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Dice Odds #1: Eight-Sided Dice Summing to 11
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Suppose you throw a pair of fair, eight-sided dice. What is the probability of the dice resulting in a total sum of 11?
Let's write all numbers printed on both dice:
Die #2: \(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8\)
We can use set theory notation to lay out its possibilities:
Since we are playing with a pair of fair dice, each with eight (8) sides, all results are equally likely. Guaranteeing exactly one outcome out of eight, we can write \(1 \over 8\). With two dice at play, we multiply over total outcomes by itself; in this case, it's \(8 \cdot 8 = 64\).
Let \(\textbf{P(E)}\) represent the probability function for the event \(\textbf{E}\).
This function equals our possible outcomes, represented by set \(\textbf{E}\), to the total outcome-sample space. Thus,
Per our set \(\textbf{E}\), there are six (6) possibilities resulting in a sum of \(11\). Thus,
Therefore, the probability of rolling a pair of fair, eight-sided dice resulting in a sum of \(11\) is \(9.38%\).
We hope you learned something new! Feel free to contact us, if you have any questions or would like to add to this article teaching about the terms we've discussed here!
(NOTE: This marks as our very first, official tutorial post here on KCU Network of Mathematics!)
Credits and a thank you to MathJax for this beautiful LaTeX library!
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