What is the probability that 3 will be a temporary maximum? 
A series of independent tosses of a fair dice is performed. The highest result until the $n$th toss is called “temporary maximum”.

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*What is the probability that 3 will be a “temporary maximum” at least once?


My solution:

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*I concluded that the length of a series can be infinite.


*It's enough to count, for every possible length of a series, how many series there are which contains a prefix of $\{1,2\}$ and then the number $3$ and then any suffix with any numbers.
I think that I make it too complicated because the answers show that it's enough to count, for every possible length of a series, how many prefixes of $\{1,2\}$ there which after every prefix, the number $3$ exist.

Also, there is a second solution:

My questions are:

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*[First approach] The question asked What is the probability that 3 will be a “temporary maximum” at least once?. This approach calculates the chances that 3 will be a “temporary maximum” only once. Why?


*[First approach] They mention the law of total probabilities. I don't see any use of it in the solution.


*[Second approach] This approach assumes that every series contain the number $3$. Why?


*[Both approaches] Both approaches do not take into account that a series may contain the numbers $\{ 1, 2, 3, 4, 5, 6\}$ in any order. It assumes that it contains only the numbers $\{1, 2, 3\}$ in a specific order. Why?
 A: 
[First approach] The question asked What is the probability that 3 will be a “temporary maximum” at least once?. This approach calculates the chances that 3 will be a “temporary maximum” only once. Why?

Three will be the temporary maximum at least once if it ever happpens at all.   The probability that it will be a tempory maximum only once is $3/16$ (that is: that happens once and not again).

[First approach] They mention the law of total probabilities. I don't see any use of it in the solution. 

That is what the summation is.  $\sum_{n=0}^\infty \mathsf P(n\text{ occurances of }\{1,2\})\mathsf P(\text{a subsequent }3\mid n\text{ occurances of }\{1,2\})$

[Second approach] This approach assumes that every series contain the number 
  $3$. Why?

You want the probability that you will throw a $3$ before any of $\{4,5,6\}$.   This will be decided on the first toss that is not $\{1,2\}$.   So we can safely ignore the arbitrary many results of $\{1,2\}$ which happen before that decisive toss.   (There is an immeasurable probability that the decisive toss will never eventually happen.)
The probability that you will throw a $3$ before any of $\{4,5,6\}$ is $1/4$, unless you quit before the decision is made.

[Both approaches] Both approaches do not take into account that a series may contain the numbers $\{1,2,3,4,5,6\}$ in any order. It assumes that it contains only the numbers $\{1,2,3\}$ in a specific order. Why?

They don't assume that.   They model the outcomes as an arbitrary many $1$ or $2$ (in any sequence) followed by a decisive toss of $3$ or higher.
