Hairs on head

gondorff · 23-01-2008 5:53am #1

How is it that I can say, with complete certainty, that there are in Ireland at least two people with exactly the same number of hairs on their heads?

Jrad · 23-01-2008 6:24am

Cause they are bald?

hivizman · 29-01-2008 1:40am

This puzzle is a well-known illustration of what mathematicians call the 'pigeonhole principle'. If you are sorting n items into m categories, and m is less than n, then at least one of the categories must have more than one item in it. For example, if you have five people in a family and only four bedrooms, at least two people must share a bedroom (I'm assuming of course that someone isn't sleeping in the kitchen!). The population of Ireland (the n items) is about 4 million, and the number of hairs on a person's head (the m categories) is much less than this (most people have between 100,000 and 150,000 hairs), so even if you can find 150,000 people each with a different number of hairs, each person in the rest of the population must have the same number of hairs as at least one of these 150,000 people.

That's the logic, but I'd be hesitant about claiming this with complete certainty, because the argument depends on the truth of the assertion that there are about 4 million people in Ireland and that the maximum number of hairs is much less than this. It may be that people living in Ireland are exceptionally hairy, so having several million hairs is no exceptional event. The pigeonhole principle only works if the number of hairs is less than the number of people, and this is a contingent fact about the world in which we live rather than a certainty.

The pigeonhole principle only works if the number of categories is finite. The mathematician David Hilbert used the example of 'Hilbert's Hotel'. This hotel has an infinite number of rooms, numbered 1, 2, 3, 4, etc. The hotel is full. So there is at least one person in each room. A new guest arrives. Even though the hotel is full, the manager is able to find a room for the new guest without the new guest having to share. How?

mathias · 30-01-2008 3:27am

With the hilbert thing , everyone moves next door freeing up room number one,

I still prefer the bald answer to the first question , its pretty damn definite that baldness is not a condition unique to one person , and 0 hairs is = to 0 hairs !

Most likely not the answer the OP was looking for , but clever nonetheless.

gondorff · 31-01-2008 7:23pm

An infinite amount of hotel rooms would have no room number one. It would be a continuum no? Therefore even if everyone moves to the room next door, all the rooms would still be occupied. There would be no room to squeeze anyone in.
Flawed logic?

I don't know the exact population of Ireland, but I'm sure a consensus would show that it's in the millions.
I'm prepared to wager that there isn't, nor has there ever been anyone in the world who has more than one million hairs on their heads.

JoeB- · 01-02-2008 4:26pm

Bald people have eyebrows, eye lashes and nostril hair... eh? So very few people would have 0 hairs on their head...

I like the pigeon hole thing better....

In the hotel if everyone goes to the room which has a room number which is double the existing room number.. then we have loads of spare rooms... (see Wiki)

mathias · 01-02-2008 4:52pm

An infinite amount of hotel rooms would have no room number one.......

It wouldnt necessarily have a room number 1 at all depending on how they were numbered ,

The Hilbert thing is an attempt to explain the concept of infinity , remember with numbers no matter how big , you can always add 1 , so that explains the room bit ,

But you could just as easily have all rooms numbered even , an infinity , or all rooms numbered odd , an infinity again , or all rooms a multiple of 3 , and so on , there is an infinity of infinities on the number line so to speak.

You can go big , and you can go small , between any two numbers or any two fractions , you can divide the division and infinite amount of times , there is no limit on how big the denominator can be. So no matter where you look on the number line , or how big a section of it you look at , there are an infinity of infinities there.

hivizman · 02-02-2008 2:14am

There are different types of infinity. In Hilbert's hotel, and in the various modifications (such as only even-numbered rooms), you have an infinite set which is either the set of natural numbers {1, 2, 3, . . .} or can be put into one-to-one correspondence with the set of natural numbers (so in the set of even numbers {2, 4, 6, . . .}, each element n can be mapped onto the natural number n/2 - alternatively, each natural number n can be mapped onto the even number 2n). This is sometimes referred to as a 'countable infinity'.

However, the 'continuum' - the set of all real numbers - can't be put into one-to-one correspondence with the set of natural numbers, so this is a sort of 'infinity of infinities', or 'uncountable infinity'.

Back to hair. Although there must be at least two people in Ireland with the same number of hairs, it's logically possible that there are towns in Ireland where everyone has different numbers of hairs (and by the way, even bald people are likely to have a few hair follicles). For example, Limerick has a population of about 90,000, so the pigeonhole principle won't rule out the possibility that everyone in Limerick has a different number of hairs (assuming that the maximum number of hairs on a person's head is 100,000 or more).

This puzzle reminds me of the 'birthday' paradox. How large a group of people do you need to have a better than 50% chance that at least two of the members of the group share a birthday? I won't answer that, but if you have a group of around 500 people, then there is over 50% chance that at least two of them have the same number of hairs.

Xhristy · 03-02-2008 12:55pm

This post has been deleted.

hivizman · 03-02-2008 8:03pm

Xhristy wrote: »

This post has been deleted.

That's because there are only 366 possible birthdays (allowing for 29 February, as we should because 2008 is a leap year). A group of up to 366 people could conceivably all have different birthdays, but in a group of 367 or more people, there must be at least two people with the same birthday.

However, as the maximum number of hairs on one's head is somewhere around 150,000, you would need a group with more than this number to guarantee that at least two people have the same number of hairs.

The surprising thing, though, is that you would need only around 500 people in your group to have a better than 50% chance of having at least two people in the group with the same number of hairs.

cookiequeen · 06-02-2008 11:49am

identical twins have the same genetic makeup and therefore the same number of hair folicles and thus the same number of hairs????

Gone West · 15-02-2008 3:10pm

cookiequeen wrote: »

identical twins have the same genetic makeup and therefore the same number of hair folicles and thus the same number of hairs????

Nope.

rgiller · 03-03-2008 6:15pm

Think the birthdays answer is 23 people:

Think about the odds that all people have different birthdays

- first guy 365/365
- second guy 364/365
- third guy 363/365

Multiplying these odds together gives you just over 49% chance of all different birthdays when there are 23 people, and so just over 50% of at least two having same birthday

yahoo_moe · 03-03-2008 7:00pm

if you believe amarillo slim's book (which, in fairness, is a biggish if), the birthday thing has won him many a bet in a room of c. 25-30 people - apparently most people don't realise the number of 2-person combinations that kind of group offers...

Mellor · 26-03-2008 1:56pm

rgiller wrote: »

Think the birthdays answer is 23 people:

Think about the odds that all people have different birthdays

- first guy 365/365
- second guy 364/365
- third guy 363/365

Multiplying these odds together gives you just over 49% chance of all different birthdays when there are 23 people, and so just over 50% of at least two having same birthday

Numbers slightly wrong, and making the maths overly complicated
Its 28 people.

The number of two person pairs from 28 people is 378 (28*27/2), this is the first above 366.
27 people gives 351.
You need 367 people to be certain, but at 28 people it becomes more likely than not.

As for the hairs, the number of people is the country is far higher than even 10 times the average. Two people the same for certain.

As for baldness no being correct due to eyebrows, there are plenty of people in ireland with alopecia. Zero hairs there.

Sir Gallagher · 01-04-2008 7:14pm

hivizman wrote: »

This puzzle is a well-known illustration of what mathematicians call the 'pigeonhole principle'. If you are sorting n items into m categories, and m is less than n, then at least one of the categories must have more than one item in it. For example, if you have five people in a family and only four bedrooms, at least two people must share a bedroom (I'm assuming of course that someone isn't sleeping in the kitchen!). The population of Ireland (the n items) is about 4 million, and the number of hairs on a person's head (the m categories) is much less than this (most people have between 100,000 and 150,000 hairs), so even if you can find 150,000 people each with a different number of hairs, each person in the rest of the population must have the same number of hairs as at least one of these 150,000 people.

That's the logic, but I'd be hesitant about claiming this with complete certainty, because the argument depends on the truth of the assertion that there are about 4 million people in Ireland and that the maximum number of hairs is much less than this. It may be that people living in Ireland are exceptionally hairy, so having several million hairs is no exceptional event. The pigeonhole principle only works if the number of hairs is less than the number of people, and this is a contingent fact about the world in which we live rather than a certainty.

The pigeonhole principle only works if the number of categories is finite. The mathematician David Hilbert used the example of 'Hilbert's Hotel'. This hotel has an infinite number of rooms, numbered 1, 2, 3, 4, etc. The hotel is full. So there is at least one person in each room. A new guest arrives. Even though the hotel is full, the manager is able to find a room for the new guest without the new guest having to share. How?

Just because the new guest doesn't have to share doesn't mean the others can't, therefore move an existing guest into another room freeing up a room for the new guest who'll have a lovely room all to him/herself??

Hairs on head

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