The ladies mentioned in Five beautiful ladies have five brothers: Adam, Bob, Charlie, David, and Edward. Three have hazel eyes and are honest; two have lime-colored eyes and always lie.

You are tasked with finding out who among the brothers is honest by submitting one multiple-choice question in writing to each of Adam, Bob, and Charlie, and examining the responses; all questions must be submitted before any responses are received.

Multiple-choice questions should be written so as to have one correct answer; assume nothing good will come of asking ambiguous questions(*). Honest people will identify and report the correct choice; liars will honestly identify the correct choice, but falsely report some other choice, selected in whatever fashion will be most vexatious. If a liar is given a 4-way A/B/C/D choice where the correct answer was B, he may arbitrarily answer A, C, or D; he will not answer B or Q). By inference, if someone answers B, that will imply that either he was truthful and the correct answer was B, or he was lying and the correct answer is A, C, or D.

(*) For simplicity, it should suffice to say that a liar is only forbidden from responding with choices which are not offered, or any choice which is clearly the unique correct answer. There is no requirement that a liar pick the "most dishonest" choice--merely that there exist some honestly-correct choice other than the one reported by the liar.

Note that as with the Ladies' puzzle, there are ten possible combinations of eye colors to resolve using three questions, which implies the need for questions with more than two possible answers. Unlike the Ladies' puzzle, however, all questions must be submitted in advance. The puzzle would be unsolvable with three liars, but is solvable with two.

  • $\begingroup$ Do we know who each answer came from? $\endgroup$
    – tfitzger
    Dec 2, 2015 at 18:34
  • $\begingroup$ @tfitzger: Yes. If nothing else, one could ask Adam a question with choices denoted A1-A5, ask Bob a question with choices B1-B5, etc. $\endgroup$
    – supercat
    Dec 2, 2015 at 18:38
  • $\begingroup$ If the choices were a) 1=1 and 2=3, or b) 3=4 and 4=5, would an honest person always pick a because it is "more truthful" (half of it is true)? Or do they look at the statement as a whole, determine that both a and b are false, and pick one at random? $\endgroup$ Dec 2, 2015 at 18:55
  • $\begingroup$ @GentlePurpleRain: If you want information from a question, ask it unambiguously. If you ask Adam "Identify one of the honest brothers A-E" and the honest brothers are Bob and Charlie, Adam could decide that a "correct" answer was Bob and thus answer C, or decide that the "correct" answer was Charlie and thus answer B; he could also answer A, D, or E, since any of those would clearly be wrong as well. By contrast, if asked "Identify the alphabetically-first honest brother" in that case, Adam would not have been able to say B. $\endgroup$
    – supercat
    Dec 2, 2015 at 19:24
  • $\begingroup$ @GentlePurpleRain: Another way of expressing the rules is to say that a liar is not required to pick the "least honest" answer, but is free to pick any answer that is not unambiguously the most honest. $\endgroup$
    – supercat
    Dec 2, 2015 at 20:21

1 Answer 1


We will ask each of A, B and C a three choice question, whose options are numbered (1), (2), (3). The questions are given in the below table:

$\begin{array}{l|l|l}\text{To Adam...}&\text{To Bob...}&\text{To Charlie...}\\\hline(1) \text{ The liars are B and E} & (1) \text{ The liars are C and E} & (1) \text{ The liars are A and E} \\(2) \text{ The liars are C and E} & (2) \text{ The liars are A and E} & (2) \text{ The liars are B and E} \\(3) \text{ None of the above} & (3) \text{ None of the above} & (3) \text{ None of the above} \\\end{array}$

Notice that if a person is a liar then the correct answer is (3), so a liar will never answer (3).

We then deduce the liars as shown below. The symbol "_" means any answer which isn't (3). $$ \begin{array}{c|c|c} \_\,33\implies AD & \_\,\_3 \implies AB & \_21\implies AE\\ 3\_\,3\implies BD & \_\,3\_ \implies AC & 1\_2\implies BE\\ 33\_\,\implies CD & 3\_\,\_ \implies BC & 21\_\implies CE\\ 333\implies DE \end{array} $$ For example, the entry "$3\_\,\_\implies BC$" is shorthand for "If Adam responds (3), and neither of Bob and Charlie respond (3), then the liars are Bob and Charlie."

Here's why this works:

  • If everyone answered (3)...

    Then none of A,B,C are liars, so the liars are DE.

  • If exactly two people answered (3)...

    For definiteness, suppose A and B answered (3), and Charlie didn't. Then both A and B must be honest, so the liars are either CD, CE or DE. CE is impossible, because in that case A would have said (2). Also, DE is impossible, because in that case C would answer (3). Thus, the liars are CD, as claimed. The other two cases work similarly.

  • If exactly one person said (3)...

    Suppose it was Adam who said (3), so that Adam is honest. D must be honest as well, otherwise at least one of B or C would answer (3). Thus, the only possibilities are BC, BE and CE. However, BE and CE were ruled out by Adam saying (3), so the true answer is indeed BC.

  • If no one said (3)...

    Since at least one of A,B or C is honest, the only three possibilities are AE, BE and CE. Thus, the majority of A, B and C are truthful, so we can simply go with the majority answer. For example, in the 212 case, both Adam and Bob are claiming the liars are CE, so the liars are indeed CE.

  • $\begingroup$ Very nice. I used a very similar strategy, though with slightly different questions. Minor nit: who's Alice? Did you like the puzzle? I can't think of any "liar" puzzles I've seen where one can ask multiple-choice questions but liars can arbitrarily select all but one of the possible answers. $\endgroup$
    – supercat
    Dec 3, 2015 at 0:29
  • $\begingroup$ @supercat I liked this a lot! Adding multiple choice questions is a nice twist on the classic theme. Also, I was very confused for a while, but you can look at this problem in a geometric way which makes it easier to come up with a solution (namely packing various sizes bricks, representing sets of possible answer profiles, into a 3x3x3 cube representing all possible answer sets, being sure the bricks are disjoint so you can unambiguously deduce the solution). $\endgroup$ Dec 3, 2015 at 2:11
  • $\begingroup$ I had the same "bricks" visualization. The problem would be unsolveable with 3/5 liars because the "first three are liars" brick would be (N-1)^3, but "two are liars" would require six bricks of (N-1)^2. Since N-1^3 = N^3-3N^2-3N+1, no value of N will allow for six bricks of the required size. It would be nice if the puzzle could be extended so the player asks a question of all five brothers but an arbitrary three respond, but I can't think of a solution for that. $\endgroup$
    – supercat
    Dec 3, 2015 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.