Probability of dice thrown

What is the probability of getting

-atleast 1 six when 6 dice are rolled?
- atleast 2 six when 12 dice are rolled? - atleast 3 six when 18 dice are rolled?

"At least 1 six" is $1-(5/6)^6$. How can I do the next 2?

• @MTurgeon: How do you know? Did you assign this exercise? – dfeuer Aug 28 '13 at 22:58
• linkIn the book, its atleast one 6. – Jaswanth Aug 28 '13 at 23:10
• @dfeuer One of the earlier edits wasn't referring to "at least" – M Turgeon Aug 29 '13 at 0:08
• link @Turgeon, so will it stay the same for atleast or will there be a change from the present – Jaswanth Aug 29 '13 at 0:19

At least two 6 when 12 dice are rolled is

1 - (probability that no 6 was rolled + probability that one 6 was rolled).

Using this, can you do the last one?

Edit: The probability of no 6 is $(5/6)^{12}$ (there are 12 dice). The probability of (exactly) one 6 is $12\times(1/6)\times(5/6)^{11}$: you have one die which is fixed at 6, all the other ones can be anything but 6, and you have 12 choices for which die has 6.

• [link]Thanks@Turgeon, Is it prob atleast one 6 or one 6, so will it become 1-[(5/6)^6+(1/6)^6]?? ANd the last, I just cant get it – Jaswanth Aug 28 '13 at 22:15
• @Jaswanth Your solution is not correct; I added the probabilities. Try to understand this one computation, and if you have questions please ask. But doing the last problem should be very similar. – M Turgeon Aug 28 '13 at 22:30
• The edit is not correct for the chance of exactly one. You have 12 choices which die is the 6, so the chance should be multiplied by 12. – Ross Millikan Aug 28 '13 at 22:31
• @RossMillikan Right, thanks. – M Turgeon Aug 28 '13 at 22:33
• link So for the last one, it will be like the present answer-(no 6 was rolled+one six was rolled) ?? – Jaswanth Aug 28 '13 at 23:06

The chance of no sixes in 12 dice, extending your result is $\left( \frac 56 \right)^{12}$. The chance of exactly one six is $12 \cdot \frac 16 \cdot \left( \frac 56 \right)^{11}$, where the $12$ is the number of ways to choose which die will be the $6$, the $\frac 16$ is the chance it is a $6$, the the $\frac 56$ is the chance that all the rest are non-$6$'s

• [link]but its two 6 right, so atleast 2 6 when 12 dice are rolled, – Jaswanth Aug 28 '13 at 22:43
• I thought so, too. If you have the probabilities of zero and one, subtract from $1$ to get the probability of at least two. – Ross Millikan Aug 28 '13 at 23:11
• link@ Ross Is chance of no 6 is (5/6)^12 or 1-[(5/6)^12] – Jaswanth Aug 28 '13 at 23:19
• Yes, a typo. Fixed. Thanks. – Ross Millikan Aug 28 '13 at 23:38