what will be the probability? Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
my approach was,the number of likely combinations are (1,6),(2,6),(3,6),(2,5),(1,5),(3,5),(2,4),(3,4),(3,3) and total combinations will be 3(for 1,2,3 in first roll)*6( in second roll).so probability should be 9/18.but answer was not accepted.please help me understanding the correct approach. 
 A: Alicia is playing the game. She wins if the sum total of values is at least $6$.
If the first toss results in a $4$ or a $5$, she won't get a chance to retoss, and she won't win.
Instant win: If she gets a $6$ on the first toss, then she has won quickly. This has probability $\dfrac{1}{6}$.
We now examine what happens if the first toss is a $1$, $2$, or $3$. Examine the cases one at a time. 
First toss is a 1: If at first she gets a $1$, (probability $\frac{1}{6}$), then she gets to retoss. To win, she needs a $5$ or a $6$ on the second toss.  The probability this sequence of events happens is $\dfrac{1}{6}\cdot \dfrac{2}{6}$.
First toss is a 2: In this case, to win she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is   $\dfrac{1}{6}\cdot \dfrac{3}{6}$.
First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is   $\dfrac{1}{6}\cdot \dfrac{4}{6}$.
Thus the probabilility Alicia wins is $\dfrac{1}{6}+ \dfrac{1}{6}\cdot \dfrac{2}{6}+ \dfrac{1}{6}\cdot \dfrac{3}{6}+ \dfrac{1}{6}\cdot \dfrac{4}{6}$. Simplify.
A: You forgot 4, 5, 6 rolls.  And, you need to take into account the fact that the odds of a 4, 5, or 6 is not the same as, say, (1, 5).  Getting a 4 happens 1/6th of the time.  Getting a (1, 5) happens 1/18th of the time.
A: This is the same as rolling the die twice (no matter what the first roll shows) and winning if the sum is $\ge 6$ and either the first roll is $\le 3$ or the sum $\ge6$ is already achieved with the first roll. Out of $36$ possible and equally likely outcomes, there are $9+6$ (don't forget $(6,1),\ldots,(6,6)$!) successfull outcomes, so the probability is $\frac{15}{36}=\frac5{12}$.
