Solution by Mark Thornquist:

Suppose the first card is red, giving me a
current score of 1. Should I stop now, and take a payoff of 1? I can
quickly see that the answer is 'no' by the following: Suppose I try to
get a score of 2. The probability that I get to 2 in the next draw is
slightly less than 1/2. The probability that I get to 2 for the first
time in three draws is the probability of drawing black, red, and red in
the next three cards, and that probability is (a little bigger than 1/2)
x 1/2 x (a little less than 1/2), or about 1/8. Thus, the probability
that I get to 2 sometime in just the next three draws is about 5/8.
Clearly there will be further future chances of reaching 2, so overall
my expected payoff if I try to reach 2 is at least 2 x 5/8 = 5/4, which
is better than the payoff if I stop now. A similar quick calculation
would indicate that if I reach 2 fairly early, my payoff is better if I
try to reach 3. On the other hand, late in the game, I should stop if
my score is positive. So the strategy of whether to continue playing or
stop must depend on both one's current score and the number of cards
remaining.
(1) I see no easy way to solve this other than by computer. The optimal
strategy is:
- 52-47 cards remaining in the deck: continue regardless of score
- 46-42 cards remaining: stop if your current score is 6
- 41-29 cards remaining: stop if your current score is 5
- 28-18 cards remaining: stop if your current score is 4
- 17-9 cards remaining: stop if your current score is 3
- 8-4 cards remaining: stop if your current score is 2
- 3-1 card(s) remaining: stop if your current score is 1.
With this strategy, I find the expected payoff to be 2.624476. I
checked it with 1,000,000 simulations and found an average payoff over
the simulations of 2.624330.
I was surprised that even if your first five cards are red, the optimal
strategy says to continue playing and, at least for the next five cards,
to try to get it up to 6.