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    \$\begingroup\$ I like your first approach. I'm having trouble understanding how a bias towards the equator is avoided if z, from a uniform distribution, is left unmodified. \$\endgroup\$ Commented Sep 9, 2019 at 8:12
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    \$\begingroup\$ @Jitse The spherical distribution is in factor uniform over each coordinate. This is something special for dimension 3. See for instance this proof that the surface area of a slice of a sphere is proportional to its height. Regarding the intuition that this is biased to the equator, note that while slices near the equator have a bigger radius, ones near the pole are titled inward more which gives more area, and it turns out these two effects exactly cancel. \$\endgroup\$ Commented Sep 9, 2019 at 8:14
  • \$\begingroup\$ Very nice! Thanks for the clarification and the reference. \$\endgroup\$ Commented Sep 9, 2019 at 8:23
  • \$\begingroup\$ @Jitse Thanks, I added it to the body. I realized though that I was only sampling positive z though, and fixed that for a few bytes. \$\endgroup\$ Commented Sep 9, 2019 at 8:26
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    \$\begingroup\$ @Jitse Indeed, the surface area of a sphere equals the lateral surface area of the enclosing cylinder! \$\endgroup\$ Commented Sep 9, 2019 at 8:50