2 ms·
[ed: no, that's not right. With two cards, you can have red-back or black-red. I guess that'd translate to N bits for 2 n cards, but it's not quite trivial...
by e12e 9y ago
[ed: no, that's not right. With two cards, you can have red-back or black-red. I guess that'd translate to N bits for 2 n cards, but it's not quite trivial...
For four, we have: rrbb, rbrb, rbbr, bbrr, brbr, brrb etc]
I think I'm misunderstanding you - but say you have 2n cards, n red, n black. If you shuffle them, you have 2n random bits, read from first to last? (this is assuming it's feasible to actually randomly shuffle cards).
- HelloNurse 9y agoNo, in this application cards are all distinct, not merely half red and half black. There are n! permutation of a deck of n cards, and a uniformly chosen random permutation contains entropy equal to the base 2 logarithm of n!. Both choosing a random permutation given a supply of random bits and, as noted in the parent post, extracting random bits from a given random permutation are nontrivial computational problems leading to interesting algorithms; shuffling cards or tiles is a good practical shortcut to obtain a random permutation more directly.
- e12e 9y agoIndeed. I was attempting to induct from basic principles - as in two distinct cards.. But gp's right - it's not trivial. Unsure if it's simple :)