# If I roll two fair dice, the probability that I would get at least one 6 would be…

$$11$$ 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?

• Yes.---------------- – Eric Auld Dec 8 '13 at 22:54
• Thank you ! Just wanted to double check :) – FiguringItOut Dec 8 '13 at 22:55
• 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. – NovaDenizen Dec 8 '13 at 22:56
• Yes. That $11/36$ is a number every backgammon player knows very well... – David C. Ullrich Aug 28 '18 at 23:28
• 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. – user247327 Sep 18 '19 at 11:40

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.

• So, then is rolling one or two in two tries 20/36 a little under 60%? – jwize Aug 9 '15 at 7:00

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!

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.

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.

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 $11/36$!!!

• You count the event $(6,6)$ twice! – user251257 Aug 11 '15 at 13:52