Ok so I mean greater than either of the other 2 not the combination. I created a grid and found that there are 15/36 possibilities that it would be greater than one of the dice, so was thinking that it would be 30/72 that it would be greater than one or the other?
Thanks for the clarification. Rolling three dice given \(\displaystyle 6^3=216\) outcomes in the form of triples.
Now must consider the numbers, look at this list:
\(\displaystyle \begin{array}{*{20}{c}}{\text{max value}}&{\text{#}}&{\text{total}}\\\hline{6|}&{25}&{75}\\{5|}&{16}&{48}\\{4|}&9&{27}\\{3|}&4&{12}\\{2|}&1&3\end{array}\)
The table tells us that there are seventy-five triples in which six it the greatest value.
That that is: the triple \(\displaystyle (6,1,5) \) has a die with a value greater than either of the other two.
BUT that happens in three ways. Look at the set of pairs \(\displaystyle \{1,2,3,4,5\}\times\{1,2,3,4,5\}\) there are twenty-five pairs.
Each of those pairs can go with the six in three ways. (Can you tell why we din't to use the one?)
Note that we counted the triple \(\displaystyle (4,5,4)\) but not \(\displaystyle (4,4,1)\) WHY?
Add up the third column. What do you get?
That is, we count the triples having a an unique maximum.
What is the answer to this problem?