The smallest largest element can be 10 when the subset is {1, 2,..., 10}. The probability of this occurrence is (1/2)^10 = 1/1024. (Each element from 1 to 10 can either be included in the subset or omitted.)
The largest element can also take on the value of 11. In this case, every number in the subset must be from 1 to 10, necessitating the exclusion of one number, yielding the probability that the largest element is 11 as C(10,1)*1/1024.
The maximum element could also be 12. Here, all subset numbers must be selected from 1 to 11, requiring the omission of two numbers, resulting in the probability for the largest element of 12 as C(11,2)*1/1024.
This reasoning can be extended similarly for additional cases:
If the largest element is 13, then the probability is C(12,3)*1/1024.
If it's 14, the probability equals C(13,4)*1/1024.
If it's 15, it corresponds to C(14,5)*1/1024.
If it's 16, then we have C(15,6)*1/1024.
If the largest element is 17, it corresponds to C(16,7)*1/1024.
If it’s 18, it translates to C(17,8)*1/1024.
If it’s 19, it equals C(18,9)*1/1024.
If the largest element is 20, it concludes with C(19,10)*1/1024.
Summing these probabilities leads to (1 + C(10,1) + C(11,2) +... + C(19,10))*1/1024. Since 1 can be treated as C(9,0), we can state (C(9,0) + C(10,1) + C(11,2) +... + C(19,10))*1/1024.
Using the Hockey Stick Identity, it holds that C(9,0) + C(10,1) + C(11,2) +... + C(19,10) equals C(20,10), so the expected value for the largest element is 1/11*C(20,10)*1/1024 = 4199/256.
