# which arrangement gets favoured in naive shuffling for a given numbers from 1 to n according to the algorithm given?

algorithm: a=[1,2,3.....n] for i in (1,n+1): j=rand(1,n+1): swap(a[i],a[j])

let us say that n=3: 132 213 231 these 3 are 5 times possible and 123 312 321 are 4 times possible....

similarly for n=4 2143 occurs maximum number of times = 15 times and 4231 and 4123 occurs min times i.e 8 times only....

i figured it out for n=1,2,3,4; but i am not able to figure out the permutation which is maximum possible and minimum possible.

• I don't see how this can be answered, since it's entirely dependent on the behaviour of the random number generator. – David Richerby Sep 9 '18 at 17:54
• @DavidRicherby Assume that the PRNG is perfect. Note that to get a uniformly random permutation, rand(1,n+1) needs to be changed to rand(i,n+1). – Yuval Filmus Sep 12 '18 at 6:38
• For $n \geq 5$, the unique answer is $n,1,2,3,\dots,n-1$, as experiments readily show. It shouldn't be too hard to figure out why. – Yuval Filmus Sep 12 '18 at 6:47