note moritz <p>You'd need a floating-point number with a mantissa of about 128 bits to hold the value of (1-1/2**128)</p> <p>Luckily there are other ways to evaluate the expression. If we set <c>x = 1/2**128</c>, we can do a [http://www.wolframalpha.com/input/?i=Series%5b%281-x%29^n%2C+{x%2C+0%2C+3}%5d|taylor series] to evaluate <c>(1-x)**n</c> around x = 0 (and since x is very small, it's a good approximation: <code> (1 - x)^n = 1 - n * x + 1 / 2 * n * (n - 1) * x ** 2 + ... </code> <p>Where higher orders can be neglected. If you chose an <c>n</c> that is high enough, you can get non-1 values as the result. <p>Though I'm not sure if that's really helpful. What you'd really need is the expectation value of the number of collisions (the lambda symbol in [wp://Poisson Distribution], then you could easily calculate everything you wanted. <!-- Node text goes above. Div tags should contain sig only --> <div class="pmsig"><div class="pmsig-616540"> [http://perl6.org/|Perl 6 - second systems done right] </div></div> 962802 962876