The Two Lying Brothers

There were once two brothers who always lied. They were fairly clever, though, and they soon realized that by always lying, people could destill truth from their statements fairly easily anyway. So, they invented a new scheme; they would each always carry a single die, and when called upon to make a statement, they would first roll the die; if it came up on a 2 or a 5 they would tell the truth; otherwise they would lie.

The first brother, after rolling his die, makes the statement "I am older than my brother". The second brother, after rolling his die, makes the statement "He’s telling the truth". What is the probability that the first brother actually is older?

Solution:

The probability that a brother tells the truth in any given statement is 1/3; the probability he tells a lie is 2/3.

Consider the general case where the brothers make any two arbitrary statements. These can be either both true, both false, or the first true and the second false, or the first false and the second true. There are 4 possible outcomes in general, and we can make a table of the probability for each of these 4 outcomes.

 TF
T(1/3)(1/3)(1/3)(1/3)
F(2/3)(1/3)(2/3)(2/3)
=
 TF
T(1/9)(1/9)
F(2/9)(4/9)

So for two arbitrary statements, the probability they are both true is 1/9, and the probability they are both false is 4/9.

But in our case the two statements made by the brothers are not arbitrary. The second statement refers to the first. Of the four possible outcomes in general, TT, TF, FT, FF, which are possible in this case? If the 1st statement is T, then the 2nd must also be T. But if the 1st statement is F, then the 2nd must also be F. Thus the only possible outcomes are TT or FF; the outcomes TF and FT cannot happen with this pair of statements. The probability that the pair of statements are both true is therefore the ratio of the probability for that outcome TT, to the sum of probabilities of all the outcomes possible in this case, i.e. TT + FF. So we get:

   (1/9)   
(1/9)+(4/9)
= 1 
5