Expectation over sequencial random shuffles I'm trying to understand this concept with this following problem:

Logan is cleaning his apartment. In particular, he must sort his old
favorite sequence, , of  positive integers in nondecreasing order.
He's tired from a long day, so he invented an easy way (in his
opinion) to do this job. His algorithm can be described by the
following pseudocode:
while isNotSorted(P) do {   
    WaitOneMinute();
    RandomShuffle(P)
}

Can you determine the expected number of minutes that Logan will spend
waiting for  to be sorted?

Let's say we need to evaluate this situation:
array with 2 items:

5,2

Solution:

There are two permutations possible after a random shuffle, and each
of them has probability . The probability to get the sequence sorted
after the first minute is . The probability that  will be sorted after
the second minute is , the probability  will be sorted after the third
minute is , and so on. So, the answer is equal to the following sum:
$\sum_{i=1}^\infty i*2^{-i} = 2.000000$

I understand that this is an expectation problem.
But I can't get the correct situation for a list different of 2 as size. As explained above
Can some one explain me what happens with 3 as size of the list?
 A: Assuming they're all distinct, the process succeeds with probability $1$ divided by the number of possible permutations, which is $n!$ for a sequence of $n$ positive integers. So it takes $n!$ steps on average.
A: Suppose he must sort the numbers $\{1,2,\dots,n\}$
If $\mu$ denotes the expected number of minutes then:
$$\mu=\frac1{n!}1+(1-\frac1{n!})(1+\mu)$$
Here $\frac1{n!}$ is the probability that he is ready after one minute. 
If he is not then he must start over again and has lost one minute.
This equality tells us that : $$\mu=n!$$
More formally:
If $E$ denotes the event that he succeeds in one minute then:
$$\mathbb EX=\mathbb E(X\mid E)P(E)+\mathbb E(X\mid E^c)P(E^c)$$
and this with $$\mathbb E(X\mid E^c)=1+\mathbb EX$$
A: The probability that $n$ distinct items are ordered  in the favorite order after one minute is
$\frac1{n!}$
The probability that $n$ distinct items are ordered in the favorite order after k minutes is
$P(X=k)=\frac1{n!}\cdot \left(1-\frac1{n!} \right)^{k-1}$
The expected value is related to the geometric distribution. 
$$E(X)=\sum_{k=1}^{\infty} k\cdot P(X=k)=\sum_{k=1}^{\infty} k\cdot  \frac1{n!}\cdot \left(1-\frac1{n!} \right)^{k-1}=n!$$
