If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 or more 1s, much fewer chains of 3 ones than would be expected from truly random number generation, and no chains of 8 or more 0s. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be $2^n$ (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. Wouldn't a deck of cards (1-10) have the same problem?

If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 or more 1s, much fewer chains of 3 ones than would be expected from truly random number generation, and no chains of 8 or more 0s. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be $2^n$ (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. On https://www.swansontec.com/bitcoin-dice.html suggests using d6 as an alternative to Hex dice, is this insecure? Wouldn't a deck of cards (1-10) have the same problem?

If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 or more 1s, much fewer chains of 3 ones than would be expected from truly random number generation, and no chains of 8 or more 0s. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be 2^n (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. Wouldn't a deck of cards (1-10) have the same problem?

If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 or more 1s, much fewer chains of 3 ones than would be expected from truly random number generation, and no chains of 8 or more 0s. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be $2^n$ (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. Wouldn't a deck of cards (1-10) have the same problem?

If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 1s, and much fewer chains of 3 ones than would be expected from truly random number generation. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be 2^n (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. Wouldn't a deck of cards (1-10) have the same problem?

If the mapping table looks like:

1: 000

2: 001

3: 010

4: 011

5: 100

6: 101

Then you are guaranteed that your key will never have a chain of 4 or more 1s, much fewer chains of 3 ones than would be expected from truly random number generation, and no chains of 8 or more 0s. Unless you're using an unintuitive mapping from d6 to binary (e.g. alternating 0 and 1), your dice would have to be 2^n (2, 4, 8, 16, etc.) sided in order to produce unpredictable binary results. Wouldn't a deck of cards (1-10) have the same problem?