Here's the question: A casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer's pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties) you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game?
Here's the solution: This surely is an insidious casino. No matter how the cards are arranged, you and the dealer will always have the same number of cards in your piles. Why? Because each pair of discarded cards have one black card and one red card, so equal number of red and black cards are discarded. As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same. The dealer always wins! So we should not pay anything to play the game.
What I don't understand about the solution is this part: "As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same". I know that for each pair of discarded cards, one black and one red cards are discarded. But how is this statement enough to conclude that 'the number of red cards left for you and the number of black cards left for the dealer are always the same'? Isn't this completely ignoring obtaining a pair of black cards or a pair of red cards???