In the excercise below I have come up with two reasonings that lead to two different results... I will be grateful to anybody that can describe where I am wrong or where the problem lies.

A dice is rolled three times. What is the probability that the product of the three throws is greater than 80 given that at least two throws are equal to 6?

  1. Reasoning: Conditional Probability with Combinatorics

We use the definition of conditional probability with the two events $A, B$ where $$A \hat{=}\text{ Product is greater than 80},$$ $$B \hat{=}\text{ Two throws are 6},$$ i.e. we must calculate

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\cap B]}{\mathbb{P}[B]}.$$

Now with combinatorical arguments we can argue that $$\mathbb{P}[A\cap B] = \frac{10}{6^{3}}$$ and $$\mathbb{P}[B] = \frac{16}{6^{3}}.$$ This leads to $$\mathbb{P}[A|B] = \frac{10}{16} = \frac{5}{8} = 0.625$$

  1. Reasoning: Using Distinction of Cases

There are three possibilities when the "non-conditioned" number appears: The first, the second or the third throw.

We only consider the first case as the others are dealt with similarly. We need at most a $3$ in the first throw hence the probability that the first throw will be favorable for our result is $\frac{4}{6} = \frac{2}{3} \approx 0.66$. The other throws do not matter as they must be $6$ by our condition.

I would expect the probabilities to be equal. However, the two results differ by about $0.04166$ which is almost $4.2\%$. The first reasoning seems more plausible to me but the second reasoning was proposed in the sample solution.

Is there a flaw with one of these two reasonings? (Or is this just because we took two different methods to model the problem?)


I realized that the 2. reasoning is not well formulated for my question. I do another attempt:

We know that we have two $6$. The third throw must - in order for the product to be above 80 - be either a $3,4,5$ or a $6$. The probability for that to happen is $\frac{4}{6}$

  • $\begingroup$ Your wording of the problem is fatally ambiguous: "two throws are equal to 6" could mean "exactly two throws are equal to 6" or "at least two throws are equal to 6". The first interpretation seems more natural to me, but your calculations suggest that you assume the second interpretation. $\endgroup$
    – TonyK
    Jun 4, 2016 at 0:18
  • $\begingroup$ You might be right with your interpretation (in fact the original problem was in German - but there was also no specification in "exactly" or "at least"). However, in my "natural understanding" it would rather be the "at least" possibility. Anyway, I can assure you that the second interpretation is thought of. $\endgroup$
    – AndreasS
    Jun 4, 2016 at 0:48

3 Answers 3


Your first reasoning is completely correct. The second reasoning has a flaw in that it assumes that "the others are dealt with equally." Indeed, you are overcounting. We can correct the second approach if we continue to use conditional probability but couple this with complementary counting. Let me clarify.

We agree that there are $16$ ways to satisfy the condition - at least two die roll $6.$ Now, what would need to happen if we were unsuccessful? Consider just the first roll, as this setup will, in fact, be symmetric with respect to the other rolls. If we roll either a $1$ or a $2,$ we are doomed. Multiplying by $3$ in order to account for the other two rolls, this results in $6$ bad ways. This leaves us with the probability of success $P = 1 - \frac{6}{16} = 1 - \frac{3}{8} = \boxed{\frac{5}{8}}.$


What is the probability of rolling 3 sixes given that that 2 of the 3 dice are sixes? It is 1/16.

In secenario 2. You have a 3/16 chance of rolling any number other than a 6.

Your chance of getting a 3,4,5,6 = (3+3+3+1)/16 = 10/16

  • $\begingroup$ So you mean that in the second reasoning I got the wrong probabilities and did not take account of the different positions the "non-6" could be? Also I added a reformulation to the 2. reasoning. $\endgroup$
    – AndreasS
    Jun 4, 2016 at 0:12
  • $\begingroup$ It would helpful to explain where the 16 came from. It took me a bit to work that out. $\endgroup$
    – Readin
    Jun 4, 2016 at 4:59

Your first approach takes account of the fact that one of the remaining throws can be a six.   There are $9$ ways to roll two sixes and one other number selected from $\{3,4,5\}$, and $1$ way to roll three sixes.   This is correct.

Your second approach neglects this and counts $3$ ways to select two dice to be sixes multiplied by $4$ ways the remaining die can be one of $\{3,4,5,\mathbf 6\}$.   This over counts the outcomes.

  • $\begingroup$ The first remark makes me more confident that the first approach is the correct one. Thanks for that. However, I do not think I forgot the possibility to have a six. My probability is $\frac{4}{6}$ so I am considering the four cases in which the third number is greater than $2$. Maybe also look at my edit. $\endgroup$
    – AndreasS
    Jun 4, 2016 at 0:30
  • $\begingroup$ @AndreasS It's not that you haven't included the possibility, it's that you over counted how often it happens. You've counted the ways to select two sixes, $(6,6,x), (6,x,6), (x,6,6)$ then said the x can be replaced with any one of four values, including $(6,6,6), (6,6,6), (6,6,6)$ . $\endgroup$ Jun 4, 2016 at 0:34
  • $\begingroup$ Ah, yes. Now I understand what you mean. However, the second approach does not count those three ways but rather give the probability that the third dice (so only one) has a number in $\{3,4,5,6\}. $\endgroup$
    – AndreasS
    Jun 4, 2016 at 0:45
  • $\begingroup$ Which is the third die when two are six and the third is a six? $\endgroup$ Jun 4, 2016 at 2:01
  • $\begingroup$ You are given that two of the dice are 6. So assume you have already two 6. Then you look at the probability that the third one, which we can now consider to be only a single die, is either $3,4,5,6$, i.e. 4/6. So the third die "when two are six and the third is a six" is the last one. $\endgroup$
    – AndreasS
    Jun 4, 2016 at 8:00

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