Probability of tie game in the insidious casino Consider the following game:
Cards are drawn by pairs. If both are red, they are placed in pile 1. If both are black, they are placed in pile 2. If it is one red/one black, they are placed in the discard pile. After all 26 pairs have been drawn, you win if more cards are in pile 1 than 2. I win if more cards are in pile 2 than 1. What is the probability of a tie?
I can envision a case wise solution to this problem using combinatorics for when there is one discarded pairs, two discarded pairs and so forth. But is there a cleaner, & more wieldy solution to this problem?

The probability of a tie is 1, since you discard only pairs of red and black. The number of black cards and the number of red cards in the red and black piles are both always $26-t$, where $t$ is the number of red and the number of black cards in the discarded pile.
