As several others have pointed out, $\pi$ obviously isn't actually random, since its digits are fixed in place forever.

But are the digits "as good as random"? The short answer is that, as far as anyone can tell, the answer seems to be empirically yes (See Marsaglia's _[On the Randomness of Pi and Other Decimal Expansions](http://www.yaroslavvb.com/papers/marsaglia-on.pdf)_). That is, there are no known ways in which the digits of $\pi$ behave systematically unlike a random number.

There are several other answers here that I think are confusing on this point. Let me explain why some of the claims made by other answers (despite being technically correct and in certain ways very insightful) don't disprove the claim that the digits of $\pi$ are as good as random.

 - One answer points out that there exists $n_0$ such that for all $n > n_0$, the $n$-th through $8n$-th bits of $\pi$ are not all zero. However (and perhaps somewhat unintuitively), this property is <i>also<i> true of any random bit string (with probability $1$). So the fact that the property holds is actually evidence *for* the randomness of $\pi$. 
 - Another answer points out that $\pi$ [doesn't seem to equidistribute](http://mathworld.wolfram.com/EquidistributedSequence.html) as uniformly as, for example, $e$. This just means that if you look at the numbers $i \pi - \lfloor i\pi \rfloor$ for $i \in \{1, 2, \ldots, 10000\}$ and you round each number down to the nearest multiple of $0.05$, then the result is less smooth then if you do the same thing for the number $e$. But the quantities being graphed here depend only on the first six digits of $\pi$. They're not meant to tell us anything meaningful about how random the digits of $\pi$ are. 
 - Finally, another answer correctly points out that the digits of $\pi$ do not fulfill the standard complexity-theory definition of a pseudorandom number generator. But that's mostly just because of a type mismatch: no individual bit string $x$ can fit the definition of a pseudorandom number generator. (Indeed, any algorithm that knows the first, say, 10 bits of $x$ can easily distinguish $x$ from a random bit string). In general, to construct a pseudorandom number generator, what you formally need is a set $S$ of bit strings with the property that no fast algorithm (i.e., no algorithm with running time substantially smaller than $|S|$) can distinguish a random element of $S$ from a truly random bit string (with probability at least, say, $2/3$). We could easily splice $\pi$ into multiple bit strings $S = \{x_1, \ldots, x_k\}$ where each $x_i$ consists of the $i$-th, $(i + k)$-th, $(i + 2k)$th, etc., bits of $\pi$; and I see no reason a priori to think that this wouldn't give a perfectly good pseudorandom number generator.