Math problem: New Game In Town

A brute force method with Excel gave me an EV of 2.01898101897439.
But I bet there's an elegant way to do the math.

Correct, but how about some details?

Before I begin, note that I use (n)C(k) to denote the number of combinations of n things taken k at a time; this is also C(n,k) or Combin(n,k).

This puzzle involves something called the Hockey Stick Principle:
(n)C(n) + (n+1)C(n) + (n+2)C(n) + ... + (n+k)C(n) = (n+k+1)C(n+1)
The name comes from the shape of the line of numbers when this is done on Pascal's Triangle.

Anyway, I will solve the generic case of subsets of K numbers out of {1, 2, 3, ..., N}
There are a total of (N)C(K) subsets
Let S be the smallest number in the subset
There are N-S numbers greater than S, and K-1 of them make up the rest of the numbers in the subset, so there are (N-S)C(K-1) subsets with the smallest number S.
There are (N-1)C(K-1) with smallest number 1, (N-2)C(K-1) with smallest number 2, and so on through (N-(N-(K-1)))C(K-1) with the smallest number N-(K-1), which is the one subset {N-K+1, N-K+2, ..., N}.
The sum of the smallest numbers = 1 x (N-1)C(K-1) + 2 x (N-2)C(K-1) + 3 x (N-3)C(K-1) + ... + N-(K-1) x 1
= ( (N-1)C(K-1) + (N-2)C(K-1) + (N-3)C(K-1) + ... + (K-1)C(K-1) ) + ( (N-2)C(K-1) + (N-3)C(K-1) + (N-4)C(K-1) + ... + (K-1)C(K-1) ) + ... + ( (K)C(K-1) + (K-1)C(K-1) ) + (K-1)C(K-1)
Apply the Hockey Stick Principle to each element in the sum
The sum = (N)C(K) + (N-1)C(K) + (N-2)C(K) + ... + (K+1)C(K) + (K)C(K)
Apply it again to the sum to get (N + 1)C(K + 1)
The expected return = (N + 1)C(K + 1) / (N)C(K)
= ( (N + 1)! / ( (K + 1)! (N - K)! ) ) / ( N! / K! (N - K)! )
= (N + 1)! / ( (K + 1)! (N - K)! ) ) x K! (N - K)! / N!
= (N + 1) N! K! (N - K)! / ( (K + 1) K! (N - K)! ) N!)
= (N + 1) / (K + 1)