Probability of two sequential cards out of three drawn I wish to caluclate the probability that two out of three cards drawn from an ordinary 52 card deck are in sequence (for example a two and a three, or a ten and a jack, and so on). This excludes any situation with three cards in sequence but not situations that include a pair (so four - five - six would not count as a two-card sequence but four - four - five would). Ace is only high, so it can be used in the sequence king - ace but not in the sequence ace - two.
The way I reason is: There are 12 different possible positions for the two-card straight. For each position there are 4 * 4 = 16 different combinations of the two cards. For the two "end positions" there are 46 potential third cards (any card except the two in sequence, including those same cards in different suits, and the four adjacent ones). For the ten other positions there are 42 potential third cards (same but excluding eight adjacent ones). Thus, the probability should be:
$${2 \times 4^2 \times 46 + 10 \times 4^2 \times 42 \over {52\choose 3}} = 0.371$$
The problem is, when I ran a simulation I got a probability of .345, which leads me to believe that the above calculation is wrong. Is it? If so, what would be the correct way of doing it?
 A: Consider two consecutive numbers on the lower $(1, 2)$ and higher end $(K, A)$. In this case, combinations with one specific third value must be discarded ($3$ and $Q$, respectively). Assuming three different values, there are ${4 \choose 1}{4 \choose 1}{40 \choose 1} = 640$ valid combinations. Assuming two different values, there are ${2 \choose 1}{4 \choose 2}{4 \choose 1} = 48$ valid combinations.
Consider two consecutive numbers $(2, 3)$ up to $(Q, K)$. In this case, combinations with two values must be discarded (e.g., $1$ and $4$). Assuming three different values, there are ${4 \choose 1}{4 \choose 1}{36 \choose 1} = 576$ valid combinations. Assuming two different values, there are again ${2 \choose 1}{4 \choose 2}{4 \choose 1} = 48$ valid combinations.
Combining both together, we find:
$$\frac{2(640 + 48) + 10(576 + 48)}{52 \choose 3} = \frac{7616}{22100} \approx 0.3446$$
This is confirmed by the following Python code:
import itertools

v = list(range(1, 14)) * 4
v.sort()
j = 0
for c in itertools.combinations(v, 3):
  if c[0] == c[1] - 1 and c[1] != c[2] - 1:
    j += 1
  if c[0] != c[1] - 1 and c[1] == c[2] - 1:
    j += 1
print(j, j / 22100)

A: It can be done as $$  $$CASE 1  Selection of two in sequence and third any other denomination other than two drawn= 12*16*44=8448  $$ $$
CASE 2 IS ALL THREE ARE IN SEQUENCE = 11 * 64 = 704 $$ $$ Hence Required probability will be = (8448-704)/22100=7744/22100=0.35
