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A fair dice is thrown three times, what is the probability of getting each number larger than the previous number?

I am aware that the sample space of this event is $6 \times 6 \times 6=216$

I am also aware that the second number must be greater than $1$. So if $i$ be the second number, then $i-1$ is the maximum possible first number, and $6-i$ is the least possible last number. However I couldn't put it in form of some equation. Please help me in this regard. Thanks

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    $\begingroup$ The probably that which number is bigger than which previous number? $\endgroup$
    – arkeet
    Oct 10, 2016 at 22:52
  • $\begingroup$ @arkeet Probably first<second<third $\endgroup$ Oct 10, 2016 at 23:19
  • $\begingroup$ OK, that would make the last paragraph make sense. $\endgroup$
    – arkeet
    Oct 10, 2016 at 23:22

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Following your idea, if we fix the second number to be $i$, then there are $i-1$ possibilities for the first number, and $6-i$ possibilities for the third number. So you want to find the sum $$ \sum_{i=1}^6 (i-1)(6-i) = \sum_{i=1}^6 (-i^2 + 7i - 6). $$ This can be done rapidly using formulas for $\sum_1^n i$ and $\sum_1^n i^2$ and so on. (And then of course you want to divide the number of possibilities by $6^3$.)


But there is a much nicer solution: The possibilities for increasing dice rolls correspond one-to-one with the ways of choosing $3$ distinct numbers out of $\{1,2,3,4,5,6\}$. How many such choices are there?

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