If you roll a dice four times, what is the probability to get two consecutive threes?

Question

I was reviewing brainteasers and this one is stumping me for some reason.

Let X be a non-3.

you have 3 cases.

33XX, XX33, and X33X where we have the successes. I suppose you also have 333X and X333, and also 3333.

EDIT: We also have the possibilities of 3X33, and 33X3, which again do not matter much since it is satisfied by {${33XX, X33X, XX33}$}

The probability I arrived on was .0995, but that doesn't feel right, and it doesn't seem the internet knows either.

How I arrived on it was

$P(3) = \frac{1}{6}$, and $P(not3) = \frac{5}{6}$

$P(four-consecutive-threes) = \frac{1}{6^4}$

$P(two-consecutive-threes = 2*\frac{1}{6}*(1 + \frac{1}{6} + \frac{1}{6^2}) * 3$

This can be read as the possibility of 2 threes in a row, multiplied by the 3 continuing sequences (333X, X333, 3333)

Which yields $\frac{43}{432}$ or ~$.0995$

Where am I misstepping?

Answer 1

The possibilities 3X33 and X333 etc. do matter because they are not satisfied by XX33 because of how you defined X as being non-3. So if you keep the definition of X the same you will have to count these sequences sepatately. There is one sequence with 4 3's, 4 with 3, 3 with 2 and all these cases are distinct so you can just add them: $1+4*5+3*25=96$ out of $6^4$ so $2$ out of $27$.

Answer 2

Method One: Direct counting

Let's count the number of good dice rolls. Of course there are $6^4=1296$ dice rolls in total.

As the OP has done, let $X$ denote a non-$3$. Of course there are $5$ possible values for $X$.

Type A: $XX33$. There are $25$.

Type B: $X33X$ There are $25$.

Type C: $33XX$ There are $25$

Type D: $33X3$ There are $5$

Type E: $3X33$ There are $5$

Type F: $333X$ There are $5$

Type G: $X333$ There are $5$

Type H: $3333$ There is $1$.

We note that $$3\times 25+4\times 5+1=96$$

Making the final answer $$\frac {96}{1296}=\frac 2{27}=.\overline {074}$$

Method Two: recursion

Let $a_n$ be the number of strings of $n$ dice rolls without $33$.

Then $a_1=6$, $a_2=35$.

For $n$ at least $2$, a string of this type of length $n$ must either start with a non-$3$ and then be followed by a good string of length $n-1$ or it must begin $3X$ and then be followed by a good string of length $n-2$. We deduce that $$a_n=5(a_{n-1}+a_{n-2})$$

This quickly gives $a_4=1200$

So there are $6^4-a_4=1296-1200=96$ strings of the type you want which do contain $33$, and this confirms the count.

Answer 3

I found your calculation hard to follow.

It might be easier to consider the complement of the given event ($E^c$ - we don't get $2$ consecutive $3$s).

This can happen in the following ways.

1. We don't get any $3$s. The probability would be

$$\left( \frac{5}{6} \right)^4 = \frac{625}{1296}$$

2. We get exactly one $3$. That would have a probability of

$$4 \cdot \frac{1}{6} \cdot \left( \frac{5}{6} \right)^3 = \frac{500}{1296}$$

3. We get exactly two $3$s but they are not together.$^{[1]}$

$$3 \cdot \left( \frac{1}{6} \right)^2 \cdot \left( \frac{5}{6} \right)^2 = \frac{75}{1296}$$

So the required probability would be

$$\begin{align*} P(E) &= 1 - P(E^c) \\[0.3cm] &= 1 - \left( \frac{625}{1296} + \frac{500}{1296} + \frac{75}{1296}\right) \\[0.3cm] &= \frac{96}{1296} \end{align*}$$

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$^{[1]}$ There are three ways for this to happen - $\times \, 3 \times 3$, $3 \times 3 \, \times$, or $3 \times \, \times \, 3$

Answer 4

Chiming in, I would have organized my cases as follows... letting $X$ represent non-3's and $A$ represent any outcome

• The first occurrence of two threes in a row occurs in the first two positions: $33AA$

• The first occurrence of two threes in a row occurs in the second and third positions: $X33A$

• The first occurrence of two threes in a row occurs in the third and fourth positions: $AX33$

These have probabilities $\frac{1}{6}\times \frac{1}{6}\times 1\times 1$ for the first case, $\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}\times 1$ for the second case, and $1\times\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}$ for the third case.

This gives a combined probability of $\frac{1}{36}\left(1+\frac{5}{6}+\frac{5}{6}\right) = \frac{2}{27}$

By breaking into these cases, we get to group subcases together like X333 and X33X into the same category as X33A rather than having to treat them separately and without any ambiguity.

Answer 5

Alternative approach: Inclusion Exclusion

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

I will following the syntax in the second link, and I will express the probability as

$$\frac{N}{D} ~: ~D = 6^4.$$

Let $~S~$ denote the collection of all possible outcomes from rolling 4 dice. For $~k \in \{1,2,3\},~$ let $~S_k~$ denote the subset of $~S~$ that represents having a three on rolls $~k~$ and $~k+1,~$ without any regard for whether a three was rolled on either of the two other rolls.

Then,

$$N = |S_1 \cup S_2 \cup S_3|.$$

By Inclusion-Exclusion theory,

$$N = \{ ~|S_1| + |S_2| + |S_3| ~\}\\ - \{ ~|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3| ~\}\\ + \{ ~|S_1 \cap S_2 \cap S_3| ~\}$$

$$ = [3 \times 36] - [6 + 1 + 6] - [1] = 96.$$

Therefore,

$$\frac{N}{D} = \frac{96}{1296} = \frac{2}{27}.$$