6 ms·
"In fact, any time you shuffle a deck to the point of randomness, you have probably created an arrangement that has never existed before." Eh... Maybe not. My
by Hasz 9y ago
"In fact, any time you shuffle a deck to the point of randomness, you have probably created an arrangement that has never existed before."
Eh... Maybe not. My bet is it is somewhat analogous to the birthday problem. I have no doubt that it's possible to create many new arrangements, but the possibilities are probably not normally distributed and there's been a hell of a lot of card games played.
All of theese come together to say that a new shuffle is probably not unique in the universe.
- theoh 9y agoNot normally distributed? Why would they be? For a "random" shuffle you would want a uniform distribution -- each permutation of the cards being equally likely. Can't think what bias could possibly consistently distort a random shuffle, but maybe you have one in mind. Edit: "to the point of randomness" effectively establishes a uniform possibility of getting any one of the possible orderings...
- deleted 9y ago[deleted]
- njohnson41 9y agoEven with the birthday problem, sqrt(52!) is still about 10^34, which is still huge. It's unlikely that any two (sufficiently random) shuffles in history have been the same.
- taneq 9y ago> (sufficiently random) Isn't that a true Scotsman? Edit: I guess the GP post uses the same criterion, though.
- benchaney 9y agoNo, that isn't what the No True Scottsman Fallacy is. Putting qualifiers on statements is completely valid. A No True Scottsman Fallacy is when you use qualifiers in a hand wavy way to move the goal posts.
- taneq 9y agoWhat makes it a No True Scotsman is the term 'sufficently random', which kind of makes it unfalsifiable. After all, if any two shuffles are the same, then (by this logic) they're not sufficiently random.
- joshuamorton 9y agoExcept its unlikely that "sufficiently random" refers to "do not equal each other" in this context. Its likely that "sufficiently random" means "passes one of a variety of proofs of being statistically random (for example: nlogn repetitions of fisher-yates)". That implies randomness, but doesn't imply "doesn't equal any other deck of cards" except with high probability.
- benchaney 9y agoThat doesn't make it unfalsifiable at all. Sufficiently random is not defined as "not the same". you also clearly misunderstood the logic of the post in question. The poster started with a clear definition of random and used that to determine if shuffles are likely repeated. If there were fewer cards in a deck, that post would have reached a different conclusion.