CareerCup

Let x be the random variable of days to reach heaven. we need expected value of that rv

x = 1/3*1 + 1/3(1+x) + 1/3(2+x)

3x = 1 + 1+x + 2+x

3x = 4 + 2x

x = 4

why not x = 1/3*x + 1/3(1+x) + 1/3(2+x) ? since the first term means after x days when the person goes back, the first gate is chosen then the person can go to the heaven immediately. but there is no solution then if changed the first term.

I actually should be:
x = 1/3*0 + 1/3(1+x) + 1/3(2+x)
3x=1+x+2+x
x=3

The expected number of days is 3.

Is there any other probable solution??

i agree with espresso answer should be 3

I have a different opinion on this.

Trying to summarize theq question again.
E(x) = Sumation(i) pi * xi = p0x0 + p1x1 + p2x2;

p0 = p1 = p2 = 1/3 ( Equally likely event to choose gate)

x0 = 0 days in Hell = 0
x1 = 1 days in Hell = 1
x2 = 2 days in Hell = 2

Objective is to find how many days in hell?

Expected value for a person in days to reach heaven is
E(x) = (1/3)*(0) + (1/3)*(1) + (1/3)*(2) = 1

Person would take one expected day to reach Heaven.

Please provide more details on earlier posted solution if you think 1 is not a solution.

Thanks
Ankush

This would have been altogether different question "if after staying one time in hell a person ".

Please provide solution to support your comments.

The three doors are shuffled every morning. So what if the person always ends up opening the 2day hell door.

The Solution seems correct to me.

As the probability of choosing a door is same.
p0=p1=p2=1/3
Even if they are shuffled.

Now if the person chooses the door to heaven the value of xi becomes 0.

Else, if he chooses 1 day @ Hell and then heaven the value of xi becomes 1.

Finally if he chooses 2 days @ Hell and then Heaven the value of xi becomes 2.

So,
Expected value for a person in days to reach heaven is
E(x) = (1/3)*(0) + (1/3)*(1) + (1/3)*(2) = 1

This is correct :)

He should follow islam.. He'll always go to heaven... with probability 1...
OBL {Allah hu akbar allah}

I think expresso is right.