Dice probability puzzle What is the probability of a run of at least 3 sixes when a die is thrown 5 times?
I think I have the answer but from what I have been told its not the correct answer. Would someone like to help?
 A: Presumably a run means consecutive.
The run might begin with the first toss. This has probability $(1/6)^3$.
The run might begin with the second toss. This has probability $(5/6)(1/6)^3$.
The run might begin with the third toss.  This also has probability $(5/6)(1/6)^3$.
Add these three numbers.
A: I simply count all the cases:
Possibilities for three consecutive $6$s but no four consecutive $6$s:
The sequence be of the one of these three forms:
$ 666XY  , X666X , YX666$ 
where the $X$'s can be chosen from $1,\dots ,5$ and the $Y$'s are also allowed to be $6$. So we get 
$5\cdot6 + 5\cdot5 + 5\cdot6$
possibilities for this case.
Possibilities for four consective digits:
$6666X$ and $X6666$
So those are $5 + 5 $ possibilites.
And $1$ case where all the dice are $6$s. 
So the probability is
$$p= \frac{5\cdot6 + 5\cdot5 + 5\cdot6 + 5 +5 +1}{6^5} = \frac{16}{6^4} = \frac{1}{3^4} $$
A: Forget about 6 and just look for the probability of at least three consecutives.
In the sequel s stands for 'toss has same result as preceding toss', d stands  for 'toss has different result as preceding toss' We start with the second toss. Possibilities are:
ss        with probability (1/6)x(1/6)
sdss      with probability (1/6)x(5/6)x(1/6)x(1/6)
dss       with probability (5/6)x(1/6)x(1/6)
ddss      with probability (5/6)x(5/6)x(1/6)x(1/6)
summation gives 2/27. If you do this for 6 (or any other) specifically you get (2/27)x(1/6)=1/81.
At this place you will find a formula for this. 
