If a deck has n cards, r red and b black, the best strategy wins with a probability of r/n. Thus, you can say "red" on the first card, the last card, or any other card you wish. Proof by induction on n. The statement is clearly true for one-card decks. Suppose it is true for n-card decks, and add a red card. I will even allow a nondeterministic strategy, meaning you say "red" on the first card with probability p. With probability 1-p, you watch the first card go by, and then apply the "optimal" strategy to the remaining n-card deck, since you now know its composition. The odds of winning are therefore: p * (r+1)/(n+1) +

(1-p) * ((r+1)/(n+1) * r/n + b/(n+1) * (r+1)/n).

After some algebra, this becomes (r+1)/(n+1) as expected. Adding a black card yields: p * r/(n+1) +

(1-p) * (r/(n+1) * (r-1)/n + (b+1)/(n+1) * r/n).

This becomes r/(n+1) as expected.

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