How can I calculate the probability of getting the same number from rolling 3 8 sided dice? I know there are similar questions but I have been out of study for a long time and I need to get a firm understanding of how to tackle probability problems like this so a good breakdown of the steps would be appreciated so I can apply it to similar problems.
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2 Answers
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We note that whatever the first number, the same number has a chance of coming up on the next two rolls. Since the events (rolls) are independent of each other, we can multiply the probabilities that they are the same as the first, so we get $1 \cdot \frac{1}8\cdot\frac{1}8=\frac{1}{64}$
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$\begingroup$ So only the second and third roll probability count, I kept including the first roll but it makes sense that it is irrelevant. Would there be any difference for consecutive numbers, like 1,2,3? $\endgroup$ Commented Mar 20, 2014 at 0:33
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1$\begingroup$ @PadraicCunningham : There are $36$ ways to roll three consecutive numbers (ie: 123, 234, 345, 456, 567, 678 with 6 permutations of each) out of a total of $8^3$. So we have $\frac{36}{512} = \frac{9}{128}$. $\endgroup$ Commented Mar 20, 2014 at 1:01
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$\begingroup$ @GrahamKemp, thanks, so our sample space is 8^3 = 512, there are 36 different ways the consecutive numbers could come, order does not matter so it is 6 * 3!, is that where the numbers are coming from? $\endgroup$ Commented Mar 20, 2014 at 1:25
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$\begingroup$ @GrahamKemp, I know it is pretty basic stuff, it has just been a while. Thanks again. $\endgroup$ Commented Mar 20, 2014 at 1:39
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You're rolling $3$ dice each with $8$ sides. For the first die, you want to list the number $1$ and for the other two dice, you want to list the number $1\over 8$ and then multiply every number to get $1\times{1\over 8}\times{1\over 8}={1\over 64}$.
answered Feb 1, 2015 at 20:58