<s>[INST] Probability-Tossing of die What's the probability of tossing a die 4 times, and getting a higher value than the previous value in all the 4 outcomes ? How can we I the method for such a case in the case of 3 tosses?
 [/INST] There is no method. you have to list all the possibilities.
The first thought is that the first result cannot higher that $3$. So, there are three groups of favorable outcomes:
$$\begin{matrix}
\color{red}1&2&3&4\\
\color{red}1&2&3&5\\
\color{red}1&2&3&6\\
\color{red}1&2&4&5\\
\color{red}1&2&4&6\\
\color{red}1&2&5&6\\
\color{red}1&3&4&5\\
\color{red}1&3&4&6\\
\color{red}1&3&5&6\\
\color{red}1&4&5&6\\
\color{white}1\\
\color{blue}2&3&4&5\\
\color{blue}2&3&4&6\\
\color{blue}2&3&5&6\\
\color{blue}2&4&5&6\\
\color{white}2&\\
\color{green}3&4&5&6
\end{matrix}$$
As a total, we have these $15$ possibilities with the common probability of $\frac{1}{\ 6^4}.$ So, the probability sought for is$$\frac{15}{\ 6^4}.$$
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