1
$\begingroup$

Three dice are rolled. What is the probability of obtaining at least one 6 if it is known that all three dice showed different faces? The answer is 0.5. Could you give a hint?

$\endgroup$
4
  • $\begingroup$ I calculated the probability of the three dice showing different faces and got 5/9. $\endgroup$ – Mary Oct 22 '16 at 22:26
  • $\begingroup$ How many ways are there to choose three different values? How many of them contain a $6$? $\endgroup$ – lulu Oct 22 '16 at 22:27
  • $\begingroup$ Don't worry about the probability of the three dice showing different faces. You are given that that occurred. Just focus on the situation after given all three faces are different. $\endgroup$ – turkeyhundt Oct 22 '16 at 22:28
  • 3
    $\begingroup$ Well, this might not sould convincing but there are 3 numbers showing and 3 numbers not showing. Any outcome is equally likely. So six being one of the numbers showing is equally likely as six not being one of the numbers showing. $\endgroup$ – fleablood Oct 22 '16 at 22:33
1
$\begingroup$

To be unnecessarily thourough. There are $6*5*4=120$ ways for the faces to be different. (An arbitrary first die can have any face, the arbitrary second can have any of the five remaining, etc.)

There are $3(1*5*4)=60$ ways for one of the faces to be a $6$. (The face that is a 6, must be a six, the second can be any of 5 and the third any of 4, and there are 3 chooses for which die is $6$. So Probability is $60/120 = 1/2$.

2) The probability of the first die being a six is $1/6$. The probability of the second die being a six, and the first die not being a six, given the dies are different is $5/6*1/5 = 1/6$. So probability of one of the first two dice is six is $1/6 + 1/6 = 1/3$ The probability of the third die being six and neither of the first two, given that all there are different is $2/3*1/4 = 1/6$. So the probability of one the faces being six given they are all different is $1/3 + 1/6 = 1/2$.

3)All combinations of different numbers are equally likely. $3$ numbers appear. $3$ do not. Each is equally likely so that a six appears (or not) is 1/2.

$\endgroup$
2
$\begingroup$

To justify the answer formally you could use what you know about conditional probability. In particular:

Let $S$ be the event that at least one six occurs, and $D$ that all three dice show different faces. The outcome of our experiment is a triple $(a,b,c)$, where $a$, $b$, $c$ are the faces showed by the first, the second and the third dice respectively. There are $N = 6^3$ such triples. In addition, there are $6 \cdot 5 \cdot 4$ ways of choosing $(a,b,c)$ such that no face is repeated therein, so $P(D) = \frac {6 \cdot 5 \cdot 4}{N}$. Let us now consider the event $S \cap D$ (i.e. one dice shows six, and all of them have different outcomes); there are 3 ways to choose which dice shows six, and $5 \cdot 4$ to select the faces of the other two, in total: $3 \cdot 5 \cdot 4$ cases, which yields $P(S \cap D) = \frac{3 \cdot 5 \cdot 4}{N} > 0$. Finally, we obtain the probability in question: $$P(S|D) = \frac {P(S \cap D)}{P(D)} = \frac {3 \cdot 5 \cdot 4}{6 \cdot 5 \cdot 4} = 0.5.$$

$\endgroup$
2
  • 1
    $\begingroup$ I edited my post, providing more details. Is it OK? $\endgroup$ – Pythagoricus Oct 23 '16 at 0:34
  • $\begingroup$ Yes, it's OK now. I haven't checked whether it is correct, but it's definitely an answer now. $\endgroup$ – Daniel Fischer Oct 23 '16 at 9:54
1
$\begingroup$

Hint: if the three dice show different faces after being rolled, then surely three different numbers are obtained. What is the probability that these include the number $6$?

$\endgroup$
1
$\begingroup$

If all three dice are different, imagine the 6 numbers from 1 to 6. Three of them have been rolled and three have not. So each number has a 50% chance of being one of the three rolled since no number has an inherent advantage over the others.

$\endgroup$
2
  • $\begingroup$ But can I calculate it using a formula? $\endgroup$ – Mary Oct 22 '16 at 22:43
  • $\begingroup$ One way would be $\frac{\text{# of ways to pick 3 numbers with one being a 6}}{\text{# of ways to pick 3 numbers}}$ which would be $\frac{5\choose2}{6\choose3}$ = $\frac{1}{2}$ $\endgroup$ – turkeyhundt Oct 22 '16 at 22:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.