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May 5, 2019 at 2:49 history edited Paul Uszak CC BY-SA 4.0
Hmin increase per die decreases with more dice!
May 5, 2019 at 2:40 history rollback Paul Uszak
Rollback to Revision 1
May 5, 2019 at 2:18 history edited Paul Uszak CC BY-SA 4.0
Dropping Hmin didn't feel right... Doh!
May 5, 2019 at 2:05 comment added Future Security Entropy is unaffected by ignoring fixed bits. It is the system that has entropy, not individual bits. In fact, if you were to concatenate the four variables and could ensure that it's always possible to extract the original four values from the combined string, then the result would have as much entropy as you could possibly get from the result of any function of those four variables. Even adding a constant prefix wouldn't lower entropy. Entropy depends only on the probability of each outcome. Not how you represent the outcome.
May 5, 2019 at 1:50 comment added Future Security Min-entropy is not dropping with more dice. The maximum probability of any given sum decreases with more dice. Min-entropy goes up as max probability goes down. For fair two dice, the most like outcome (sum 7) has $0.1\overline{6}$ chance of occurring. $-\log_2(1/6) \approx 2.58$. For three dice, sums 10 and 11 tie for 12.5%. Therefore the min-entropy of this 3-dice-sum distribution is $-\log_2(0.125) = 3$ bits.
May 5, 2019 at 1:32 comment added Future Security No. Entropy is not dependent on variance or vice versa. Entropy depends only on the individual probability of each possible outcome. Variance depends on the value of each outcome and its likelihood in relation to other outcomes. A random variable which is 1 20% of the time, 2 30%, 3 50% has the same entropy as one that is 1, 10, or 1000 with the respective probabilities 0.5, 0.2, and 0.3, but each variable has a different variance. Variance will not tell you what the entropy is. Entropy will not tell you the variance.
May 4, 2019 at 22:52 history answered Paul Uszak CC BY-SA 4.0