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?
 A: There are $6^4$ possible sequences of four die rolls, all of which we assume are equally likely.  The number of those in which the sequence of values is strictly ascending is the same as the number of ways to choose 4 elements from the set $\{1,2,3,4,5,6\}$.  So the probability that the sequence of die rolls is strictly ascending is
$$\frac{\binom{6}{4}} {\ 6^4} = \frac{15}{1296}$$
A: 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$$\frac{15}{\ 6^4}.$$
