# Roll 10 fair die. What is the probability that the number 1 appears exactly four times, on four consecutive rolls

I am asked:

Roll $10$ fair die. What is the probability that the number $1$ appears exactly four times, on four consecutive rolls?

The answer I was given is $$\frac{1}{6^{10}}\cdot7\cdot5^6$$ with a comment saying we multiply by 7 "for the first roll".

However, this doesn't make sense to me since there are only $6$ possible sides to the die. So it's not counting sides of the die.

The only reason I can come up with for the 7 appearing is the case when the first $6$ rolls do not produce a sequence of consecutive $1$'s, so the $7$th item must begin the sequence of consecutive $1$'s

So where exactly is this $7$ coming from?

• "for the first roll" refers to the position (out of the ten) of the first roll which is a 1. Only 7 positions for the first 1 are possible: position one through seven. – Ian Miller Oct 24 '16 at 5:47
• Do you mean "Roll 10 fair dice", or "Roll a fair die 10 times"? – psmears Oct 24 '16 at 11:38

## 2 Answers

For a better understanding, the seven alternative cases are:

 [1 1 1 1] * * * * * *
* [1 1 1 1] * * * * *
* * [1 1 1 1] * * * *
* * * [1 1 1 1] * * *
* * * * [1 1 1 1] * *
* * * * * [1 1 1 1] *
* * * * * * [1 1 1 1]


The six stars are $2,3,4,5,6$ (five categories) and can take $5^6$ combinations. Hence the numerator is $(7)5^6$. The total number of combinations is $6^{10}$.

• This was really helpful for visualizing the problem. – user225477 Oct 24 '16 at 6:07
• So you could say that the four consecutive "1" are one single contagious unit and together with the 6 other free spaces that makes 7 permutable objects in total? – QBrute Oct 24 '16 at 11:29
• Yes @QBrute, you can think of the sequence as a "unique" element among $6$ other elements. – msm Oct 24 '16 at 11:34

Consider a shifting window where you record the beginning position of consecutive $1$.

It can start from position $1$ to $7$.

Starting from position $1$ means roll $1$ to roll $4$ are $1$'s, the rest are not.

Starting from position $7$ means roll $7$ to roll $10$ are $1$'s, the rest are not.