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First I made a mistake of not taking into account that this event is dependent.

So to get two '$6$' the probability would be:

$$ P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3 $$

Where

  • $p = \dfrac16$
  • $q = \dfrac56$
  • $N= 5$ (the number of elements of the system)

Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:

$$ P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2 $$

But then the professor told me that $p=\dfrac15$ and not $p=\dfrac16$ That's what I don't get.

Of course at the end we multiply those two probabilities to get the final probability that we want.

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    $\begingroup$ The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s. $\endgroup$ Oct 12, 2018 at 4:50

2 Answers 2

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I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.

Method 1

You can solve this using the joint pmf for a multinomial distribution.

$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \frac{5!}{2!1!2!} \left( \frac{1}{6} \right)^2 \left( \frac{1}{6} \right) \left( \frac{4}{6} \right)^2 \approx .062$$


Method 2

Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.

Let $E$ be the event you get two sixes.

Let $F$ be the event you get one five.

Let $G$ be the event you get two of the others.

$$P(EFG) = P(E)P(F \mid E)P(G \mid EF) $$

$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3 $$

$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$

$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$

$$P(EFG) \approx .062$$

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  • $\begingroup$ The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces. $\endgroup$
    – Nuz
    Oct 12, 2018 at 5:14
  • $\begingroup$ In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space. $\endgroup$ Oct 12, 2018 at 5:38
  • $\begingroup$ Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something? $\endgroup$
    – Nuz
    Oct 12, 2018 at 7:11
  • $\begingroup$ Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F \mid G)P(E \mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting. $\endgroup$ Oct 12, 2018 at 7:36
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But then the professor told me that $p=1/5$ and not $p=1/6$ That's what I don't get.

You have correctly evaluated the probability for the event that two from the five dice show 6.

$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$

Now, of the three dice which don't show 6, each may show faces $\{1,2,3,4,5\}$ with equal probability.   Thus the conditional probability for a die showing 5, when given that it does not show 6, is $1/5$.   So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:

$$\mathsf P(N_5{=}1\mid N_6{=}2)=\binom 31\dfrac{1^14^2}{5^3}$$

Of course at the end we multiply those two probabilities to get the final probability that we want.

And the $5^3$ factors will cancel, so:

$$\mathsf P(N_6{=}2,N_5{=}1)=\binom{5}{2}\binom{3}{1}\dfrac{1^21^14^2}{6^5}$$

PS: This is called a multinomial distribution.

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