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Am I right in thinking there are 90! possibilities. Would it be random? Would you trust this technique?
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Am I right in thinking there are 90! possibilities. Would it be random? Would you trust this technique?
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The address space has 2160 possible addresses, so to fully benefit from that space, you'd need to surpass a space of
2160 = 1.46E48 combinations.
This is achieved by picking 26 numbers from your toy bingo machine as
90!/64! = 1.17E49
while you keep track of the order in which they are picked.
To achieve at least 64 bit of entropy, you'd need 10 numbers (order recorded) from your toy bingo machine:
90!/80! = 2.07E19 > 1.84E19 = 264
Which was suggested as a minimum on the related question: How many throws of a dice are necessary to defeat a brute force attack?
All numbers are assuming that the toy bingo machine acts reasonably random.
But only if you always fill up all 90 fields. If you also allow shorter combinations there are even more! ;) – Murch – 2016-02-15T23:01:14.803