3 ms·
No, 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
by HelloNurse 9y ago
No, 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 :)