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$11$ out of $36$? I got this by writing down the number of possible outcomes ($36$) and then counting how many of the pairs had a $6$ in them: $(1,6)$, $(2,6)$, $(3,6)$, $(4,6)$, $(5,6)$, $(6,6)$, $(6,5)$, $(6,4)$, $(6,3)$, $(6,2)$, $(6,1)$. Is this correct?

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  • $\begingroup$ Yes.---------------- $\endgroup$
    – Eric Auld
    Commented Dec 8, 2013 at 22:54
  • $\begingroup$ Thank you ! Just wanted to double check :) $\endgroup$ Commented Dec 8, 2013 at 22:55
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    $\begingroup$ It's easier to calculate the probability that you roll no 6's, which is ${(\dfrac{5}{6}})^2 = 25/36$, leaving $11/36$ probability that you roll at least one. $\endgroup$ Commented Dec 8, 2013 at 22:56
  • $\begingroup$ Yes. That $11/36$ is a number every backgammon player knows very well... $\endgroup$ Commented Aug 28, 2018 at 23:28
  • $\begingroup$ Or argue that the probability of "6, non 6" is (1/6)(5/6)= 5/36, the probability of "non 6, 6" is (5/6)(1/6)= 5/36, and the probability of two 6's is (1/6)(1/6)= 1/36 so the probability of "at least one 6" is 5/36+ 5/36+ 1/36= 11/36. $\endgroup$
    – user247327
    Commented Sep 18, 2019 at 11:40

5 Answers 5

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That's right. The easier approach would be to calculate the chance of not rolling a $6$ - that's just $\frac56$ for the first die, and $\frac56$ for the second die, so by the product rule (as the events are independent), the probability is $\frac56 \cdot \frac56 = \frac{25}{36}$.

Then the probability of rolling a $6$ is $1$ minus the probability of not rolling a $6$, which we just calculated: so it is $1-\frac{25}{36}=\frac{11}{36}$.

If calculating the probability of an event not occurring to calculate the probability of it occurring feels weird to you, you might want to read up on complementary events. The idea here is that the product rule can sometimes make probabilities smaller when that wouldn't make any sense --- in our example, if you had just multiplied $\frac16 \cdot \frac16 = \frac1{36}$, then that would've been obviously wrong. This is because when we work with probability, we're dealing with quantities in the $[0,1]$ interval, so multiplication usually makes things smaller, rather than larger.

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  • $\begingroup$ So, then is rolling one or two in two tries 20/36 a little under 60%? $\endgroup$
    – jwize
    Commented Aug 9, 2015 at 7:00
  • $\begingroup$ How would you expand this to work out, say, the chance of getting 3 dice of the same side in 10 dice? So for example, 3 6's in 10 dice. $\endgroup$
    – alexr89
    Commented Feb 24, 2023 at 11:10
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Yes indeed, you've got them all. So counting them, we get $11$ of the possible $36$ outcomes of which at least one $6$ is rolled. Now simply express this probability as a fraction!

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http://www.wolframalpha.com/input/?i=chance+of+throwing+1+6%27s+with+2+dice

X = # of occurrences of 6 with 2 dice

P(X=1) = P(the first dice has 6, the second hasn't) + P(the second dice has 6, the first one hasn't) = 1/6 * 5/6 + 5/6 * 1/6 = 5/ 36 + 5/36 = 5/18 = 0.2778.

Which is also what Wolfram Alpha computes.

Then P(X=2) = 1/6 * 1/6 = 1/36.

5/18 + 1/36 = 11/36, which is indeed the answer.

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The probability of rolling a certain number, can be easily calculated by listing the options available, or listing the probability of not rolling that specific number and subtracting the answer from 1. The probability of a certain event occurring is always a number between 0 and 1.

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The chance of getting a $6$ with the first dice is $1/6$ and the chance with the second dice is $1/6$ therfore the chance of getting a $6$ with either dice is $1/6 + 1/6 = 1/3$....not $11/36$!!!

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    $\begingroup$ You count the event $(6,6)$ twice! $\endgroup$
    – user251257
    Commented Aug 11, 2015 at 13:52

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