Probability and dice rolls 
Two fair six sided die are rolled fairly and the scores are noted. What is the probability that it takes 3 or more rolls to get a score of 7?

Here is what I have got so far:
The probability that it takes $3$ or more rolls to get a score of $7$ is equal to $$1-P(\text{you get a score of $7$ on less than $3$ rolls})$$
So $$P(\text{get a $7$ on zero rolls})=0\\P(\text{get a $7$ on $1$ roll})=\frac16\\P(\text{get a $7$ on the $2$ roll})=\frac16$$
So $$1-\left(0+\dfrac16+\dfrac16\right)=1-\dfrac26=\dfrac46$$
 A: #Pairs that sum 7 = 6
#total pairs       = 36
$P($getiing a 7 after the third roll or later$)$ = $P($getting it at the third roll$)$ and $P($getting it at the forth roll$)$ and  ... etc
getting that each probability is exclusive (that is you get the roll on the $n$th is because you could'nt get the roll between $1$ and $n-1$th) you get
$P($getiing a 7 after the third roll or later$)$ = $\sum_{i>2} P($ getting the roll on the $i$th$)$
$P($Getting the roll on the i-th$)$ =  $P($not getting the roll$)^{i-1}P($getting the roll$)$ = ${(\dfrac{5}{6})}^{i-1}\dfrac{1}{6}$
therefore 
$$P = \sum_{i=3}^{\infty}{(\dfrac{5}{6})}^{i-1}\dfrac{1}{6} = \sum_{i=0}^{\infty}{(\dfrac{5}{6})}^{i}\dfrac{1}{6} - \dfrac{1}{6} -\dfrac{5}{6}\dfrac{1}{6} = \dfrac{1}{6}\dfrac{1}{1-\dfrac{5}{6}}-\dfrac{1}{6}-\dfrac{5}{36}=1-\dfrac{11}{36}=\dfrac{25}{36}$$
A: One of the two dices has red eyes, the other one blue eyes. On a single throw of the two dices, the probability that the number of blue eyes is $7$ minus  the number of red eyes is ${1\over6}$. Therefore with probability ${5\over6}$ you don't make a total of $7$ in the first throw, and with probability ${25\over36}$ you don't make a total of $7$ in the first two throws. It follows that the answer to your problem is ${25\over36}$.
A: 1) If you mean 'to get an $overall$ score of 7, then the only way to get less than 7 (i.e. 6) in the first three rolls is to get 1 on each die every time you roll them, hence the probability is $1-\frac{1}{6^6}$.
2) If you mean 'to get a score of 7 on $each$ roll', then you need to find the probability that on the first three rolls the sum is strictly less than 7, i.e. 6 or less. You have 11 outcomes, 5 of them are less than 7. Can you handle from here? 
EDIT: this is for the case when you need to $toss$ exactly 7: there's a total of 36 outcomes in each toss. 6 of can get you 7 (why?). Hence 30 of the don't. What you need is the probability to not toss 7 on all three first tosses.  
A: The basic approach in the question is a good one,
except for an error in combining the 
probability that the first roll is a $7$ and the
probability that the second roll is a $7$.
In general, for two events $A$ and $B$,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$
If $A$ is the event of rolling $7$ on the first roll,
and $B$ is the event of rolling $7$ on the second roll,
then $P(A) = P(B) = \frac 16$ but $P(A \cup B) < \frac 13$
because $P(A \cap B) > 0$.
You need to find the probability that both the first and second rolls will be $7$,
and then you can apply the formula above.

A similar but slightly different approach is to let $A$ be the probability
that the first roll is a $7$
and let $B$ be the probability that the first roll is not $7$
but the second roll is a $7$.
You can confirm that any sequence of rolls that gets $7$ before the third roll
has either event $A$ or event $B$.
But now $A$ and $B$ are disjoint, that is, $P(A \cap B) = 0,$
so you can simply add the probabilities to get $P(A \cup B),$
and the solution is found by evaluating $1 - (P(A) + P(B))$.
This is similar to the attempted solution, except that $P(B) < \frac16.$
