Given that no 1 turned up at the first two throws, what is the probability that at least three throws will be necessary for a 1 or 6 to appear? Question: A die is thrown as long as necessary for a 1 or a 6 to turn up. Given that no 1 turned up at the first two throws, what is the probability that at least three throws will be necessary?
Answer: $\frac{16}{25}$
what I did: 
Since the first 2, nothing turned up I assumed the probability was $\frac{4}{6}$ and for the 3rd try I assume it will show up so the probability is $\frac{2}{6}$ I then subtracted it to $1$ to get the probability since it says at least three throws but it yields a wrong value. Thank you for any help
$$1-\frac{4}{6}\frac{4}{6}\frac{2}{6}=\frac{23}{27}$$
 A: The reason your answer is wrong is that you are ignoring the part that "no ace turned up at the first two throws". So we know that in the first 2 throws there was no ace (I assumed "ace" mean 1). So for those 2 first throws the probability should only cover no 6 appearing so it will be 4/5. 
So the overall probability will be:
4/5 * 4/5 = 16/25
Note that since the question wants "at least" 3 throws any number of throws equal or bigger than 3 is good and since the probability never occurring a 1 or 6 in infinite throws is 0 so the probability is equal to 16/25. 
A: Rephrase the question:   A die is thrown until a 1 or 6 show.   What is the probability of this happening on the third or later throw, when given that a 1 will not show on the first two throws?
Ie: What is the probability something other than a 1 or 6 show on the first two throws when guaranteed that a 1 will not?
A: To solve this question let us first find what is the probability that we don't meet the condition in three throws.
Since we know there was no 1 in first two throws(you didn't consider this) the probability of us failing in two moves is 
 $$P(Failure)= \frac{4}{6-1}.\frac{4}{6-1} $$
Now, As we need at least three throws we just have to ensure that it fails for two throws.
Hence,
      the required probability is,$$P(n)=P(Failure)=\frac{4.4}{5.5}=\frac{16}{25}=0.64$$
