A Developer E-pistle

This post is in regard to Jeff Atwood’s post on Shuffling and The Danger of Naïveté.

In his second post he used an example of shuffling three-card deck 600,000 times to explain how naive solutions can be dangerous. The post is extremely well written. I enjoyed every bit of it. However I had two questions that I couldn’t find an answer for in Jeff’s post. Why exactly 3 permutations (out of the 6 possible ones) are biased? and why those 3 specifically are biased and not any other permutations?

I spent sometime thinking about it and using a pencil and a paper I was able to get some answers.
First here is the naive method: (I reversed the loop to be as similar as possible to the correct method)

for (int i = cards.Length - 1; i >= 0; i--)
{
    int n = random.Next(cards.Length);
    Swap(ref cards[i], ref cards[n]);
}

And here is the correct one (Knuth Shuffle)

for (int i = cards.Length - 1; i > 0; i--)
{
    int n = rand.Next(i + 1);
    Swap(ref cards[i], ref cards[n]);
}

Now let’s agree on some facts real quick. For a three-card deck:

1. There are exactly 6 unique possible permutations for that deck (3!)
2. Using the Naive shuffling method, there are 27 possible sequences of random-numbers on any given run (3³)

The answer to the first question above, is that 27 mod 6 is equal to 3. What that means is there are 3 extra sequences of random numbers that have to map to 1 (or 2 or 3) of the 6 unique permutations of the deck.

To answer the second question we have to note something first about using swapping to perform a shuffle, take a look at the state of the deck at each iteration given the following random numbers:

Notice how the two different sequences of random-numbers resulted in the same permutation of the deck. Now remember there are 27 possible sequences and only 6 unique permutations. if we divide 27/6 we get 4 (with remainder 3) which means there are at least 4 paths (or sequences) for every unique permutation of the deck, the 3 extra possible sequence lead to 3 of the unique permutations giving them an advantage over the other 3.

so basically we have 6 permutations of the deck, 3 of them have 4 random-number sequences that lead to them, and the other 3 have 5 sequences that lead to them.

In this particular example if you follow the algorithm above and try the following random-number sequences you will get the same deck permutation for all 5.

• 1,1,2
• 3,1,3
• 2,2,2
• 1,3,1
• 2,3,3

Conclusion

The problem with the Naive shuffling method is swapping! The fact that swapping elements in 2 (or more) different sequences can lead to the same deck permutation is a subtle problem.

The Knuth shuffle algorithm avoids that problem by ensuring N! possible random sequences where N is the number of elements in a deck (or any container).