This is probably my favorite puzzle. It uses logic, but it's not one of those "logic table" puzzles which i dislike immensely. What is this type of puzzle called and are there more like it that are not just variants?

Three children are playing on the beach and all three have mud on their foreheads. An old man walks by and asks them, “After looking at both of your friends foreheads, raise your hand if one or both of them have mud on their foreheads.“ All three children look at both their friends and raise their hands.

He then asks, “For a dollar, without touching your own forehead or looking in a mirror or anything similar, tell me if you have mud on your own forehead, and how you know that.” All three children are bewildered. Finally, one raises his hand and says, “i have mud on my forehead.” He explains how he knew this and receives the dollar.

What was his reasoning?

Note: The old man and the children are honest, able to employ logic, and want the dollar.

  • 1
    $\begingroup$ The indecision offers knowledge on of itself. The information provided to the children isn't enough to figure it out on their own. It requires the additional knowledge that they can't figure it out on their own to provide them with the information they require to figure it out. $\endgroup$
    – Neil
    Jul 15 '14 at 10:03
  • $\begingroup$ Nowadays (six years after the question was asked) we have a tag (meta-knowledge) that describes the defining property of this puzzle pretty well. $\endgroup$
    – Bass
    Oct 9 '20 at 19:27

As with klm123, I don't know of a specific name for this type of problem (oracle riddle?). However, I've usually seen this question in the context of an island with people of two eye colors and an oracle.

In this case, the old man is the oracle, whether or not the children have mud on their face corresponds to whether the person has brown or blue eyes, and raising the hand is equivalent to leaving the island.

This is a vaguer retelling in a more natural setting, however. Only one child raises his hand, even though all of them should be able to reach the conclusion at the same time. Likewise, the oracle problem has a defined amount of time that has to pass, rather than waiting for some arbitrary "Oh, he didn't react fast enough, which gives me information" point.

Gilles' explanation of the oracle problem is a great answer, and I'll quote a bit of it here:

Suppose that only Alice has blue eyes. Before day 0 [when the oracle speaks], she never knew anyone with blue eyes. On day 0, she learns that someone has blue eyes; since nobody else does, it has to be her and only her, so she takes the ferry that night.

Now suppose that only Alice and Bill have blue eyes. Before day 0, Bill already knew that there was someone with blue eyes, but he did not know that Alice knew. If Bill had had green eyes, Alice would have been the only blue-eyed person and would not have known. On the first night after the guru, Alice doesn't leave; this tells Bill that Alice did not know the color of her eyes, so learn that she was the only blue-eyed person. Since Bill knows that either Alice is the only blue-eyed person or Bill and Alice are the only two, Bill now knows that both he and Alice have blue eyes.


For informational purposes, I'll give the solution to the quoted problem.

There are two possible scenarios that would cause all three children to raise their hands at the first question:

  • All three children have mud on their foreheads.

  • Two children have mud on their foreheads and one doesn't.

Each of the children know this. Now, without loss of generality, let's pick a specific child; say, we'll name him Andy. Andy sees that the other two kids have mud on their foreheads. He can't deduce immediately whether he has mud or not, since both those two scenarios are part of the initial possibilities.

But, Andy does see that neither of the other two kids are raising their hands and saying they have mud on their foreheads. If Andy's forehead was clean, each of the other two kids would only see one muddy forehead, and would therefore immediately know that their forehead was muddy. So he deduces that his forehead must also be muddy from this information, as well as the rather protracted pause after the old man asks the second question.

Eventually all three kids could deduce this conclusion, but Andy simply got to it first.

  • 5
    $\begingroup$ +1. Or, Andy figures he's got a 50% shot at getting a free dollar and guesses. $\endgroup$
    – Neil
    Jul 15 '14 at 10:07
  • $\begingroup$ "each of the other two kids would only see one muddy forehead, and would therefore immediately know that their forehead was muddy." Can you clarify that? I think a step is missing. $\endgroup$ Jul 15 '14 at 11:18
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    $\begingroup$ Since each kid knows that at least two foreheads are muddy, if they only see one other muddy forehead, they can deduce that theirs must also be muddy. $\endgroup$
    – user88
    Jul 15 '14 at 11:39
  • 1
    $\begingroup$ @Neil: That would fail the "tell me how you know" test. $\endgroup$
    – user88
    Jul 15 '14 at 18:03
  • $\begingroup$ @JoeZ. Then we also have to weigh the fact that a child who understands why he has mud on his forehead in the traditional way may not be able to properly explain the why. ;) $\endgroup$
    – Neil
    Jul 16 '14 at 7:32

There is no special name for such type of puzzles as far as i know.
But usually they are formulated in terms of hats worn by logicians.

If you google you can find a lot of puzzles of this type, for example, this:

The Logical Hats Puzzle. Logicians A, B and C each wear a hat with a positive integer on it such that the number on one hat is the sum of the numbers on the other two. They can see the numbers on the other two hats but not their own. They are given this information and asked in turn if they can identify their number. In the first round A, B and C each in turn say they don't know. In the second round A is first to go and states his number is 50. What numbers are on B and C?

Or this:

The jailer puts three of the men sitting in a line. The fourth man is put behind a screen (or in a separate room). He gives all four men party hats (as in diagram). The jailer explains that there are two red and two blue hats; that each prisoner is wearing one of the hats; and that each of the prisoners is only to see the hats in front of them but not on themselves or behind. The fourth man behind the screen can't see or be seen by any other prisoner. No communication between the prisoners is allowed.

If any prisoner can figure out and say to the jailer what color hat he has on his head all four prisoners go free. If any prisoner suggests an incorrect answer, all four prisoners are executed. The puzzle is to find how the prisoners can escape, regardless of how the jailer distributes the hats.

Also there was a lot of them on this site:

Guessing hat colors
Hats and Aliens
N logicians wearing hats of N colors In the 100 blue eyes problem - why is the oracle necessary? (suggested by Bobson)

  • $\begingroup$ It's really the same puzzle as the hats? I thought the hats puzzle were removing impossible combinations, whereas the mud is a form of reverse-logic. Regardless, i never thought of the connection between the two. $\endgroup$ Jul 13 '14 at 21:42
  • $\begingroup$ @BrianTkatch, Which hats puzzle exactly are you talking about? For example, first puzzle by the link uses reverse logic to solve it. I mean the one, which tells "They are given this information and asked in turn if they can identify their number". $\endgroup$
    – klm123
    Jul 13 '14 at 22:08
  • $\begingroup$ @kim123 truthfully, i do not know. But, for some reason, i want to say they are different. I'm going to sit on this for a day or two to see if i can understand whatever it is that is bothering me. $\endgroup$ Jul 13 '14 at 22:15
  • $\begingroup$ I see, the logic is indeed the same. Thanx for pointing that out! If you will, please bear with me. What leads to the logic is different. Hats has an order and requires other answers 1st. Mud has no order. On 2nd thought, Mud has an implicit order, as the one who answered translates the others' non-answering as each saying "i don't know". So, there actually is an order. Hmm... In Mud the order is: 1st round all 3 answer knowing, 2nd round 1st 2 do not know, then #3 knows. In Hats: 1st round everyone says they do not know, 2nd round they do know. That's the exact opposite, isn't it? $\endgroup$ Jul 14 '14 at 13:32
  • $\begingroup$ @BrianTkatch, sorry, what you are saying is too complicated for me. Ether my english knowledge is not enough or you skip a lot of words\phrases here, or your formulation is simply not made enough (is not clear enough)... In "Guessing Colours Hats" problem hats have 2 colours and no order, similarly to Mud. May be you can reformulate the Question more precisely now? I think my Answer fully answers what you asked in the Question in its current formulation. $\endgroup$
    – klm123
    Jul 14 '14 at 19:24

Here a completely logical solution with no shortcuts involved :)

There are 4 scenarios

  1. None have mud.
  2. One has mud.
  3. 2 have mud.
  4. All of them have mud.

Case 1: If no one of them have mud on their foreheads, then no one would raise their hands. However everyone raised their hands so this is not the case. (only 3 cases left)

Case 2: If one person had mud on his head, he would see that both of his friends raised their hands while they didn't have mud on their foreheads. That must mean he has mud and he would immediately get the dollar. However this is not the case because all three raised their hands. (only 2 cases left)

Case 3: Imagine people A, B, and C. People A and B have mud on their foreheads. Person A would see that person B raised his hand while person C didn't have mud on their forehead. That means person A would know he is the only one person B could be raising his hands for, and he would immediately know he had mud on his head and get the dollar. The same logic goes for person B.

However, "All three children are bewildered." so this cannot be the case.

That leaves us with case four being the only possible answer through process of elimination!!


I've seen this puzzle with men and coloured hats. The way it was framed was in stages, which led the reader (by the nose) to the solution of the most complex case.

The puzzle was simple - there are three men, all wearing hats, and a diabolical king. The hats are either white or black. The men can see each other's hat, but not their own. The king tells them that if they can guess the colour of their hat they need to simply raise their hand and (if they are right) they will live. The king tells them that there is at least one white hat, and the men are not allowed to communicate in any way.

For each scenario, your task is to determine who can figure out what they are wearing.

  1. One man is wearing a white hat.
    Solution: The two men in black hats see one white and one black, but cannot determine their own hat colour, so they do not immediately raise their hands. The man in the white hat sees two black hats, and since he knows at least one hat is white, he raises his hand immediately. When the other two see this, they know he has seen two black hats, so then they know the colour of their own hats and are able to raise their hands as well.
  2. Two men are wearing white hats.
    Solution: The two men wearing white see one white and one black, and the man wearing black sees two white. If only one man was wearing white, he would have raised his hand immediately as in the first scenario. The men know this, and since no one raised their hand immediately, they determine that there is more than one white hat. The two men who see one white and one black correctly deduce that they must be wearing white and raise their hand. The last man in black sees that they hesitated before raising their hands, so is also able to rule out the first scenario. Since they were able to figure out their hat colour relatively soon after, he then correctly guesses that his hat must be black.
  3. All three are wearing white hats.
    Solution: All three men see two white hats. If it had been the previous scenario, then two of them would have raised their had after some initial hesitation. Since this doesn't happen, they can all infer that they must be wearing white. Thus after a second hesitation, they all raise their hand.

In a twist, the question is sometimes asked as follows (same premise as above).

After a lengthy pause, all three men finally raise their hands. What colour were their hats?

Using the above logic, we can infer that they were all wearing white.


I would suspect the child knew he had mud on his forehead because he could feel it there.

They don't know enough to to be able to employ other logic here.

In addition to:

The old man and the children are honest, able to employ logic, and want the dollar.

Each child must know that the old man, and the other children are are honest, able to employ logic, and want the dollar. The fact that the children are all honest does not imply that that know their fellow playmates are honest or that the old man is. This knowledge is required for them to be able to make this logical leap.

Also, what tends to hang me up about these puzzles (and I haven't yet seen mentioned) is that each child would have to know how fast the other children would react, if that child was the only one with mud on their forehead. The fact that only one wins (and there isn't a three-way draw) implies that they are not simply purely logical clones of each other.

I apologies if I am splitting hairs here.

  • $\begingroup$ I think the puzzle is quite clear. If you retell the puzzle, please change the details to suit your fancy. :) $\endgroup$ Jul 20 '14 at 11:09
  • 1
    $\begingroup$ A more common version of this problem is each looks at each other (no old man) and starts laughing. When none stops, one stops because he realizes that he must have mud too. This would likely work better with drunk frat guys and a sharpie. $\endgroup$
    – kaine
    Jul 21 '14 at 13:18

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