Puzzle Of The Muddy Children

[michaellevenson]

A father tells a group of children that at least one of them has a muddy face. " Step forward if you have a muddy face" he tells them, and repeats this request until those with muddy faces steps forward. If there are 'x' muddy children then 'x' will step forward at the ' xth' request, ie if 2 then 2 will step forward at 2nd request, or 4 at 4th request.
But if the father doesn't tell them at the outset that at least one has a muddy face, this will not happen, even if there is more than one muddy face, so that everyone can see there is at least one ,and knows what the father doesn't tell them. Why?

 

[michaellevenson]

Spoiler: muddy children answer

First scenario- Father tells them at least one has a muddy face. They can all see the other faces but not their own. We assume the children are all logically minded, honest and assume the others are too.
Suppose just one has a muddy face, that child will know her face is muddy since she will see that no one else is, so she will step forward. The others will see her face is muddy and not knowing about their own face will stand still.
Suppose two have muddy faces, at the first request to step forward no one will know if their face is muddy. Some will see two muddy faces, two of them will see one, but they all know at least one has a muddy face, so they themselves may or not have one. When no one has stepped forward after the first request they all know there is more than one muddy face. The ones that see one muddy face will conclude their faces must be muddy too. The others will see two muddy faces, not knowing about their own will stand still. If there are three muddy faces then no one will have stepped forward after two requests because everyone will see at least two muddy faces. So they all know more than two are muddy, so therefore the three that only see two muddy faces will know their face is muddy and step forward.

Scenario 2 - Suppose now there is more than one muddy face, so they all know there is at least one. The father though doesn't tell them at least one has a muddy face. Then surprisingly no child will ever step forward even though they know exactly the same as the father told them in scenario 1.
The reason is that in scenario 2 there is no common knowledge. For example, if there is two muddy children, Angela and Mary, neither knows their own face is muddy. At the second request Angela cannot reason that Mary didn't step forward at the first because she saw Angela 's face was muddy, so she cannot reason that there are just two muddy faces when she sees only one other which is muddy. There is UNIVERSAL knowledge that there is one muddy face but there isn't COMMON knowledge. For common knowledge everyone must know that everyone knows that everyone knows. It was achieved in the first scenario because the father's statement that there was at least one muddy face was public, they all heard it and knew everyone heard it.
To explain the difference in UNIVERSAL and COMMON knowledge: traffic lights work only because there is common knowledge that people will stop at red. It is not enough for me to know that red means stop, if I am not confident that other people know this and that THEY know others know etc. Universal knowledge is not sufficient, everyone may know how traffic lights work but they need to know others know. In scenario 1 the children do not need infinite common knowledge. If there is exactly 'x' muddy children then the common knowledge required is that everyone knows that everyone knows that there are at least one muddy child, with " everyone knows " iterated 'x' times.