A pair of six-sided dice is thrown. If the dice are fair, what is the probability that the dice show doubles or a sum less than 10?

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Feb 8, 2018

$\frac{28}{36} = \frac{7}{9}$

Explanation:

The possible rolls of a pair of fair dice is:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

The number of rolls that are 10 and above is 6, which means the number of rolls that sum to less than 10 is $36 - 10 = 26$. To this, we need to add double 5 (sum 10) and double 6 (sum 12), for a total of 28 rolls that meet the conditions. This is from a possible 36 rolls, giving a probability of:

$\frac{28}{36} = \frac{7}{9}$

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