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What is the probability that if we roll one six-sided die for n times - whatever the number we see in the first roll, no other following roll has a number higher than that (that is all the numbers we see in the following rolls are less than or equal to the first number)?

Example:

We can consider some sample cases to better explain the question. We roll ONE 6-sided die 5 times

1 1 1 1 1 -> Success

5 4 3 2 1 -> Success

4 4 3 2 4 -> Success

5 4 3 2 6 -> Fail

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If the first number is $1$, then the probability is $\left(\frac{1}{6}\right)^{n-1}$

If the first number is $2$, then the probability is $\left(\frac{2}{6}\right)^{n-1}$

If the first number is $3$, then the probability is $\left(\frac{3}{6}\right)^{n-1}$

And so on.

So the required probability is $\sum_{k=1}^6 \frac{1}{6}\left(\frac{k}{6}\right)^{n-1}$, or more simply $\frac{1}{6^n}\left(1^{n-1}+2^{n-1}+\cdots+6^{n-1} \right)$.

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  • $\begingroup$ But what if the first number is $6$ and the second number is $4$? $\endgroup$ – Yiyuan Lee Apr 5 '14 at 3:51
  • $\begingroup$ I had started a list of cases, and at a certain stage said and so on. If the first is a $6$, then the probability is $\left(\frac{6}{6}\right)^{n-1}$, which, if we feel like it, simplifies to $1$. $\endgroup$ – André Nicolas Apr 5 '14 at 3:54

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