What am I doing wrong in this probability exercise?

Question

In the experiment of throwing a die $n$ successive times, what's the probability of observing two consecutive 3's?

The consecutive threes can be observed in the first and the second throw, in the second and third throw, in the third and fourth throw, in the fourth and fifth throw, ..., and in the (n-1)-th and n-th throw. I would say these combinations account for $(n-1)*6^{n-2}$ favorable outcomes out of a total of $6^{n}$ different possibilities.

Then, I concluded the answer had to be something like $(n-1)*6^{n-2}$ over $6^{n}$.

Where does my argument go astray?

Thanks a bunch for your help

Answer 1

Suppose that thre are only $3$ dice. In how many cases can we have $2$ consecutive $3$'s? In the following cases (I shall use $*$ to denote something that is not a $3$):

• $33*$;
• $*33$;
• $333$.

Do you see the problem now? You missed the $333$ possibility.

Answer 2

Yes, the probability for the "EXACTLY two $3$s" case is correct: $$\underbrace{(n-1)}_{\text{positions of $33$}}\cdot\underbrace{\left(\frac{1}{6}\right)^2}_{\text{probability of $33$}}\cdot\underbrace{\left(\frac{5}{6}\right)^{n-2}}_{\text{probability of the rest}}=\frac{(n-1)5^{n-2}}{6^n}.$$ As regards the "AT LEAST two $3$s" case, note that in $\{3,*\}^n$ there are $\binom{n-k+1}{k}$ strings with $k$ $3$s where $33$ never occurs. Hence the probability is $$1-\frac{1}{6^n}\sum_{k=0}^{\lceil n/2\rceil}\binom{n-k+1}{k}5^{n-k}.$$ For example, for $n=3$, the above formula yields: $11/6^3$ (note that $33*\to 5$,$*33\to 5$,$333\to 1$, $5+5+1=11$).