Optimal strategy in a simple card game

Question

So some friends and I are playing this simple card game. We choose 15 random cards from a normal deck of card and lay them on a table in a pyramid shape. 5 cards in the bottom row, 4 in the next row and so on.(5+4+3+2+1=15) Next the player is to choose one of the cards from the bottom row. If it is a 2,3,4...10 he proceeds to choose a card from the next row. The same rules applies here, and if he reaches the top row without drawing knight, queen, king or Ace once, he wins the game. If he in any row draws one of those cards, he has to start at the bottom again and all the cards he chose are discarded and replaced with cards from the shuffled deck. Now one of the players made the statement that if you loose the game once, on your second try you should pick a card that was NOT replaced. Since there would be about 4 or 5 knights, queens, kings or Aces on average in the original pyramid, this number has now decreased after you picked one of those cards. Hence the ratio of "loosing cards" is higher in the deck of cards, and you should avoid those cards that was newely replaced. According to the other players it does not matter which card you pick, the probability is all the same. So who is correct? Answers appreciated

Answer 1

The other players are right; it doesn't make a difference which cards you choose. The information you gain from turning over losing cards is that all remaining cards are less likely to be losing cards. This applies equally to all remaining cards, the ones already lying on the table and the ones still in the deck. They're all equivalent, and all permutations among them are equiprobable. The physical fact that you've placed some of them on the table and some not doesn't affect the information you have about them, which is the same.