Including the 2-elements array.
There is 50% possibility that no elements will land in its position,
and 50% possibility both of the elements are in the right position.
So the expected is 50% * 2 + 50% * 0 = 1

2011/5/8 Jin Mingjian <[email protected]>:
> "So we can expect that in any random permutation, one element will land in
> its correct sorted pos." except for the 2-elements array?
>
>
> On Sun, May 8, 2011 at 3:20 PM, rajatag12 <[email protected]> wrote:
>>
>> Yes. First thing to see is that Goro would never hit the table with open
>> elements that *could not* land in their correct position for any
>> permutation. Once it is understood, it will be easy to see that solving the
>> prob for each permutation cycle is an independent one. Answering the main
>> question, observe that each element in a sequence of N elements appears in
>> its correct place (N-1)! times out of the N! permutations. So we can expect
>> that in any random permutation, one element will land in its correct sorted
>> pos. So after each hit, Goro should include the last element placed
>> correctly in his set of elements to hold before he hits. Hence, expected
>> number of hits = n.
>> Formal proof has already been provided above.
>>
>> On Sunday, May 8, 2011 7:22:39 AM UTC+5:30, Balajiganapathi wrote:
>>>
>>> I meant: Is n the expected number of hits to sort a n-element *cycle*.
>>
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