Consider a standard deck of cards. We draw cards from the top until we get either a King or an Ace.
1.Write a lower bound for the expected value of the number of Kings drawn before the first Ace.
The solution given in this problem is 1/2.
I'm not sure how they arrive at this solution. I assumed that both Kings and Aces are equally likely to be drawn so shouldn't the # of kings being drawn before first Ace appear be $n* \frac {1}{52}$ ?
2. Find the expected value of the number of Kings drawn before the first Ace.
Here since the number of kings in a deck is 4 and there are 52 cards in total shouldn't the expected number of kings drawn before the first Ace be $n* \frac {4}{52}$ ?
Best Answer
A lower bound is any value that is certainly bellow E(X), you cold say $0$, it would be a right answer. They provided a bit less obvious answer 1/2, since there is probability of 0.5 you will find an ace first and same probability you will find a king first, so 0.5*0+0.5*1=0.5. But there may be more kings after that first so expected value is certainly above 0.5
For the expected value you just have to play with the permutations of KKKKAAAA. There is total of $8\choose 4$=70 permutations.
With all 4 kings first - 1 permutation.
With 3 kings first and then an ace - 4 permutations (KKKA+KAAA or AKAA or AAKA or AAAK =$4\choose 1$).
With 2 kings first and then an ace $5\choose 2$=10 permutations.
With 1 king followed by ace $6\choose 3$=20 permutations
So E(X)=(4*1+3*4+2*10+1*20)/70=56/70=0.8