Given: A list of integers is there.Now there are 2 buckets -bucket A and bucket B.This step is repeated as long as there are numbers left in the list.Integers from start or end of the list are removed from the list and added to bucket A. The probability of picking a number from the start or the end is equal to 0.20 . Then similarly,the next number picked from start or the end is added to B. Then to A and so on.
Required: Expected values in Bucket A and Bucket B.
Eg 1 ,2m3 1 or 8 can be added to bucket A. 1) 1-> 4 or 8 can be added to B. a)B picks 4 -> A picks 8 b)B picks 8-> A picks 4 2) 8-> 1 or 4 can be added to B. a)B picks 1 ->A picks 4 b)B picks 4 ->A picks 1 So E(A)=(((8+4)/2+1)+((4+1)/2+8))/2=8.25
I tried to get a general formula for E(A) and E(B) .I was getting a tree structure with 2^n-1 nodes but I wasn't able to derive a specific formula. Can anyone derive this? I tried for a general list a,b,c,d having 4 integers.
Also,clearly the problem is symmetrical from both ends of the list.That is coefficient of a and d is same, b and c is same in E(A) Can anyone simplify this for any general list having m numbers?