Suppose x(0) = 0 and x is observed at t=1, 2, ..., N, and x(t) = x(t-1) + z(t), where z(t) is N(0,1) (normally distributed). What is the probability that x(0) is either a minimum or maximum of x(0:N), meaning that either all N later observations are positive or all N observations are negative? For N=2 the probability is 3/4, since given one step z(1), the only way x(0) is not the minimum or maximum of x(0), x(1), and x(2) is that z(2) is opposite in sign to the first step and larger in magnitude. Each of these probabilities equal 1/2, and they are independent, and 1/2^2 = 1/4. I simulated the general problem and got this result and results for N > 2, but I wonder if this question has been studied and if there is an analytical solution.

Intuitively, the probability goes to zero as N goes to infinity. On the very first move, you will have eliminated one of the options, say if the first observation is negative. Then, we know that the continuously observed process will come back above zero in finite time with probability one. I’m guessing that the discrete sampling is not going to change that in the limit. For finite values of N it sounds kind of messy.

Isn't it a product of CDFs for x1, x1+x2, ...?

So there should be a factor 1/2^n and (1 + erf[sum of x's / sqrt(2)]. Erf is bounded from -1 to 1 and the sums of x's tend to 0.

I would check out Feller, Vol. II Ch 12, or possibly Spitzer's 'Principles of Random Walk'.

p.s. Perhaps this is answered by Corollary 5, here 

It agrees with your answer for N=2, which is apparently 3/8. 

It's quite interesting, IMO, that the answer doesn't depend upon the distribution of the increments of the walk, as long it is continuous and symmetric.

BTW, somewhat related: I did an article a while back for Wilmott magazine titled "Diffusions, Jumps, and the Distribution of the Maximum." I could post it here if there is any interest.