# Probability/statistics; prisoner and lions

A prisoner is placed in a circular room with six doors around the room numbered clockwise 1 through 6. He is told that behind two consecutive doors (ie 1,2 or 2,3 or 3,4 or 4,5 or 5,6 or 6,1) are lions. If the prisoner is able to open two doors without releasing a lion, he will be set free. Otherwise he will be lunch. However, there is a catch. His first choice depends upon the roll of a die. If he survives his first choice, for his second choice he may either select the next door in line (eg if he rolled a 3 first he may subsequently choose door 4), or he may roll the die again.

Now suppose he is successful with his first door. That is he rolls the die, opens the door according to the number rolled and there is no lion behind the door. For his next choice, should he select the next door in line, or should he roll the die again, and why? Note that it is permissable for the same door to be opened twice. For example, if the prisoner rolls a 3 to begin and opens door 3 successfully he may either choose door 4 as his second choice or roll the die again in order to select another door. He is allowed to roll another 3 which of course would mean he was successful again. You should also assume that the die is fair. Each number is equally likely.

I don't understand how to get the probability of getting a safe door by choosing the next door in line. I asked my teacher and he tried to help, but then had to be summoned elsewhere. I did get a hint, I don't understand what he said. The thing he told me was that we know he is successful with the first door, so be entered one of the safe doors. So, now only 1 of the 4 safe doors has a bad door next to it, which is a 1/4 chance of getting a bad door (or 3/4 chance of getting another safe door). This is what I'm confused about, how we get down to 4 doors? I thought it was a 3/5 chance since 3 of the 5 doors are safe (still only 2 doors have lions behind them).

I did figure out that if he decided to roll the die again, he will have a 4/6 chance (2/3) of getting a safe door, and a 2/6 chance (1/3) of getting a bad door, since 2 of the 6 doors have lions behind them.

We are in the realm of conditional probability.

Here, we are asking to compare $Pr(\text{survives second turn if choosing to roll}|\text{survived first turn})$ to $Pr(\text{survives second turn if choosing adjacent door}|\text{survived first turn})$

We have the definition of conditional probability: $Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}$

Suppose without loss of generality that the lions are behind doors $1$ and $6$.

Since he survived, we know that he had picked one of the doors $2,3,4,5$. Letting $B$ be the event that he survived the initial roll of the die, assuming the die is fair that would imply that $Pr(B)=\frac{4}{6}$.

Letting $A$ be the event that the room after the one chosen in the first turn is safe, we have $A\cap B$ is the event that the initial and consecutive rooms chosen are both safe. This occurs precisely when the initial roll of the dice is $2,3,$ or $4$, but will not occur with $5,6$ or $1$. (A $6$ or a $1$ and he would have been eaten on the first turn, a $5$ and he would be eaten on the second turn). This implies $Pr(A\cap B)=\frac{3}{6}$.

Combining this information, we have $Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{3/6}{4/6}=\frac{3}{4}$.

Notice, in the special case of an unbiased sample space (such as here), the formula $Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}$ simplifies as $=\frac{|A\cap B|}{|B|}$. The idea being, we essentially restrict our sample space to include only those outcomes where $B$ occurs.

Note: That the prisoner opens the next consecutive door is different than the prisoner choosing a different door at random. If the prisoner was asked to open a different door at random, his probability of continued survival would be $\frac{3}{5}$, but that is not what the problem is asking for.

To complete the problem, we still wonder what the probability of his survival after the second turn would be if he were to roll the die a second time. It is well known that two consecutive rolls of a die are independent from one another. Let $C$ denote the event that his second roll of the die also gives him a safe option.

We have in this case, $Pr(C|B)=Pr(C)=\frac{4}{6}$ by the fact that they are independent events. If you aren't immediately aware of this fact, then we could approach the longer way as well as $Pr(C|B)=\frac{Pr(C\cap B)}{Pr(B)} = \frac{16/36}{4/6}=\frac{4}{6}$ same as before. (Having calculated $Pr(C\cap B)$ as there being $16$ safe ways of rolling two dice consecutively out of $36$ equally probable ways of rolling two dice consecutively)

Since $\frac{3}{4}>\frac{2}{3}$, his better chances of survival are by picking the consecutive door.

Note: Swept under the rug a bit here is the reasoning why we could assume where the lions were ahead of time. The reason is that the location of the lions is independent to the probability of his survival. We could have run conditional probability on the locations of the lions and broken this into several cases, which becomes rather tedious, but we didn't need to since we could tell that regardless where the lions actually were, his chances of success remained the same.