If there are \( N \) cards of each color then there are \( \binom {2N} {N} \) possible orders for the colors of the cards. If one were to bet 100% on each card it is possible to win \( $2^{2N} \), but only if you guess the exact order, any mistake leaves $0. So one strategy is to assign \( \frac{1}{ \binom {2N} {N} } \) of your original dollar to each of the potential card color orders, then use the 'bet it all' strategy separately and simultaneously with each equal part of the dollar. All except one are guaranteed to lose, but the one that wins will have \( \frac{2^{2N}}{ \binom {2N} {N }} \).

This yields 8/3, 16/5, 128/35 for N=2, 3, 4, and continues to grow unbounded with N.

We're not allowed to make multiple bets in a round, but we can execute the strategy 'virtually.' If, on any round, there are \( p \) red and \(q+p\) black cards (WLOG assume \( q \geq 0 \) ), each simulation that is still alive will have the same amount of money, and we are guaranteed that at least p will win and p will lose. So we can get the same net affect by betting \( \frac{q}{q+2p} \) of our total money on black. For example, in the first round we bet zero, and in round 2 we bet \( \frac{1}{2N-1} \) on the opposite of the first color.