# What is the probability of rolling at least one $7$, $11$, or doubles in an experiment consisting of two rolls?

The experiment is rolling a pair of dice, two times. Rolling a $7$, $11$, or doubles is a success, rolling anything else is a failure.

What is the probability of succeeding on at least one of the two rolls.

I calculate the probability of succeeding in a one-roll experiment to be $\frac{14}{36}$:

There are six ways to roll a $7$, six ways to roll doubles, and two ways to roll an $11$, out of $36$ possible outcomes.

For two consecutive rolls, I think that the sample space of outcomes would be $36\cdot 36$.

How would you figure out how many outcomes would succeed in a two-roll experiment? What is the general way to figure out things like this?

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In answering this type of question, it is often useful to answer the opposite question!

"At least..." immediately suggests multiple routes, with multiple calculations, while "never" is more easily dealt with

So, there is a probability of 22/36 of not succeeding on the first roll of two dice, and again the same probability of not succeeding on the second roll. The probability of failing is thus (22/36)^2. The probability of succeeding is then (1 - 484/1296) or around 0.6265

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I agree that in a single roll we have: $P(\mathrm{success})=\frac{14}{36}$. Now, instead of counting outcomes of two rolls, we can calculate the probability of succeding on at least one of the two rolls (assuming independence of the rolls):

$$P(\text{at least one success in two rolls})=1-P(\text{no successes in two rolls})\\ =1-P(\text{no success in a single roll})^2=1-\left(\frac{22}{36}\right)^2.$$

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+1 for arithmetic accuracy :-) –  robjohn Jan 14 '13 at 7:49

The probability of failure each roll is $1-\mathrm{P}(7)-\mathrm{P}(11)-\mathrm{P}(\text{pair})=1-\frac{6}{36}-\frac{2}{36}-\frac{6}{36}=\frac{11}{18}$. The probability of failing twice is $\left(\frac{11}{18}\right)^2$. Thus the probability of succeeding at least once is $$1-\frac{121}{324}=\frac{203}{324}$$

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