Lies, Lies

Let us have 3 cases: Case 1: Prisoner 1 is let go first Case 2: Prisoner 2 is let go first, Prisoner 1 is let go second Case 3: Prisoner 3 is let go first, Prisoner 1 is let go second

Clearly, the sum of the probabilities of all 3 of these cases will yield the answer.

The probability of Case 1 is $0.2 + 0.8 \cdot 0.5 \cdot 0.1 \cdot 0.2 + 0.8 \cdot 0.5 \cdot 0.1 \cdot 0.8 \cdot 0.5 \cdot 0.1 \cdot 0.2 +\cdots = \dfrac{0.2}{1 - 0.04} = \dfrac{20}{96} = \dfrac{5}{24}$ .

Case 2 has two parts. The probability that Prisoner 2 is let go first is $\dfrac{0.4}{0.96} = \dfrac{40}{96} = \dfrac{5}{12}$ , and the probability that Prisoner 1 is let go after that is $0.1 \cdot 0.2 + 0.1 \cdot 0.8 \cdot 0.1 \cdot 0.2 + \cdots = \dfrac{0.02}{0.92} = \dfrac{2}{92} = \dfrac{1}{46}$ , and the probability of Case 2 is $\dfrac{5}{12} \cdot \dfrac{1}{46} = \dfrac{5}{552}$ .

Case 3 also has two parts. The probability that Prisoner 3 is let go first is $\dfrac{0.36}{0.96} = \dfrac{3}{8}$ , and the probability that Prisoner 1 is let go after that is $0.2 + 0.8 \cdot 0.5 \cdot 0.2 + 0.8 \cdot 0.5 \cdot 0.8 \cdot 0.5 \cdot 0.2 + \cdots = \dfrac{0.2}{1-0.4} = \dfrac{0.2}{0.6} = \dfrac{1}{3}$ , and the probability of Case 3 is $\dfrac{3}{8} \cdot \dfrac{1}{3} = \dfrac{1}{8}$ .

Now, the sum of the probabilities is $\dfrac{5}{24} + \dfrac{5}{552} + \dfrac{1}{8} = \dfrac{5}{23} + \dfrac{1}{8} = \dfrac{63}{184}$ , and $63 + 184 = \boxed{247}$ .

It is useful to first describe and solve for a general game as follows: Let Player $P$ be the first in line, who tells the truth with probability $p$ . Furthermore, given Player $P$ does not tell the truth, then let $q$ be the probability Player $P$ gets another try. We can use this model to describe games where there are 2 or 3 (or any number of) players.

Let $x$ be the probability that Player $P$ is the first player released from this game. First step analysis gives us that $x = p + (1-p)qx.$ That is, we can break this down into two exclusive events. First, Player $P$ can tell the truth on the first turn (probability $p$ ). Second, Player $P$ does not tell the truth on the first turn and the game gets around to Player $P$ again. Since this is the exact same situation as the start of the game, the probability for this case is $(1-p)qx$ .

Solving for $x$ in terms of $p$ and $q$ gives us $x = \frac{p}{1 - (1-p)q}.$ Note that the denominator looks messy, but it is simply 1 minus the probability that everyone lies once.

We now consider the two cases where Prisoner 1 is not released:

Case 1: Prisoner 2 is the first released and Prisoner 3 is the second released. The probability of this is the product of the following sequence of events:

• Prisoner 1 lies on the first turn, giving Prisoner 2 a turn: $4/5.$

• Prisoner 2 is first released: using $p = 1/2, q = \frac{1}{10}\frac{4}{5} = \frac{2}{25}$ in our model, we get $25/48.$

• This leaves Prisoner 1 and 3 with Prisoner 3 going first. Prisoner 3 is released first with probability (using $p = 9/10, q = 4/5$ ) of $45/46.$

So the probability of Case 1 is $75/184.$

Case 2: Prisoner 3 is the first released and Prisoner 2 is the second released. The probability of this is the product of the following sequence of events:

• Prisoners 1 and 2 lie on their first turn: $\frac{4}{5} \frac{1}{2} = \frac{2}{5}.$

• Prisoner 3 is first released: using $p = 9/10, q = \frac{4}{5}\frac{1}{2} = \frac{2}{5}$ in our model, we get $15/16.$

• Prisoner 1 lies on their next turn, giving Prisoner 2 another turn: $4/5.$

• Prisoner 2 is released next: using $p = 1/2, q = 4/5$ , this is probability $5/6$ .

So the probability of Case 2 is $1/4$ .

The total probability of these two cases is $121/184.$ The probability that Prisoner 1 is released is the complement of this, $63/184,$ giving an answer of $63 + 184 = \boxed{247}.$