- https://github.com/bitcoinbook/bitcoinbook/blob/develop/ch05.asciidoc (this one talks about deterministic wallets as well)
- https://learnmeabitcoin.com/technical/checksum (already shared above, great site for learning the technical parts)
Fun fact: Throwing it a hundred times would suffice, even if the dice is very biased. Proof:
Suppose the probability of resulting 6 is 1/2, instead of 1/6. This means that on average, in every two rolls, you get a 6. We know that:
for i from 1 to 6: Σpi = 1
=> p1 + p2 + p3 + p4 + p5 + p6 = 1
=> p1 + p2 + p3 + p4 + p5 = 1 - p6 = 0.5
=> p1, p2, p3, p4, p5 = 1/10
Equation becomes: H(X) = - (p(1)*log2(p(1)) + p(2)*log2(p(2)) + ... + p(6)*log2(p(6))). For i=5, p(i) = 0.1 (for i < 6), that's equal with: -Σip(i)log2(p(i)) - p(6)log2(p(6)) = 1.660964 + 0.5 = 2.160964.
Rolling it a hundred times would give you about 216 bits of entropy. Please correct me if I'm wrong somewhere.
Went a little bit off-topic, but it does good to refresh your math knowledge once in a while.