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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?

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    "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 05:47
  • Do you mean "Roll 10 fair dice", or "Roll a fair die 10 times"? – psmears Oct 24 '16 at 11:38

2 Answers2

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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 $6^{10}$.

msm
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    This was really helpful for visualizing the problem. –  Oct 24 '16 at 06:07
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    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
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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.

Siong Thye Goh
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