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2

Like Ray, I'd like to point out that if the PINs are not chosen randomly but selected by humans and there is no rejection of the easiest pins, the same rules as for passwords apply: some are very, very, very common. This analysis of 4-digit pins shows that 3 tries will allow you to break over 18% of 4-digit pins, not the 0.03% you would expect from the ...

0

Maarten and Max correctly analyze the probabilities for cases in which the PIN is selected uniformly from all 8 digit numbers. But if the user is permitted to choose their own PIN, the distribution will be biased towards those numbers that are easily memorable. (Naturally, the exact distribution will differ from person to person and thus can't be calculated ...

6

The chance to guess right is just $$\frac{3}{10^x}$$ where in your case, $x$ is $8$. You could write it as $$1-\frac {99999999}{100000000}\cdot\frac {99999998}{99999999}\cdot\frac {99999997}{99999998}=1-\frac {99999997}{100000000}=\frac 3{100000000}$$ To simplify it, just imagine a one digit PIN. You have ten possibilities, from 0 to 9. With three tries, ...

23

If a computer is doing the selection of PIN numbers, then you would be very lucky indeed to guess a PIN in three times. The entropy - assuming that all numbers are valid - is of course $\log_2{10^8} \approx 26.57$ bits. The chances of guessing the PIN correctly in 3 tries is 1 - \frac{x-1}{x} \cdot \frac{x - 2}{x-1} \cdot \frac{x-3}{x-2} = 1 - \frac{x-3}{...

1

Yes it most certainly is, on the assumption that the hashing algorithm used is cryptographically secure and uniform in its output. As a thought experiment, consider that after Elliptic Curve Diffie Hellman, the shared secret's (curve point) x co-ordinate is recommended by good practice to be passed through a hashing algorithm to derive a key. This is ...

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