Interesting puzzle. I'm getting some non-intuitive results...

Spoiler for

Let S

_{0}be the strategy where we do not choose any card (default choice is the 52nd card).There are 52! different possible ways that the deck can be arranged. Let this entire set of 52! card arrangement be denoted as D.

Let S be any strategy where we divide D into two sets: a set B where we do not choose any card (default choice is the 52nd card), and a set A where for each card arrangement we bet on some card before the 52nd card.

Let's consider set A. Suppose for some arrangement a

_{i}in A we bet on the c_{i}-th card, where c_{i}< 52. By symmetry, the chance of the c_{i}-th card being red is the same as the last card (52nd) card being red. Therefore for all arrangements in A we would have the same winning chance if we default to the 52nd card.For instance, suppose we opened 30 cards and they turned out to be 18 blacks and 12 red, and our strategy calls for betting on the 31 card. The 31st card has 14/22 chance of being red, but the 52nd card also has a 14/22 chance of being red. So you might as well as not choose the 31st card at all and let the game default to the 52nd card.

Consequently, any such strategy S would have the same winning chance as S

_{0}, which is 50%.