Given n numbers (each of which is a random integer, uniformly between 1~n), what is the expected number of increasing subsequences?

closed as too localized by Bill Johnson, Emil Jeřábek, Did, Yemon Choi, Bruce Westbury Feb 8 '12 at 19:07
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This is not a researchlevel question and should be closed, but since someone already gave a wrong answer, I’ll write the correct one. Given independently uniformly random $x_1,\dots,x_k\in\{1,\dots,n\}$, the probability that the $x_i$’s are pairwise distinct is $n(n1)\cdots(nk+1)n^{k}$, hence the probability that $x_1< x_2<\dots< x_k$ is $\binom nkn^{k}$. By linearity of expectation, the expected number of increasing subsequences in a randomly chosen sequence $(x_1,\dots,x_n)$ is thus $$\sum_{k=0}^n\binom nk^2n^{k}.$$ Note that this includes the empty sequence (which is vacuously increasing). 


According to: MR0626437 (84e:05012) Lifschitz, V.; Pittelʹ, B. "The number of increasing subsequences of the random permutation. " The answer for permutations on $n$ elements is: $\sum_{k = 1}^n C_{n,k}/k!$ The reason is that for each subset $I$ of $k$ indices from $1$ to $n$ there is a $1/k!$ chance that a random permutation will be increasing when restricted to $I$ (for each set of values only $1$ out of the $k!$ yields an increasing sequence). 

