Question 1089658
I am interpreting your question to ask what the probability is that the thief guesses the PIN precisely on the 3rd guess.

Since there are 10000 PINs (0000, 0001, …, 9999) and he/she presumably guessed two other unique PINs already, the probability of guessing correctly on the 3rd attempt is: 
 {{{ highlight(1/9998) }}} or about <b> 0.00010002</b>

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As an aside/related note:
If you wanted to know the probability that he/she guesses the PIN using <em>at most</em> 3 guesses, then that is
1 - P(no guesses are correct) = 1-P(first guess is incorrect)*P(2nd is incorrect)*P(3rd is incorrect) 
= 1 - (9999/10000)(9998/9999)(9997/9998) 
= 1 - (9997/10000)
= 3/10000 
= <b>0.0003</b>

Although I calculated it as pick1 followed by pick2, then pick3, notice how the result is equivalent to 3 picks out of 10000 (if there were 10000 slips of paper, each with a unique PIN written on it, you could take 3 out all at once), as you'd expect intuitively.

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