TITLE: Chances of being picked last in a hat
QUESTION [5 upvotes]: A couple of friends and I are struggling with coming up with an answer to this question.  It's seemingly simple but I need a little help. 

6 people put their name into a hat.  One by one, names are pulled
  without replacement.  What are the odds that your name will be picked
  last?

My thoughts are: in each pick you're looking at the odds that your name is not getting picked. So first it's $\frac{5}{6}$, then $\frac{4}{5}$, then $\frac{3}{4}$, $\frac{2}{3}$, $\frac{1}{2}$. And it's the chance you don't get picked first AND you don't get picked second, third ... so
$$\frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$$
Agree/Disagree?
Thank you in advance for your help.

REPLY [2 votes]: Here might be simplest way to understand the problem:
The list of people is shuffled randomly. What is the probability of your assignment to a given position? There are six positions, so 1 in 6.
The fact that the position in question just happens to be last fades from view.