You can simplify this quite a bit if you use a radius of 1, rather than a diameter of 1000. When R = 1, the test becomes $x * $x  + $y * $y <= 1.

(Very) Old Habits Die Hard. Hence my use of $x*$x rather than $x**2. I doubt there's much difference on modern processors.

Update: s/\^/**/. Thanks to kyle for pointing it out

In reply to Re: probabilistic pi algorithm by FunkyMonk
in thread probabilistic pi algorithm by spx2

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