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A coin flip by any other name (2023)
sider the following graph. If we remove each edge with independent probability what is the probability that there is still a path from the top to the bottom vertex? Since the removal of each subset of edges is equally likely, this kind of question is equivalent to counting the subsets of edges that we can remove without disconnecting two vertices.
Finally, we return to the problem of finding a "duality-like" rearrangement to explain the special fact that n n n points in R 2 n\R^{2n} R 2 n have the origin in their convex hull with probability 1/ 2. To mix things up, let's also look at condition (1)(1)(1) another way: it is the event that kkk randomly chosen hemispheres cover the entire unit sphere in Rn\R^nRn. Let's do the same kind of analysis for the dual condition ∃y∈Rn,−MTy>0,(2)\exists y \in \R^n, - M^T y > 0 \tag{2},∃y∈Rn,−MTy>0,(2) which we will view as the probability that nnn random vectors in Rk\R^kRk admit a strictly positive linear combination.
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