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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?

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  • $\begingroup$ The question is a little confusing. In the example are you asking about 11 on either throw or on both? $\endgroup$
    – herb steinberg
    Nov 5, 2021 at 19:21
  • $\begingroup$ @herbsteinberg 11 means not eleven. It means the first die rolls 1 and the second die rolls 1. So, after the 2nd rolls, getting 1156 means I got 1 in the first die on the first roll, 1 in the second die on the first roll, 5 in the first die on the second roll and 6 on the second die on the second roll. I agree this is confusing and I should've explained this notation I used $\endgroup$
    – Abdul Muhaymin
    Nov 5, 2021 at 19:26

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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 $.945.. \gt \frac{17}{18}$.

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I'm not following the question. Are you asking the probability of not getting the same specific number (i.e. 1) or not rolling the same number (i.e. rolling any number twice). If it is a specific number like 1 then you are right that after one roll there is a 35/36 chance. If you are looking at any number then you have to multiple by 6 and get 5/6 chance. Personally, when I read "What is the probability of not getting the same number on either of the 2 rolls of a pair of dice?" I think that it is for any number not just a specific one like 1.

For the two rolls you are very close but you actually only have 71 cases since you are double counting the "1111" case. An easier way to do this is by noting that the two rolls are independent which means you can multiple the probabilities together. So either $(5/6)^2$ or $(35/36)^2$.

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  • $\begingroup$ Answer of your first question: I'm asking about the probability of not getting the same specific number (i.e. 1). I am trying to edit my question to be more specific. $\endgroup$
    – Abdul Muhaymin
    Nov 5, 2021 at 19:29
  • $\begingroup$ To be more specific, let's say I want to know that what is the probability that I won’t roll double sixes on any of my two rolls, or 100 rolls for that matter? (This should be the same for any number of roll) $\endgroup$
    – Abdul Muhaymin
    Nov 5, 2021 at 19:33
  • $\begingroup$ In that case, you are right that it is 35/36 and since the different rolls are all independent you can multiply them together and get $(35/36)^n$, where $n$ is the number of rolls. $\endgroup$
    – Pedro Amaral
    Nov 5, 2021 at 19:35

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