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We throw three dice and want to find the probability of getting one 6 on the condition that no two dice display the same number.

$A$ - getting 6

$B$ - getting a different number on every die $$P(A \cap B) = 1 \cdot 5 \cdot 4/ 6^3$$ since we want one 6 and something different on the other two $$P(B) = 6 \cdot 5 \cdot 4/ 6^3$$ since we want something different on every two dice $$P(A|B) = P(A \cap B) / P(B) = 1/6$$

Is this answer ok?

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Your calculation for the probability of obtaining a different number on each die is correct. However, your calculation for the probability of obtaining a six while obtaining a different number on each die is incorrect. You calculated the probability of obtaining a six on the first die and two other numbers on the other two dice. Since there are three dice on which the six could appear, you need to multiply your answer by three.

Say the dice are blue, green, and red. The probability of obtaining a six on the blue die, a number different from six on the green die, and a number different from both of those numbers on the red die is $$\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{4}{6}$$ By symmetry, the probability of obtaining a six on the green die, a number different from six on the blue die, and a number different from both of those numbers on the red die is the same, as is the probability of obtaining a six on the red die, a number different from six on the blue die, and a number different from both of those numbers on the green die. Hence, the probability of obtaining exactly one six while obtaining three different numbers is $$\Pr(A \cap B) = \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} + \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{4}{6} + \frac{5}{6} \cdot \frac{4}{6} \cdot \frac{1}{6} = 3 \cdot \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{4}{6}$$ Therefore, $$\Pr(A \mid B) = \frac{\Pr(A \cap B)}{\Pr(B)} = \frac{3 \cdot \dfrac{1}{6} \cdot \dfrac{5}{6} \cdot \dfrac{4}{6}}{\dfrac{6}{6} \cdot \dfrac{5}{6} \cdot \dfrac{4}{6}} = \frac{1}{2}$$

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At each [legal] throw you get 3 different outcomes out of 6 possible different dice throw outcomes.

Let split all six numbers into two tuples of three numbers, one tuple that we have got in a throw and another tuple, that we did not.

The probability, that $6$ made it to the right tuple, is exactly $\frac{1}{2}$. It either did, or did not [outcomes are symmetric: for each pair of tuples, where $6$ made it to the right tuple, there is exactly the same tuple with the roles reversed, where the $6$ did not made it].

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  • $\begingroup$ I am not sure I follow the "three different outcomes". Do you mean, perhaps, that the odds to get one six, given the entries are unique is (by symmetry $3pq$ where $p$ is the odds of getting a six in the first throw ($1/6$) and $q$ is the odds of getting a six given that you have unique entries and you rolled a six in the first entry ($1$)? $\endgroup$
    – Pedro ♦
    Commented Sep 28, 2023 at 13:54
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    $\begingroup$ @Pedro dEmigOd means that three different numbers are obtained. There are $\binom{6}{3}$ ways to obtain three distinct numbers. If six is one of those three numbers, then two of the other five numbers must be obtained, which can occur in $\binom{5}{2}$ ways. Hence, the probability of obtaining a six when three different numbers is obtained is $\frac{\binom{5}{2}}{\binom{6}{3}} = \frac{1}{2}$. $\endgroup$
    – N. F. Taussig
    Commented Sep 28, 2023 at 15:40
  • $\begingroup$ Ah, I see. That's neat. I was thinking about it from a different perspective. $\endgroup$
    – Pedro ♦
    Commented Sep 28, 2023 at 15:52

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