TITLE: Probability of a given name be picked
QUESTION [0 upvotes]: Had a probability discussion about the following context:
10 names were put in a box. 2 names will be picked up randomly from it.
What's the probability of a given specific name be picked?
My opinion: 1/10 + 1/9. 
Other opinion: 1/10 + 1/10. 
Small side-question: Does the probability is different when picking 2 names or picking one name and then another (with no reposition)? 
If the probability equal in the above methods, what's the difference between those methods and the https://en.wikipedia.org/wiki/Monty_Hall_problem (Monty Hall problem).
Thank you!

REPLY [1 votes]: Others have answered with the specific probability for the question you asked. For how is it different from the Monty Hall problem, the difference is that you are not gaining additional knowledge about the remaining names in the box. Let's alter your problem to make it more similar to the Monty Hall problem. Let's say you pick a name from the box. Without looking at it, Monty looks at all of the remaining names in secret, then reveals one of the slips of paper that does not contain the specific name. You can either keep your name or choose a new one from those remaining.
The slip of paper you chose has a $\dfrac{1}{10}$ chance of being correct while any of the other slips has a $\dfrac{1}{9}$ chance of being the correct name (because when you chose, you did not know about the incorrect slip, so there were 10 possible choices, but once Monty removed one, now there are only 9 possible choices--this includes the one you already picked--so any one you change to will have a greater probability of being correct).
Next, picking two names and checking if one of them is the correct name would have the same probability as picking one name, checking if it is the specific name, then picking a second name, and checking if it is a specific name. This is because the outcome winds up being the same. You have picked two names and the success condition is still one of them is the specific name.
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