Katrine has a bag containing 4 buttons with distinct letters M, P, F, G on them (one letter per button).  She picks buttons randomly, one at a time, without replacement, until she picks the button with letter G.  What is the probability that she has at least three picks and her third pick is the button with letter M? 

To solve this problem, we need to determine the probability that Katrine picks the button with the letter M on her third pick, given that she picks the button with the letter G on her fourth pick. We will break down the problem step-by-step.

1. **Identify the total number of ways to arrange the buttons:**
   Since Katrine picks buttons without replacement, the total number of ways to arrange the 4 buttons is given by the factorial of 4:
   \[
   4! = 24
   \]

2. **Determine the specific arrangement we are interested in:**
   We are interested in the arrangement where the third pick is M and the fourth pick is G. This can be represented as:
   \[
   \_ \_ M G
   \]
   The first two positions can be filled by the remaining two letters (P and F) in any order. There are \(2!\) ways to arrange these two letters:
   \[
   2! = 2
   \]
   Therefore, there are 2 favorable outcomes: PFMG and FPMG.

3. **Calculate the probability of the favorable arrangement:**
   The probability of the favorable arrangement is the number of favorable outcomes divided by the total number of possible outcomes:
   \[
   \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{24} = \frac{1}{12}
   \]

4. **Verify the probability of having 4 picks:**
   Since Katrine stops picking once she picks the button with the letter G, the only way she can have exactly 4 picks is if G is the last button picked. The probability of G being the last button picked is:
   \[
   \frac{1}{4}
   \]

5. **Combine the probabilities:**
   The probability that Katrine has at least three picks and her third pick is the button with the letter M, given that she has 4 picks, is:
   \[
   \frac{1}{12} \cdot \frac{1}{4} = \frac{1}{12}
   \]

The final answer is \(\boxed{\frac{1}{12}}\).