Which outcome will occur first [explanation] Suppose I have a 100 sided die:
24 sides have the letter A
3 sides have the letter B 
What is the percentage likelihood to roll at least 6 As before 1 B. 
I believe the answer is (24/27) ^6 = 0.49327
But this doesn't make sense to me, since the expected number of rolls to get 
6 As = 6/(.24) = 25
1 C = 1/(.03) = 33.33 
Shouldn’t the likelihood to roll 6 As first be much greater than 50%?  
 A: Your example would raise the same issues if you had a $9$-sided die with $8$ As and $1$ B
There is a difference between comparing the expectations of initial occurrence of the two events measured separately and a race between them, as in a race they can interfere with each other
As an illustration, based on Penney's game, suppose you toss a fair coin until you get a consecutive sequence of THT or THH. In a race using the same tosses they are equally likely to occur first (you need to get TH at some stage, and then the equally likely next result will finish the race), even though in non-race conditions the former has an expectation of $10$ tosses needed while the latter has an expectation of $8$ tosses needed
Contrast this with  tossing a fair coin until you get a consecutive sequence of THH or HHT. In a race using the same tosses the former is three times as likely to occur first as the latter (if you start with HH then the latter must win; otherwise the former must win), even though in non-race conditions they each have an expectation of $8$ tosses needed  
To make things worse, the interference can be non-transitive, with races where: 


*

*THH is more probable to happen first than HHT 

*HHT is more probable to happen first than HTT 

*HTT is more probable to happen first than TTH 

*TTH is more probable to happen first than THH   

A: I think your result is correct.  
However the variance of the number of rolls for getting a C is much greater than the variance of the number of rolls for getting 6 A's (variance for the 6 A's is $\frac{27}{32}$, while the variance for the C is $72$).  Thus the higher expectation for the C is due to occasional very long waits.
Another way to look at it is that on the average, for every eight A's there is one C, and so it is likely that among eight A's and a C, the C will come somewhere in the first $6$ positions. 
A: We don't care how many times we get any other results.   We only care about the rolls on which the result in either A or B.
So on any of the rolls which count, the probability of obtaining an $A$ is: $\mathsf P(A\mid A\cup B)= \dfrac{24}{24+3}$
Now the expected number of rolls which count until we get six A is: $\tfrac{9\times 6}{8}$ or $6.75$.   But clearly if it takes more than six such rolls to get six A we must have gotten some B first.
Likewise the expected number of rolls which count until we get one B is: $9$.   But again clearly only if it takes more than six such rolls to get that B will we have gotten at least six A first.
So a simple comparison of expected times until the first B and sixth A isn't going to give an estimate of probability.   The matter of interest is the probability of getting 6 A on the first six rolls which count.
The probability of obtaining at least 6 A before any B is : $\left(\dfrac{8}{9}\right)^{6}$, as you had found.
