Azure / Carmine Balls Problem. This Problem is already asked but not answered properly...

i believe there must be a tricky short answer for it, but even though i couldn't solve it through induction or recursive formula...
Can someBody Help Me To find either short or long answer for it?
How do I solve this probability problem of randomly drawing balls from a urn?

it could be a miss Understanding about this problem: 
notice that the problem said "Restarted"
it means if after a run of Azure You pick a Carmine, if the next ball was an Azure it will be acceptable. i think the answer in the above link is wrong because of this miss understanding
Thanks...
 A: The probability is clearly $\frac12$ by symmetry if the number of azure and carmine balls is equal, and in particular if there are $2$ balls, one azure and one carmine  
Suppose as an inductive hypothesis that the probability that the last ball drawn is azure is $\frac{1}{2}$ if you start with $n-1$ or fewer balls not all the same colour  
Then if you start with $n=a+c$ balls, you might 


*

*draw all the carmine balls initially, so the last ball drawn will be azure, and the probability this happens is $\dfrac{1}{{n \choose c}}$ 

*draw all the azure balls initially, so the last ball drawn will be carmine,  and the probability this happens is $\dfrac{1}{{n \choose a}}$

*some other case, in which case you restart with fewer balls not all the same colour, and we know from the inductive hypothesis this gives a probability of $\frac12$, and the probability this happens is  $1-\dfrac{1}{{n \choose c}} -\dfrac{1}{{n \choose a}}$


So the probability that the last ball drawn is azure is $$1 \times \dfrac{1}{{n \choose c}} + 0 \times \dfrac{1}{{n \choose a}} + \dfrac{1}{2}  \times \left(1-\dfrac{1}{{n \choose c}} -\dfrac{1}{{n \choose a}}\right) = \dfrac12$$ since ${n \choose a}={n \choose c}$
This means that by strong induction the assertion is true    
