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What are the odds of rolling triples with $3d6$ like this: $111, 222, 333, 444, 555$, and $666$ perfectly in a row? I know that the odds of rolling triples in a row without any specific order is $36^3 = 1/46656$. But I don't know how to calculate it when order matters.

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    $\begingroup$ Shouldn't the odds of rolling (six) triples in a row without any specific order be $1/36^6$, not $1/36^3$? $\endgroup$ Jan 5, 2015 at 21:06
  • $\begingroup$ The chance of rolling any triple in a row is $\frac{1}{36}$, as the first roll is "free" and the next two must match the first. Therefore the odds of rolling six triples in a row should be, as @BarryCipra says, $\left(\frac{1}{36}\right)^{6}$. $\endgroup$
    – Avraham
    Jan 6, 2015 at 1:31

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$$ \left(\frac{1}{6}\right)^{18} $$

You need to get 18 dice right, each having $\frac{1}{6}$ probability.

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the probability of a specific triple is $\frac{1}{216}$. Since you require six specific triples in a row, the prob is $$ \frac{1}{6^{18}} $$

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The above answers are correct, but perhaps the simplest way to consider it is that when rolling the set of triples you suggest in specific order, you MUST get the following sequence: 111222333444555666—nothing else will suffice. This is equivalent to rolling one die, eighteen times, and getting exactly the one number you need each time. The probability of getting the one number you need on a given roll is $\frac{1}{6}$ (assuming the die is fair) so the overall probability (assuming rolls are independant) is: $$ \large \underbrace{\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\ldots\cdot\frac{1}{6}}_{18\;\textrm{times}} $$ or $$ \left(\frac{1}{6}\right)^{18} $$

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