Crowns of the Minotaur 

Image by Matt Collins puzzle

We can approach the solution to this problem in two stages - the obvious and the thinking man's version. Whether a prisoner would be able to fathom out the final solution within ten seconds is perhaps questionable.

Assumption: Since the King chooses which color crown to place on the heads of the prisoners by tossing a coin, there is a equal probability that any one of eight configurations can exist:

Prisoner 1Prisoner 2Prisoner 3
blue
blue
blue
blue
blue
red
blue
red
blue
blue
red
red
red
blue
blue
red
blue
red
red
red
blue
red
red
red

Solution 1:

Without looking at the other prisoners, a prisoner can arbitrarily choose a color and have a 50% chance of being right. But of course that does not help, since the King's rules say that those who do not pass must be right. Thus the possibility of getting the right combination is 1 in 8. Choosing between red, blue or pass evenly, improves the odds to about 1 in 3, because the pass counts counts as true in most cases.

Solution 2:

But since each prisoner can see what is on the heads of the other two prisoners, and in six out of the eight possible cases, two prisoners will wear crowns of the same color. Thus we can develop a strategy that is correct in 75% of the time:

If the crowns that you can see are both the same color,
then choose the other color as your answer
else (the crowns are different colors) say "pass".

This produces the following table:

Prisoner 1Prisoner 2Prisoner 3Response 1Response 2Response 3Result
blue
blue
blue
red
red
red
They die
blue
blue
red
pass
pass
red
They live
blue
red
blue
pass
red
pass
They live
blue
red
red
blue
pass
pass
They live
red
blue
blue
red
pass
pass
They live
red
blue
red
pass
blue
pass
They live
red
red
blue
pass
pass
blue
They live
red
red
red
blue
blue
blue
They die

An improvement?:

If all prisoners are required to register their answers simultaneously, then the last solution above is as good as can be expected. However, if the prisoners are allowed to respond in their own time, then the third one to answer can use the data from the other responses to improve his odds to 100% -- that is provided that the others follow the above algorithm! The truth table for this case is as follows:

Prisoner 1Prisoner 2Prisoner 3Response 1Response 2Response 3ReasoningResult
blue
blue
blue
red
red
blue
The others are seeing all blues, so you must be the same.They die
blue
blue
red
pass
pass
red
Others are both seeing one of each color, so yours must be the one other color to theirs.They live
blue
red
blue
pass
red
blue
Prisoner 1 is blue and is seeing one of each color; Prisoner 2 saw two blues, so you must be the other blue.They live
blue
red
red
blue
pass
red
Prisoner 2 is seeing one of each color; Prisoner 1 is blue and is seeing two reds, so you must be the other red.They live
red
blue
blue
red
pass
blue
Prisoner 2 is seeing one of each color; Prisoner 1 is red and is seeing two blues so you must be blue.They live
red
blue
red
pass
blue
red
Prisoner 1 is red and is seeing one of each color; Prisoner 2 is blue and is seeing two reds so you must be red.They live
red
red
blue
pass
pass
blue
The others are seeing one of each color, and you see both with the same color, so you must be the odd man out.They live
red
red
red
blue
blue
red
The others are seeing all reds, so you must be the same.They die

Interestingly enough, this additional information does not improve the final overall odds. Now could you work that out in 10 seconds?


There are solutions for more than three prisoners!


Last updated 2001/12/04
© J.A.N. Lee, 2001.