# Four dice are thrown simultaneously

Four dice are thrown simultaneously. The probability that $$4$$ and $$3$$ appear on two of the dice given that $$5$$ and $$6$$ appear on the other two dice is:

a) $$1/6$$

b) $$1/36$$

c) $$12/51$$

d) None of these

Since the events are independent, I feel the probability is $$1/6 \times 1/6 = 1/36$$

• Conditional probability, namely, $P(x) = P($each dice has $3, 4, 5$ and $6$ respectively $)/P(5$ and $6$ appear on $2$ dices$)$, where x is our event we want to find probability for. – Makina Nov 16 '18 at 8:26
• Given the current list of choices, the correct answer is d). – kludg Nov 16 '18 at 9:24
• Die is singular; dice is plural; dices is the third person singular form of the verb to dice, meaning to cut into small cubes. – N. F. Taussig Nov 16 '18 at 10:39

Even if there are only two dice, the probability of observing $$4$$ and $$3$$ is not $$1/6 \times 1/6$$, but rather, $$1/18$$, since the sample space is the set of ordered pairs $$(a,b)$$ where each $$a, b \in \{1, 2, 3, 4, 5, 6\}$$. Thus there are two desired outcomes $$(4,3)$$, $$(3,4)$$ out of $$6^2 = 36$$ possible outcomes.

When there are four dice, two of which you are told are $$5$$ and $$6$$, you must reason carefully and precisely. Given the set of all outcomes of four dice rolls, select those that show at least one $$5$$ and at least one $$6$$. Of these, how many show $$3$$ and $$4$$ on the other two dice?

Doing mathematics is not about "feelings." It is about showing and justifying your reasoning. Describing your calculation without providing a sound basis for why you are doing what you are doing, is not math.

• But isn't it given that two events - 3 and 4 have surely happened? Then why to worry about that? – Archer Nov 16 '18 at 8:10
• I now think that the answer should be 1/18 – Archer Nov 16 '18 at 8:10
• Why don't you actually try the calculation in the correct way, instead of just claiming that the answer "should" be something that isn't even one of the answer choices? – heropup Nov 16 '18 at 8:24

The tricky part of the solution is to find the number of outcomes such that 2 dice land $$5$$ and $$6$$; it can be done using inclusions/exclusions.

Let $$\Omega$$ be the set of all outcomes, of size $$6^4$$. Let $$S_5$$ be the set of outcomes that contain no $$5$$'s, of size $$5^4$$. Let $$S_6$$ be the set of outcomes that contain no $$6$$'s, of size $$5^4$$. Let $$S_{5,6}$$ be the set of outcomes that contain no $$5$$'s no $$6$$'s, of size $$4^4$$.

Using inclusion/exclusion principle the number of outcomes that contain at least one $$5$$ and at least one $$6$$ is $$6^4-5^4-5^4+4^4=302$$ The rest is simple, and the answer is $$12/151$$