Question

A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1,2, and 3 with probabilities \(0.5,0.3,0.2\), what is the expected number of days until he reaches freedom? (b) Assuming that the prisoner is always equally likely to choose among those doors that he has not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door 1 , then when he returns to the cell, he will now select only from doors 2 and 3.) (c) For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.

Short answer

The expected number of days until the prisoner reaches freedom is 2.125 days for the case with given probabilities, and 1.5 days for the case with equally likely unused doors. The variances for the number of days for these cases are 1.734375 and 1.015625, respectively.

Step-by-step solution

(a) Find expected number of days using given probabilities

Let X be the random variable representing the number of days until the prisoner reaches freedom. We can calculate the expected value of X (E[X]) using the given probabilities: \( E[X] = 0.5*(2 + E[X]) + 0.3*(3 + E[X]) + 0.2*0 \) \( E[X] = E[X] + 1.7 \) Since 0.2E[X] = 0.5*(2 + E[X]) + 0.3*(3 + E[X]), we can find the expected number of days to freedom as: \( E[X] = \frac{1.7}{0.8} = 2.125 \) So, the expected number of days until the prisoner reaches freedom while using the given probabilities is 2.125 days.

(b) Find expected number of days with equally likely unused doors

Now, let's calculate the expected number of days with the other strategy that the prisoner chooses only from those doors he hasn't used before: Let E1, E2, and E3 represent the expected values if the prisoner starts from door 1, door 2, and door 3 respectively. Now: \( E1 = \frac{1}{2}(2+E2) + \frac{1}{2}(2+E3) \) \( E2 = \frac{1}{2}(3+E1) + \frac{1}{2}(3+E3) \) \( E3 = 0 \) By solving these equations, we get: \( E1 = \frac{17}{8} = 2.125 \) \( E2 = \frac{19}{8} = 2.375 \) \( E3 = 0 \) Now, the expected value of the days to reach freedom in this case is simply the average of E1, E2, and E3: \( E[X] = \frac{2.125 + 2.375 + 0}{3} = 1.5 \) So, the expected number of days until the prisoner reaches freedom while using the new strategy is 1.5 days.

(c) Find the variance for parts (a) and (b)

For each part, we will first calculate the second moment (E[X^2]) and then use it to find the variance, which is given by Var(X) = E[X^2] - (E[X])^2. Part (a) Variance: \( E[X^2] = 0.5*(2 + E[X])^2 + 0.3*(3 + E[X])^2 + 0.2*0 \) Substitute the E[X] we found in part (a): \( E[X^2] = 0.5*(4.125)^2 + 0.3*(5.125)^2 \) \( E[X^2] = 8.515625 \) Now we can find the Variance: \( Var(X) = 8.515625 - (2.125)^2 \) \( Var(X) = 1.734375 \) Part (b) Variance: In this case, we need to find E1^2, E2^2, and E3^2 first: \( E1^2 = \frac{1}{2}(2+E2)^2 + \frac{1}{2}(2+E3)^2 \) \( E2^2 = \frac{1}{2}(3+E1)^2 + \frac{1}{2}(3+E3)^2 \) \( E3^2 = 0 \) Evaluating these and finding E[X]^2: \( E[X]^2 = \frac{202.625}{8} \) \( Var(X) = \frac{202.625}{8} - (1.5)^2 \) \( Var(X) = 1.015625 \) In conclusion: a) The expected number of days using given probabilities: 2.125 days b) The expected number of days with equally likely unused doors: 1.5 days c) The variance of the number of days for parts (a) and (b): 1.734375 and 1.015625, respectively.