You'll want to basically perform the reverse of a Fisher-Yates shuffle. Instead of generating a random number, you look at which card needs to be swapped with

You'll want to basically perform the reverse of a Fisher-Yates shuffle. Instead of generating a random number, you look at which card needs to be swapped with the Kth card in the deck to make the top K to N positions of the deck in sorted order. If deck is a list holding a permutation of the numbers 0 to N - 1, it looks something like: random_value = 0 for i in range(len(deck)-1, 0, -1): index = deck.index(i) random_value += index random_value *= i deck[index], deck[i] = deck[i], deck[index] The above is a one-minute prototype and may or may not work. When I get some time, I'll actually test the above code, but it's close to doing the right thing. The above naive implementation is O(N^2). Your end result is a sorted deck, and random_value is a number that when repeatedly mod'd and divided by the appropriate index to generate a "random" number during a Fisher-Yates shuffle would produce your original permutation. In practice, you'd implement the algorithm in O(N) by first doing a single linear scan to build an index (actually just held in an array) of cards to their positions. Also, you wouldn't actually perform the swap, since you never look at deck[i] after performing the swap. You just do deck[index] = deck[i]. So, at the end, you won't wind up with deck being sorted.