|Think about Loose Coupling|
Fisher-Yates theory... does this prove that it is invalid?by MarkM (Curate)
|on Jul 25, 2003 at 01:00 UTC||Need Help??|
UPDATE: As pointed out by others, I made an error when translating the code. See their summaries for good explanations. Cheers, and thanks everyone. (tail between legs)
In a previous article, I was challenged for doubting the effectiveness of the Fisher-Yates shuffle as described in perlfaq.
Below, I have written code that exhausts all possible random sequences that could be used during a particular Fisher-Yates shuffle. Statistically, this should be valid, as before the shuffle begins, there is an equal chance that the random sequence generated could be 0 0 0 0 0 as 0 1 2 3 4 as 4 4 4 4 4. By exhaustively executing the Fisher-Yates shuffle, and calculating the total number of occurrences that each result set is produced, we can determine whether the Fisher-Yates shuffle has the side effect of weighting the results, or whether the shuffle is truly random, in that it should be approximately well spread out.
With the above code, I was able to determine that with a deck size of 5, and an initial set of 1 2 3 4 5, there is three times the probability that the resulting set will be 3 1 2 5 4 than the probability that the resulting set will be 2 3 4 5 1. To me, this indicates that this theory is flawed.
If anybody needs to prove to themselves that the test is exhaustive, print out "@$rand" in the shuffle subroutine.
Please analyze the code carefully, pull out your school books, and see if I have made a mistake.