Rolling 6 Dice Probability The question is 
If I roll 6 fair 6 sided dice, what is the probability that the max is equal to 5?
I have two attempted solutions and cannot determine which is correct and the flaw in the other.

Attempt 1:  The max is less than or equal 5 if there are no 6s rolled.  So probability of that is $P_5=\left(\frac{5}{6}\right)^6$ since each roll is allowed to be anything except 6.  The max is less than or equal to 4 if there are no 5s or 6s rolled which has probability $P_4=\left(\frac{4}{6}\right)^6$.  So the max equals 5 with probability $P_5-P_4=\left(\frac{5}{6}\right)^6-\left(\frac{4}{6}\right)^6\approx .25$

Attempt 2:  The max equals 5 if there are no 6s and at least one 5.  So each roll has only 5 choices.  Except one roll has one choice; it must be a 5.  So the probability that max equals 5 is $\left(\frac{5}{6}\right)^5\times \frac{1}{6}\times 6\approx .40$  The first part makes sure that 5 of them have no 6, the second piece makes sure that at least one is a 5, and the last piece accounts for the fact that there are 6 different places the 5 can occur.

In full disclosure, I have always been a little shoddy with combinatorics.  So I am pretty sure, the 2nd attempt is the incorrect one.  I would appreciate it though if your explanation used as much combinatorics lingo as possible to help me improve.  Thanks!
 A: Both of your attempts are wrong. We have to use something called conditional probabilty. As you said, there is a $(\frac{5}{6})^6$ probability that there is no $6$. But from here we have to look at the probability that there is a $5$ under the assumption that there is no $6$. This probability is given by $1-(\frac{4}{5})^6$. Together we have $(\frac{5}{6})^6\times (1-(\frac{4}{5})^6)=0.25$
This is exactly the result you got in your first attempt, but only by pure chance.
A: The second answer is incorrect indeed because of a combinatorical dillemma. You over count all the ways that there are multiple $5$'s. You could use this approach to get to the correct answer, but you must throw in some Inclusion Exclusion. 
Let's label the dice as dice $1,2,\dots 6$. Then we have the number of ways where a $5$ is the maximum number rolled to be: $$5^5(6) - 5^4\binom{6}{2}+5^3 \binom{6}{3}- 5^2 \binom{6}{4}+ 5\binom{6}{5} -1 = 11529$$
Now we find the probability: $11529 / 6^6 \approx .2471$
A: Indeed as you have guessed, the second attempt is wrong. The formula
$$\left(\frac{5}{6}\right)^5\times \frac{1}{6}\times 6$$
implies we are talking about $6$ distinct events, in each event we have chosen a position, where we are expecting to get the maximum $(5)$.
However, these aren't distinct events, since for example selecting the number $1$ position and selecting the number $5$ position both count this roll: $(5,1,3,4,5,2)$
Cases with $2$ rolls of $5$ are counted twice, cases with $3$ rolls of $5$ are counted $3$ times, etc. ... The roll of $(5,5,5,5,5,5)$ is counted a total of $6$ times in your formula.
If you'd like to count them up this way, we'd need to use the inclusion-exclusion principle:
$$\Bbb{P}(\text{The maximum roll is } 5) = \sum_{i=1}^6\Bbb{P}(\text{The maximum roll is } 5 \text{ and that appears in the } i\text{th place}) = \\ = \sum_{i=1}^6 \frac{1}{6}\left(\left(\frac{5}{6}\right)^{6-i}\left(\frac{5}{6}-\frac{1}{6}\right)^{i-1}\right) \approx 0.25.$$
