Probability a card can win a trick with a trump suit I'm working on the AI for a card game that uses a standard deck of 52 cards consisting of 13 cards in 4 (spades, clubs, diamonds, hearts) suits.  
Each player starts with 13 cards in their hand.  When a trick is started, a player must follow suit if they can, otherwise they can play any card.
I'd like to calculate the probability that a given card in my hand is a winner if I start a trick.
To do so, I'm using the following formula:
t = number of unknown cards in play
c = cards in the opponent's hand
n = number of "key" cards
$$
P_i = \frac{(t - c)! (t - n)!}{t! (t - c - n)!}
$$

If I want to play the king of clubs, only the ace of clubs can beat me, so the probability that the king can win is:

$$
P_1 = \frac{(39 - 13)!(39 - 1)!}{39!(39 - 13 - 1)!} = 0.667
$$
This result seems to make sense because the opponent only has 1/3 of the remaining deck.
Next, I'd like to determine the probability that the opponent does not have any clubs and can play a trump.  I calculate the probability that the opponent has any club.  Let's say I have four clubs in my hand and the opponent does not have the ace of clubs (five clubs total):
$$
P_2 = \frac{(39 - 13)!(39 - 8)!}{39!(39 - 13 - 8)!} = 0.025
$$
So, the opponent has about a 3% chance of not holding a club. Finally, I calculate the probability that my opponent is holding any trump.  Let's assume I'm holding four trump cards:
$$
P_3 = 1 - \frac{(39 - 13)!(39 - 9)}{39!(39 - 13 - 9)!} = 0.985
$$
I'd like to say:
$$
P(win) = P_1 + (P_2 * P_3)
$$
This would yield about a 69.1% chance to win.  
But, I don't believe these probabilities are independent because I've encountered some scenarios where the final result sometimes falls outside the range of [0, 1].
What's the correct way to account for the chance that an opponent does not have a particular suit and can play trump to win the trick so I can accurately calculate the chance a given card is a winner?
 A: For the sake of definiteness, let's say you want to play clubs; there are $k$ clubs you don't have above the one you want to play, $l$ clubs you don't have below it, and $m$ trumps you don't have.
It's the losing events, not the winning events, that are disjoint.
One of two losing events is that your opponent has a higher club, with probability
$$
1-\frac{\binom{39-k}{13}}{\binom{39}{13}}=1-\frac{(39-k)!26!}{(26-k)!39!}\;,
$$
as you correctly calculated. The second, disjoint losing event is that your opponent has no clubs but has a trump. To calculate the probability for this, we need the probabilities that she has no clubs and that she has neither clubs nor trump. The probability that she has no clubs is
$$
\frac{\binom{39-k-l}{13}}{\binom{39}{13}}=\frac{(39-k-l)!26!}{(26-k-l)!39!}\;.
$$
The probability that she has neither clubs nor trumps is
$$
\frac{\binom{39-k-l-m}{13}}{\binom{39}{13}}=\frac{(39-k-l-m)!26!}{(26-k-l-m)!39!}\;.
$$
The probability that she has no clubs but has trump is the difference between the two. Thus, the overall probability you're looking for is
$$
1-\frac{(39-k)!26!}{(26-k)!39!}+\frac{(39-k-l)!26!}{(26-k-l)!39!}-\frac{(39-k-l-m)!26!}{(26-k-l-m)!39!}\\
=1-\frac{26!}{39!}\left(\frac{(39-k)!}{(26-k)!}+\frac{(39-k-l)!}{(26-k-l)!}-\frac{(39-k-l-m)!}{(26-k-l-m)!}\right)\;.
$$
