## Puzzle – Liars and Truthtellers

December 31, 2009 – 1:27 am

This is another probability question that seems straightforward but can be deceptive…

You’re an FBI agent and have been monitoring a large gang, and you know that 1/4 of its members always tell the truth, the rest always lie.  Then you get taken hostage by the gang, and two gang members are assigned to guard you.  You ask the first: ”Am I going to live?”  He answers “Yes.”  Then you ask the second the same question.  He also answers “Yes.”  What are your odds of living?

It can be tempting to say it’s the probability the first is telling the truth multipled by the probability the second is telling the truth, which would mean you have a 1/16th chance of living.

It’s not correct, however, because there’s another piece of information you have which is that they both gave the same answer.

Let’s look at the probability tree:

Guard 1       Guard 2      probability

T                  T               1/16

F                   F               9/16

T                   F               3/16

F                   T               3/16

The only two rows that apply to this problem are the first two – they can either both be telling the truth, or both be lying.  The chances they’re both telling the truth is:

$\frac{\frac{1}{16}}{\frac{1}{16}+\frac{9}{16}}=\frac{1}{10}$

So – what would your chances of living be if the first guard told you that you would live, but the second that you would die?

Look at the table in the answer above – at the rows where the answers were different… you’ll see that your chances are 50/50.