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PODoes this simple shuffling algorithm return a randomly shuffled deck of playing cards?
 

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    <p>You have a list of 52 cards where the position of the cards in that list does not move. You have a second list of card positions. At first, the position list is the same as the first list.</p> <ol> <li><p>Iterate through the first list. </p></li> <li><p>For each card in the first list, generate a number from 1 to 52. Swap its position in the second list with the card in that position.</p></li> </ol> <p>Does a bias exist? Why?</p> <p>Update: Never one to believe pure math or logic, I decided to implement this myself. Here are the percent chance of the 5th card (position-wise) to be each number from 1 to 52:</p> <pre><code>1. 1.9346% 2. 1.9011% 3. 1.8513% 4. 1.8634% 5. 1.8561% 6. 1.8382% 7. 2.5086% 8. 2.4528% 9. 2.4552% 10. 2.3772% 11. 2.3658% 12. 2.3264% 13. 2.3375% 14. 2.287% 15. 2.2627% 16. 2.2151% 17. 2.1846% 18. 2.1776% 19. 2.1441% 20. 2.1103% 21. 2.084% 22. 2.0505% 23. 2.0441% 24. 2.0201% 25. 1.972% 26. 1.9568% 27. 1.9477% 28. 1.9429% 29. 1.9094% 30. 1.8714% 31. 1.8463% 32. 1.8253% 33. 1.8308% 34. 1.8005% 35. 1.7633% 36. 1.7634% 37. 1.769% 38. 1.7269% 39. 1.705% 40. 1.6858% 41. 1.6657% 42. 1.6491% 43. 1.6403% 44. 1.6189% 45. 1.6204% 46. 1.5953% 47. 1.5872% 48. 1.5632% 49. 1.5402% 50. 1.5347% 51. 1.5191% 52. 1.5011% </code></pre> <p>As you can see, quite un-random. I'd love a mathematician to prove why the 5th card is more likely to be a 7 than anything else, but I'm guessing it has to do with the fact that early cards, like 7, have more opportunities to swap -- which is exactly what the right algorithm prevents, it only lets cards swap once.</p>
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