9**1

8**2

7**3

6**4

As possible pin-numbers combining (1) and (2)

Combining these with (3), (4): Since the first and last number add up to 10. Because of (3) the second and third number must add up to 6.

If we combine that with (4) we find possible numbers:

9331

8602

8242

7513

7153

6424

6064

Then the probability that 2 digits are the same is: $\displaystyle \frac{4}{7}$

The possibility that it's the incorrect number is: $\displaystyle \frac{6}{7}$

Just like Soroban said.

The last question has ofcourse nothing to do with how the numbers look like: