Computing the proportion of vectors with the same sign

Let $s \propto (1, \ldots, 1)$ be a d-dimensional unit vector, so that $s_i = 1 / \sqrt{d}$. Let ${\cal V} = \{ u \in {\mathbb S}^{d-1}:u \cdot s = \cos \theta \}$ be a collection of unit vectors that form an acute angle with $s$. Suppose that $\cos \theta \rightarrow 1$ as $d \rightarrow \infty$ at a rate to be specified below.

I am interested in the following: what fraction of elements of ${\cal V}$ have the same sign as $s$? Put another way, suppose that $U$ is uniformly distributed over ${\cal V}$, then what is ${\mathbb P}[\cap_{i=1}^d\{{\rm sign}(s_i) = {\rm sign}(U_i)\}]$?

It is clear that when $d(1-\cos^2\theta) \rightarrow 0$ that all elements of ${\cal V}$ have the positive sign. However, I am not sure how to obtain a closed form solution when $d(1-\cos^2\theta) \rightarrow \infty$?

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What do you mean by the 'sign' of a vector? – Chris Taylor May 16 '11 at 15:38
@Chris Taylor: I would guess that ${\rm sign}((s_1,\ldots,s_n))=({\rm sign}(s_1),\dots,{\rm sign}(s_n))$. But that is just a guess. – mac May 16 '11 at 17:07
@Chris Taylor: I mean that sign is applied to each element of the vector and the equality holds element-wise. – mkolar May 16 '11 at 21:50