Throw a dice three times, conditional probability that the product will be greater than 80 given that 6 is thrown twice.

Question

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 $$\mathbb{P}[A|B] = \frac{10}{16} = \frac{5}{8} = 0.625$$

2. 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.

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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?)

Edit

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 $\frac{4}{6}$

Answer 1

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 $P = 1 - \frac{6}{16} = 1 - \frac{3}{8} = \boxed{\frac{5}{8}}.$

Answer 2

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

Answer 3

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.