Dice probability puzzle

Question

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?

Answer 1

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

$$p= \frac{5\cdot6 + 5\cdot5 + 5\cdot6 + 5 +5 +1}{6^5} = \frac{16}{6^4} = \frac{1}{3^4} $$

Answer 2

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.

Answer 3

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.