TITLE: Exam with $12$ yes/no questions (half yes, half no) and $8$ correct needed to pass, is it better to answer randomly or answer exactly 6 times yes?
QUESTION [58 upvotes]: In an exam with $12$ yes/no questions with $8$ correct needed to pass, is it better to answer randomly or answer exactly $6$ times yes and 6 times no, given that the answer 'yes' is correct for exactly $6$ questions?
I have calculated the probability of passing by guessing randomly and it is
$$\sum_{k=8}^{12} {{12}\choose{k}}0.5^k0.5^{n-k}=0.194$$
Now given that the answer 'yes' is right exactly $6$ times, is it better to guess 'yes' and 'no' $6$ times each? 
My idea is that it can be modelled by drawing balls without replacement. The balls we draw are the correct answers to the questions.
Looking at the first question, we still know that there are $6$ yes and no's that are correct. The chance that a yes is right is $\frac{6}{12}$ and the chance that a no is right is also $\frac{6}{12}$. 
Of course the probability in the next question depends on what the first right answer was. If yes was right, yes will be right with a probability of $5/11$ and a no is right with the chance $6/11$.  If no was right, the probabilities would change places.
Now that we have to make the choice $12$ times and make the distinction which one was right, we get $2^{12}$ paths total. We cannot know what the correct answers to the previous questions were. So we are drawing $12$ balls at once, but from what urn? It cannot contain $24$ balls with $12$ yes and $12$ no's. Is this model even correct?
Is there a more elegant way to approach that?
I am asking for hints, not solutions, as I'm feeling stuck. Thank you.

Edit: After giving @David K's answer more thought, I noticed that the question can be described by the hypergeometric distribution, which yields the desired result.

REPLY [11 votes]: A quick way to find the correct answer, which IMHO also gives insight into why this works better than 12 independent coin flips:

Let $X =$ no. of correct answers if you randomly guess exactly 6 Yes and 6 No.
As others have observed, $X \in \{0, 2, 4, 6, 8, 10, 12\}$.  
$P(X=6) = {{6 \choose 3}{6 \choose 3} \over {12 \choose 6}} = {400 \over 924} \approx 0.4329$.
By symmetry, $P(X \ge 8) = {1 - P(X=6) \over 2} = {262 \over 924} \approx 0.2835$.

IMHO, this is ultimately why this method works so much better than 12 independent coin flips.  If the passing grade were 7 correct answers, flipping 12 independent coins would have been better:

Let $Y =$ no. of correct answers if you flip 12 independent coins.
$P(Y = 6) = {12 \choose 6} ({1\over 2})^{12} \approx 0.2256.$
By symmetry, $P(Y \ge 7) = {1 - P(Y = 6) \over 2} \approx 0.3872.$
Meanwhile, of course $P(X \ge 7) = P(X \ge 8) \approx 0.2835$.

However, the passing grade is 8 correct answers, and you have invented a method which is guaranteed not to have 7 correct answers thereby eliminating the (useless-to-you) 7 correct answers situation.

Bonus: if the passing grade is 7 correct answers, then the correct strategy is to actually randomly answer 5 (or 7) of them as Yes!  

Let $Z =$ no. of correct answers if you randomly guess 5 Yes and 7 No.  (Math is same if you guess 7 Yes and 5 No.)
The only possible values are $Z \in \{1, 3, 5, 7, 9, 11\}$ 
By symmetry, $P(Z \ge 7) = {1 \over 2} > P(Y \ge 7) \approx 0.3872 > P(X \ge 7 ) \approx 0.2835$.

I bet this observation generalizes: i.e., if the passing grade is odd, guess 5 (or 7) as Yes, and if the passing grade is even, guess 6 as Yes.  What a way to game an exam!