Suppose we have a pair of dice and we roll both dice 2-times. My question is what is the probability that we will not get the same number on any of the rolls?
Let the particular number of our interest is 1. So, what is the probability that we will not roll 1 on either of my first two rolls?
For roll 1, there are a total of 36 possible outcomes such as {11,12,13,…,21,22,…,64,65,66}. So, the probability of getting 11 is 1/36. So, the probability of not getting 11 is 35/36.
For the second roll after the first roll, there are a total of 1296 outcomes (i.e., 36*36). And in only 72 cases among them, we have 11 on either the first or on the second roll (I'm not sure, this is just my 'intelligent' guess). So, the probability is 72/1296 = 1/18. So the probability of not getting 11 on either of the rolls is 17/18.
Am I correct? or is this answer 35/36 for any number of roll?
Best Answer
The probability of not getting $11$ on a given roll is $\frac{35}{36}$ so the probability of never getting $11$ on $n$ rolls is $(\frac{35}{36})^n$. For $n=2$, you get $.945.. \gt \frac{17}{18}$.