I needed 3 random bits (and threw out any overflow), which I got by checking whether arbitrary words or phrases I thought of had an even or odd number of letters. That’s the most random completely mental (heh) way I know of, I wonder if there are others.
Average is irrelevant. What’s relevant is the standard deviation.
Since standard deviation goes as the square root of the number of items being added, phrase length for any reasonably-sized phrase, so long as it wasn’t a line of poetry, should be pretty evenly distributed.
It’s not obvious to me that it’s unbiased. My gut feeling suspects that if I randomly chose a word it’d be more likely to have an odd than an even number of letters.
Is (the seconds’ figure in my watch) mod 5 random enough?
I used the least significant digit on my time-remaining-to-full-charge. And ended up propping up the most populated entry.
I needed 3 random bits (and threw out any overflow), which I got by checking whether arbitrary words or phrases I thought of had an even or odd number of letters. That’s the most random completely mental (heh) way I know of, I wonder if there are others.
… you could have done it more-reliably evenly by taking the mod 5 of the phrase/word length.
Considering that the average word length in English is about five letters, I suspect that’d be quite far from being uniformly distributed.
Average is irrelevant. What’s relevant is the standard deviation.
Since standard deviation goes as the square root of the number of items being added, phrase length for any reasonably-sized phrase, so long as it wasn’t a line of poetry, should be pretty evenly distributed.
It’s not obvious to me that it’s unbiased. My gut feeling suspects that if I randomly chose a word it’d be more likely to have an odd than an even number of letters.