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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.

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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.

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  • $\begingroup$ yes sir, i also figured it out in same way, but unfortunately not when needed. $\endgroup$ – Abhishek Pandey Aug 23 '13 at 8:30
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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.

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  • $\begingroup$ how you are saying there will be 21 possibilities. i am not getting please elaborate. $\endgroup$ – Abhishek Pandey Aug 17 '13 at 12:47
  • $\begingroup$ yes got it, i was not considering 4,5,6 and our likely case (6,0),answer should be 10/21.am i right? $\endgroup$ – Abhishek Pandey Aug 17 '13 at 13:03
  • $\begingroup$ @AbhishekPandey No, the 21 possibilities are not all equally likely. Half the time you roll a 4, 5, or 6 on the first roll. The other half the time, you roll a 1, 2, or 3. So, the 4, 5, 6 are just as likely as all 18 combinations starting with 1, 2, 3. Try making a tree diagram. The 4, 5, 6 won't have any branches extending from it. Those scenarios are each 1/6 probability. Then, (1, 3), for example, has probability 1/36 = 1/6 * 1/6, because the odds of rolling 1 on the first roll is 1/6 and the odds of rolling 3 on the second is 1/6. $\endgroup$ – GeoffDS Aug 17 '13 at 13:22
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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 $\frac{15}{36}=\frac5{12}$.

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  • $\begingroup$ but sir it is given in first throw only three faces of dies have appeared,1,2,3 then why you are considering 6,and why leaving other combinations like 4 and 5 in first throw. $\endgroup$ – Abhishek Pandey Aug 17 '13 at 12:52
  • $\begingroup$ dice is rolled second time only if in first roll we get 1,2,3.so how you are suggesting 36.please help $\endgroup$ – Abhishek Pandey Aug 17 '13 at 13:02

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