| ▲ | edent 7 hours ago |
| I'm not so sure. If you wanted to be deliberate, you'd choose a plausible value that you could control. No one is going to suspect `98761316832165974216498789746` is anything other than random. But `000000000000000123` looks suspicious. |
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| ▲ | K0balt 5 hours ago | parent | next [-] |
| But that would be obviously intentional. Much more plausible deniability with 0x4. |
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| ▲ | sscaryterry 7 hours ago | parent | prev [-] |
| Why are many of them, suspicious? |
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| ▲ | kenniskrag 6 hours ago | parent | next [-] | | The probability of getting a 0 is the same for every digit. However, in my opinion, the probability of getting a number that starts with two zeros is 1/10 * 1/10, and the probability of getting a number that starts with the same two digits is 10 * 1/10 * 1/10. But the probability to receive one number or the other are the same. | |
| ▲ | matsemann 6 hours ago | parent | prev [-] | | By random chance you probably wouldn't get that many zeros. Of course, one could say it's just as random as any other string, but.. | | |
| ▲ | sscaryterry 6 hours ago | parent [-] | | Not a mathematician, but I'm sure the probability of them all being below 384 is going to be in the order of 0.000000001% if not less :) | | |
| ▲ | betterunix2 3 hours ago | parent | next [-] | | Sure, if they were actually sampling 384 bits. Sounds like they sampled a lot less than that; maybe someone though "384 bits" meant "in the range [0, 383]" or whatever. | |
| ▲ | kenniskrag 6 hours ago | parent | prev [-] | | 10 times the amount of digits is the possible number space. smaller than 384 means 384 numbers (including 0) divided by the amount of possible numbers (see above) imho | | |
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