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As in my answer heremy answer here, the Poisson approximation is pretty accurate. $$\mathbb{P}(\text{no collision})\approx \exp(-n^2/2M)\approx 1-{n^2/2M}.$$ Setting this equal to .9999 and $M=62^{10}$, we find that we can issue around $n=12.95$ million keys without worrying.

As in my answer here, the Poisson approximation is pretty accurate. $$\mathbb{P}(\text{no collision})\approx \exp(-n^2/2M)\approx 1-{n^2/2M}.$$ Setting this equal to .9999 and $M=62^{10}$, we find that we can issue around $n=12.95$ million keys without worrying.

As in my answer here, the Poisson approximation is pretty accurate. $$\mathbb{P}(\text{no collision})\approx \exp(-n^2/2M)\approx 1-{n^2/2M}.$$ Setting this equal to .9999 and $M=62^{10}$, we find that we can issue around $n=12.95$ million keys without worrying.

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user940
user940

As in my answer here, the Poisson approximation is pretty accurate. $$\mathbb{P}(\text{no collision})\approx \exp(-n^2/2M)\approx 1-{n^2/2M}.$$ Setting this equal to .9999 and $M=62^{10}$, we find that we can issue around $n=12.95$ million keys without worrying.