear. Anyone who still disagrees is simply wrong!

And I'm sorry if this post doesn't match the rigorous intellectual argument of your quote above.

[Gavin Chipper]

You've posed two different questions here.

1. What is the probability of a random individual being in the bottom 6 of 11 individuals ranked in order of height?

2. What is the probability of a random individual being in the bottom 6 of 11 individuals ranked in order of height, given that 4 of the top 5 positions are already filled, and 5 of the bottom 6 are already filled, and John and A.N.Other will occupy the other two?

As it's 20 years since I did any stats, I've forgotten just how to formulate the difference between the two questions, but there is a difference.

The question as posed is pretty much: What is the probability of a random individual being in the bottom 6 of 11 given that John is currently in 5th place?

This is the same as: What is the probability of a random individual being in the bottom 6 of 11? By the way, the person currently in 5th is called John.

Which is the same as: What is the probability of a random individual being in the bottom 6 of 11?

OK, so maybe you think it makes a difference that we've followed John from the start rather than simply looking at 10 people (before the final comparison) and found John in 5th place. Should it make a difference? No, because the crucial issue is that nothing about the group of 10 as a whole is implied by the comparisons. We have no information telling us that the heights are likely to be above average, below average, skewed in some way or whatever. If we pick 10 people at random and bring in an 11th, then the 11th person as a 6/11 chance of being in the bottom 6. If we pick 10 people at random but then look at how the others compare to one of them in particular, that changes nothing regarding the 11th person.

And I'm sorry if this post doesn't match the rigorous intellectual argument of your quote above.

[David Roe]

Your arguments sound very plausible, but it still doesn't feel right to me. Maybe it's the artificiality of the question - the way John has made absolutely no observation about the actual heights of the people he was observing. Ah well, one of us is wrong, and it might be me (though I'll NEVER admit it!!!)

[David Williams]

I don't think anyone has really articulated the answer to this particularly well, and I'm not sure I can either. So it's probably easier to rubbish the incorrect answer! Ten people wake up in a hospital ward suffering from amnesia, but with all their logical faculties. The shortest looks around the room, and announces that there is a 9/9, or 100%, chance that the next person to enter the room will be taller than him. The tallest also concludes that there is a 9/9 chance that this person will be shorter than him. Collectively they realise that the tallest and shortest people in the world are in the room, and they are the only survivors of the human race.

The best I can do to explain the correct answer is this. Take a group of eleven people who haven't met. Pick one at random, invite the other ten to come together and line them up in order of height. Then invite the other person in. I don't think there's any doubt that there is a 1/11 chance of him being the tallest, second tallest etc. Now think of this from the perspective of the individual who found himself fifth (or first, or anywhere) in the original line-up of ten.

David

[David Roe]

There's definitely a flaw in this argument, David. And it relates to the highly specific conditions of the problem.

Your example referes to a finite group of 11 people, and it's certainly true that in that case the 11th man has a 1/11 chance of being the tallest, etc. HOWEVER, if the 11th man is suddenly spirited away and replaced by a random individual from the whole population, then the chances of the new number 11 being the tallest depend on the height of tallest of the 10. If the current tallest is 5'10, then the chances of the random 11th man being tallest are much higher than if the tallest of the 10 is 6'4".

All the solutions posited on here are based on the first puzzle, with a finite group. But the question was actually based on the infinite group, and it's a diffferent question with a different answer.

[Gavin Chipper]

You will if you make the same incorrect assumption about uniformity you've been making all along. David is correct.

STFU and take a course in probability.