5 ms·

Even 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 b

• taneq 9y ago
  > (sufficiently random) Isn't that a true Scotsman? Edit: I guess the GP post uses the same criterion, though.
  • benchaney 9y ago
    No, 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 ago
      What 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 ago
        Except 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 ago
        That 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.