A bit of fun solution - for these examples Z is any integer greater than or equal to 0

John Wood

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• Publié par: Zewyor
• Nombre de lectures: 43
• Langue: English

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A bit of fun

–

long answer, but is it correct?

This is the longest and most mathematical answer to the light-hearted article

‘

A bit of fun

’

which

appeared in our September 2009 newsletter.

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For these examples Z is any integer greater than or equal to 0.

1)

1 person with jam on their face, Z people with clean faces

The bell rings (1)

The person with jam on their face can see no one else (all Z people) with jam on their face. However they

know that someone must have jam on their face and work out that it must be them. They therefore wipe

their face.

All other people can see that one person has jam on their face and therefore don’t wipe their faces.

2)

2 people with jam on their faces, Z people with clean faces

The bell rings (1)

Both people with jam on their faces can see one other person with jam on their face (this justifies to them

the first ring of the bell). All others can see two people with jam on

their faces and therefore don’t wipe their

faces.

The bell rings (2)

Both people with jam on their faces know (given our previous example) that if there was only 1 person with

jam on their face that the person with jam on their face, that they can see, would have wiped their face.

Therefore they can deduce, that person must have been able to see someone else with jam on their face. No

one else has jam on their face, therefore that other person must be them. Therefore both people wipe their

face. The remaining Z people can see 2 people with jam on their faces therefore do not wipe their faces.

3)

This logic can be extended to the example of 3 people with jam on their face and Z people with clean

faces and as such it takes 3 rings of the bell for those with jam on their face to realise that they do and for

them and only them to wipe their faces.

Attempt at proof by induction (extension to all group sizes)

For this K is an integer greater than or equal to 1, Z is any integer greater than or equal to 0.

For K people with jam on their face and Z people with clean faces it always takes K rings of the bell before

those K people (and only those people) wipe their faces.

Already shown for the case of 1 person with jam on their face, Z people with clean faces

Assume it is true for K people with jam on their face, Z people with clean faces

If K people have jam on their face then it takes K rings for those people to realise and wipe their faces

Does it still hold for K+1 people with jam on their face and Z people with clean faces?

If K+1 people have jam on their face then for rings 1 to K they can assume that they don’t have jam on their

face given their knowledge about examples for 1 to K people. This is because they can see K people with jam

on their face.

However once the bell rings for the (K+1)th time they must deduce that they have jam on their face (as do

the remaining K people with jam on their face) and therefore they all wipe their faces. The remaining Z

people can see

K+1 people with jam on their faces and therefore don’t wipe their faces.

Therefore for K people with jam on their face and Z people with clean faces it always takes K rings of the bell

before those K people (and only those people) wipe their faces.

N.B. Assumptions

-

No cheating

-

Everyone can see everybody else

-

Everybody uses the same logic / acts rationally

For these examples Z is any integer greater than or equal to 0.

1)

1 person with jam on their face, Z people with clean faces

The bell rings (1)

The person with jam on their face can see no one else (all Z people) with jam on their face. However they

know that someone must have jam on their face and work out that it must be them. They therefore wipe

their face.

All other people can see that one person has jam on their face and therefore don’t wipe their faces.

2)

2 people with jam on their faces, Z people with clean faces

The bell rings (1)

Both people with jam on their faces can see one other person with jam on their face (this justifies to them

the first ring of the bell). All others can see two people with jam on their faces and therefore don’t wipe their

faces.

The bell rings (2)

Both people with jam on their faces know (given our previous example) that if there was only 1 person with

jam on their face that the person with jam on their face, that they can see, would have wiped their face.

Therefore they can deduce, that person must have been able to see someone else with jam on their face. No

one else has jam on their face, therefore that other person must be them. Therefore both people wipe their

face. The remaining Z people can see 2 people with jam on their faces therefore do not wipe their faces.

3)

This logic can be extended to the example of 3 people with jam on their face and Z people with clean

faces and as such it takes 3 rings of the bell for those with jam on their face to realise that they do and for

them and only them to wipe their faces.

Attempt at proof by induction (extension to all group sizes)

For this K is an integer greater than or equal to 1, Z is any integer greater than or equal to 0.

For K people with jam on their face and Z people with clean faces it always takes K rings of the bell before

those K people (and only those people) wipe their faces.

Already shown for the case of 1 person with jam on their face, Z people with clean faces

Assume it is true for K people with jam on their face, Z people with clean faces

If K people have jam on their face then it takes K rings for those people to realise and wipe their faces

Does it still hold for K+1 people with jam on their face and Z people with clean faces?

If K+1 people have jam on their face then for rings 1 to K they can assume that they don’t hav

e jam on their

face given their knowledge about examples for 1 to K people. This is because they can see K people with jam

on their face.

However once the bell rings for the (K+1)th time they must deduce that they have jam on their face (as do

the remaining K people with jam on their face) and therefore they all wipe their faces. The remaining Z

people can see K+1 people with jam on their faces and therefore don’t wipe their faces.

Therefore for K people with jam on their face and Z people with clean faces it always takes K rings of the bell

before those K people (and only those people) wipe their faces.

N.B. Assumptions

-

No cheating

-

Everyone can see everybody else

-

Everybody uses the same logic / acts rationally