Rolling a die six times in a row

How many ways are there of rolling a die six times in a row such that:

(a) there are exactly four consecutive $$3$$'s somewhere?
(b) there are at least four consecutive $$3$$'s somewhere? (my approach: take complement)
(c) there are more $$3$$'s than $$4$$'s?

Part (a) attempt: should I consider the four $$3$$'s as one block? This would mean that the possibilities would be $$R_1 R_2 R_3 R_4, R_2 R_3 R_4 R_5, R_3 R_4 R_5 R_6,$$ This would mean that there are three ways to roll four consecutive 3's.

Part (b) attempt:

Case 1: four consecutive $$3$$'s: From part (a) there are $$3$$ ways

Case 2: five consecutive $$3$$'s: $$R_1 R_2 R_3 R_4 R_5, R_2 R_3 R_4 R_5 R_6$$ $$\rightarrow$$ So there are $$2$$ ways

Case 3: six consecutive $$3$$'s: There is only one way to do this since we are rolling the die six times in a row.

Adding all these cases: $$3+2+1=6$$ ways.

• In counting the number of ways of obtaining exactly four consecutive threes, keep in mind that each of the remaining positions in the sequence can be filled in five ways (one of the other possible outcomes of a die roll). For at least four consecutive threes, I suggest using the Inclusion-Exclusion Principle. – N. F. Taussig Oct 15 '18 at 21:13

How many ways are there of rolling a die six times in a row such that there are exactly four consecutive threes?

We need to consider cases, depending on whether the string of consecutive threes begins with the first roll, second roll, or third roll. Let $$R_i$$, $$1 \leq i \leq 3$$, be the event that the string of exactly four consecutive threes begins with roll $$i$$.

$$|R_1|$$: There must be a three in each of the first four positions. The fifth position must be filled with a number other than $$3$$, which can occur in five ways. The sixth position can be filled by any of the six possible outcomes of a die roll. Hence, there are $$5 \cdot 6$$ ways to have exactly four consecutive threes in six rolls of a die if the string of four consecutive threes begins in the first position.

$$|R_2|$$: There must be a three in each of the four middle positions. Thus, the first and last rolls must be filled by a number other than $$3$$. Thus, there are $$5^2$$ sequences of six die rolls in which there are exactly four consecutive threes beginning in the second position.

$$|R_3|$$: By symmetry, $$|R_3| = |R_1|$$ since the four consecutive threes must occupy the last four positions, which means the second position cannot be a $$3$$ and the first position can be any of the possible outcomes for a die roll.

Hence, the number of ways of rolling a die six times so that there are exactly four consecutive threes is $$|R_1| + |R_2| + |R_3| = 5 \cdot 6 + 5^2 + 5 \cdot 6 = 30 + 25 + 30 = 85$$

How many ways are there of rolling a die six times in a row such that there are at least four consecutive threes?

Let $$R_i$$ be the event that there are four consecutive threes beginning in the $$i$$th position, where $$1 \leq i \leq 3$$. Notice that the word exactly was not used in the description of $$R_i$$.

Observe that to have five consecutive threes, there are two possibilities:

1. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position.
2. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the third position.

To have six consecutive threes, there must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position and four consecutive threes beginning in the third position.

Thus, the set of events in which at least four consecutive threes are rolled is $$R_1 \cup R_2 \cup R_3$$. By the Inclusion-Exclusion Principle, $$|R_1 \cup R_2 \cup R_3| = |R_1| + |R_2| + |R_3| - |R_1 \cap R_2| - |R_1 \cap R_3| - |R_2 \cap R_3| + |R_1 \cap R_2 \cap R_3|$$

$$|R_1|$$: Each of the first four positions must be filled with a $$3$$. The remaining two positions can each be filled in six ways. Hence, $$|R_1| = 6^2$$. By symmetry, $$|R_1| = |R_2| = |R_3|$$.

$$|R_1 \cap R_2|$$: Each of the first five positions must be filled with a $$3$$. The remaining position can be filled in six ways. Hence, $$|R_1 \cap R_2| = 6$$. By symmetry, $$|R_1 \cap R_2| = |R_2 \cap R_3|$$.

$$|R_1 \cap R_3|$$: Each of the six positions must be filled with a $$3$$. Hence, $$|R_1 \cap R_3| = 1$$.

$$|R_1 \cap R_2 \cap R_3|$$: Each of the six positions must be filled with a $$3$$. Hence, $$|R_1 \cap R_2 \cap R_3| = 1$$.

Therefore, by the Inclusion-Exclusion Principle, the number of ways of rolling a die six times and obtaining at least four consecutive threes is $$|R_1 \cup R_2 \cup R_3| = 6^2 + 6^2 + 6^2 - 6 - 1 - 6 + 1 = 96$$

How many ways are there of rolling a die six times in a row such that there are more threes than fours?

By symmetry, there are an equal number of outcomes with more threes than fours as there are with more fours than threes. Thus, the number of outcomes in which there are more threes than fours can be found by subtracting the number of outcomes in which the number of threes and the number of fours are equal from the total number of outcomes, then dividing the result by two.

• I obtained the same answers when doing this question. – Mathaholic24 Oct 16 '18 at 3:54

protected by Community♦Oct 15 '18 at 21:42

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