After shuffling order can be same as previous order. It is quite natural. 
As you said you can shuffle again if you don't want that case. 

On Wednesday, January 30, 2013 10:16:57 AM UTC+4, shady wrote:
>
> Yes I agree to it, it won't be random... but suppose I don't want a case 
> when all elements are at their own position........ because that case means 
> that they are not shuffled. Perhaps we can run the algorithm again, since 
> the probability of same event occurring two times in a row will be very 
> less.
>
> On Tue, Jan 29, 2013 at 12:13 AM, Carl Barton 
> <odysseus...@gmail.com<javascript:>
> > wrote:
>
>> Because then it's not a random shuffle? If you randomly shuffle something 
>> the order you currently have should be just as likely as any other
>>
>>
>> On 28 January 2013 12:29, shady <sin...@gmail.com <javascript:>> wrote:
>>
>>> Why do we use Fisher Yates 
>>> algorithm<http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle#The_modern_algorithm>
>>>  when 
>>> in the worst case there is no shuffle at all ?
>>> we can modify it by generating random number not inclusive of the 
>>> element that we are about to swap....
>>>
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