Question

Only do Q2 please!!! please Type the code in python!!!thanks

1. (10 points) (Without Python) Unfortunately, Eddard is lost in the dark dungeon of the Red Keep. He knows there are three doors, and one of the doors will lead him to freedom. If Eddard takes Door A, he will wander around the dungeon for 3 days and return to where he started If he takes Door B, he will wander around the dungeon for 4 days and return to where h<e started. If he takes Door C, he will find the exit after 2 days. If Eddard returns to where he started, he immediately picks a door to pass through, and since it's pitch black in the dungeon, Eddard picks each door uniformly at random. How long, on average, will Eddard wander in the dungeon before finding his way out? 2. (30 points) (With Python) In this Python exercise, you will estimate the number of days it will take for Eddard to escape the dungeon. The setup for the problem is the same as above. For this exercise, you may want to recall the following: » The Law of Large Numbers says that if we have a sequence of i.i.d. random variables Xi for i = 1, 2, . . . , n, then Tl That is, the average of the results found from a large number of trials tends to the expected value. Be sure to annotate your code with short explanations of what you are doing (worth 10 points) (a) Simulate Eddard's journey to freedom 10,000 times. Find the average time it takes for Eddard to escape the dungeon. (b) Print out your result using the print function in Python. Please be sure to say want you are printing/outputting. For example, print("After 10,000 simulations, the average time it takes Eddard to escape is %s days." %averageTimeInDungeon) Take a screenshot showing your code and your results side by side.

Only do Q2 please!!! please Type the code in python!!!thanks

Answer 1

Doing 1 just to establish the theoretical value

1. Let T be the time in Dungeon.

Let D={A,B,C} be the door taken with probability P(D=A) = P(D=B)=P(D=C)=1/3

The expectation of T can be written as

$\begin{align*} E(T)&=E[E(T\mid D)]\\ &=P(D=A)E(T\mid D=A)&plus;P(D=B)E(T\mid D=B)&plus;P(D=C)E(T\mid D=C)\\ &=\frac{1}{3}\left( E(T\mid D=A)&plus;E(T\mid D=B)&plus;E(T\mid D=C)\right) \end{align*}$

If D=A, then Eddard wanders around for 3 days and returns back to the start and hence would take on an average E(T) more before he gets out

E(T|D=A) = 3+E(T)

If D=B, then Eddard wanders around for 4 days and returns back to the start and hence would take on an average E(T) more before he gets out

E(T|D=B) = 4+E(T)

If D=C, then Eddard would find the exit in 2 days

E(T|D=C) = 2

Hence

$\begin{align*} E(T)&=E[E(T\mid D)]\\ &=P(D=A)E(T\mid D=A)&plus;P(D=B)E(T\mid D=B)&plus;P(D=C)E(T\mid D=C)\\ &=\frac{1}{3}\left( E(T\mid D=A)&plus;E(T\mid D=B)&plus;E(T\mid D=C)\right)\\ &=\frac{1}{3}(3&plus;E(T)&plus;4&plus;E(T)&plus;2)\\ \end{align*}$

by solving the equation we get

$\begin{align*} &3E(T)=9&plus;2E(T)\\ \implies& E(T)=9 \end{align*}$

Eddard would take on an average 9 days to get out.

2) Now the Python code with comments is below

import numpy as np #define a function to randomly choose a door def choosedoor): #get a uniform random number in the interval (0,1) u-np.random.random) # choose the door A if 0< u<1/3, B if 1/3c u<2/3, C if 2/3c u<1 if u<l/3: return ('A') return ('B') return C'C') elif u<2/3: else: # function to get the total time spent in the dungeon dei dungeon (door, days): # if door C get out in 2 days if door='c': days-days+2 #if door A take 3 days and go back to start by recursively call back the function elif door 'A days-days+3 door=choosedoor ( ) days-dungeon (door,days) #if door B take 4 days and go back to start by recursively call back the function else: days-days+4 door=choo3edoor ( ) days dungeon (door,days) return (days) # set the seed for the random number generator np.random.seed (123) # set the number of simulations N-10000 #initialize a variable to hold the time spent TimeInDungeon [] # simulate the run for N times for i in range (N) # choose a door for the first time door-choosedoor) #call the function dungeon and store the time spent TimeInDungeon.append (dungeon (door, 0)) #find the average time spent avergaeTimeInDungeon np.mean (TimeInDungeon) rint ("After 10,000 3imulations, the average time it takes Eddard to escape is %3 days . "save raaeTimeInDungeon)

Get the following output

After 10,000 simulations, the average time it takes Eddard to escape is 9.0264 days.

The result is close enough to the theoretical value calculated in 1)

The code in text format (the tabs may be lost)

import numpy as np

#define a function to randomly choose a door
def choosedoor():
#get a uniform random number in the interval (0,1)
u=np.random.random()
# choose the door A if 0<=u<1/3, B if 1/3<=u<2/3, C if 2/3<=u<1
if u<1/3:
return('A')
elif u<2/3:
return('B')
else:
return('C')

# function to get the total time spent in the dungeon
def dungeon(door,days):
# if door C get out in 2 days
if door=='C':
days=days+2
#if door A take 3 days and go back to start by recursively call back the function
elif door=='A':
days=days+3
door=choosedoor()
days=dungeon(door,days)
#if door B take 4 days and go back to start by recursively call back the function
else:
days=days+4
door=choosedoor()
days=dungeon(door,days)
return(days)

# set the seed for the random number generator
np.random.seed(123)
# set the number of simulations
N=10000
#initialize a variable to hold the time spent
TimeInDungeon=[]
# simulate the run for N times
for i in range(N):
# choose a door for the first time
door=choosedoor()
#call the function dungeon and store the time spent
TimeInDungeon.append(dungeon(door,0))

#find the average time spent
avergaeTimeInDungeon=np.mean(TimeInDungeon)
print("After 10,000 simulations, the average time it takes Eddard to escape is %s days."%avergaeTimeInDungeon)