This will appear simple for the Stat majors, but I ran a little simulation to evaluation the REDS chances to make the wild card, using the classic binomial distribution.

I'm assuming 85 wins will be required. (14 wins out of 19 games remaining).

If we consider the probability of winning any one game at 50%

P= 0.5

Prob (14 or more) = 3.18%
Prob (13 or more) = 8.35%

Factor in the weakness of schedule and put (its a stretch, I know) p at 55%

n = 19
p = 0.55

Prob (14 or more) = 7.78%
Prob (13 or more) = 17.3%

Not good odds, but hey...we've seen it worse.