3 people were playing a logic game together.
In the game, there were 3 black hats and 2 white hats.
The gamemaster randomly placed a coloured hat onto the head of each of the 3 people, such that none of them could see their own hat, but can see the other 2 hats on the other 2 people.
Let the 3 people be Mr A, Mr B and Mr C. A conversation ensued amongst them:
Mr A: "Mr B does not know his hat colour."
Mr B: "Mr C does not know his hat colour."
Mr C: "Mr A know his hat colour."
By looking at their conversation, can it be deduced that all 3 people have black hats?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
You cannot deduce that all of them are wearing black hats because B will not make his statement if they are.
Let's call the three statements SA, SB, and SC.
Without hearing anything, B can only know his own color if he sees two white hats. A will only make his claim if he is certain that B can see at most one white hat. For A to make his statement, B must be wearing a white hat and/or C must be wearing a black hat. So once A makes his statement, B and C gain knowledge based off of each other's hat colors. If B is wearing black, C now knows he must be wearing black too. If C is wearing white, B now knows he is wearing white as well.
For B to say SB he has to know his own hat color. We can see this because if he doesn't then he always has to consider the possibility that he is wearing a black hat and therefore C knows his own hat color.
In scenario All Black Hats (BBB), A will say SA. Once he does, C knows his hat color. B does not know his own, but he knows it is possible that his own hat tipped C off, so he can't say SB.
If you want to see what scenarios all three statements can be made for, take the original seven hat combinations for ABC: BBB, BBW, BWB, WBB, BWW, WBW, and WWB
SA eliminates any scenario where B is wearing a black hat and C is wearing a white one, leaving: BBB, BWB, WBB, BWW, WWB
Of those five, B can only know his hat color in one: BWW.
For SB to be true, A must be wearing Black, B must be wearing White, and C must be wearing White.
And yes, as there is only one combination that both SA and SB are true for, A will immediately know his own hat color and SC can also be stated truthfully.