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In conditional probability I'm asked to answer the question:

If two fair dice are​ rolled, find the probability that the sum of the dice is $8$​, given that the sum is greater than $3$.

I have been getting them wrong because when finding the number of events that the sum equals $8$. I count $\{4,4\}$ twice, but in a table of events its only listed once. That doesn't make sense to me because they do count $\{2,6\},\{6,2\},\{3,5\},\{5,3\}$. So why not $\{4,4\},\{4,4\}$?

Thanks. Spencer

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  • $\begingroup$ The events (2,6) and (6,2) are different : the first event corresponds to "the first dice gave 2 and the second dice gave 6" while the second is "the first dice gave 6 and the second gave 2". While (4,4) and (4,4) are the same event $\endgroup$ – Binyamin Riahi May 27 '19 at 20:07
  • $\begingroup$ I think that is where I'm confused because when I say it like that in my head the first dice gave 4 and the second dice gave 4. Ha, I just realized typing it I see exactly why its the same event. I would just be repeating the statement for the event. Thanks! $\endgroup$ – spence May 27 '19 at 20:10
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There is only one possible way to throw $\{4,4\}$ with two dices, your first dice should be $4$ and your second dice should be $4$. To throw the combination $\{2,6\}$ we have two possibilities, your first dice should can be $2$ or $6$, and your second dice should then be respectively $6$ or $2$. Does this help?

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  • $\begingroup$ yes, thanks so much! $\endgroup$ – spence May 27 '19 at 20:13
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    $\begingroup$ +1 Not that it matters, but strictly speaking the plural is dice (as is the adjective) and the singular is die $\endgroup$ – Henry May 27 '19 at 21:06
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You've got a red die and a blue die. For convenience's sake, we'll say the red one always gets shown first. When we do this, swapping which die is which changes the order of the results, so we have to count it twice: $(\color{red}3,\color{blue}5)$, $(\color{red}5,\color{blue}3)$. With doubles, $(\color{red}4,\color{blue}4)$, swapping which die is which gives... the same thing, so it doesn't count separately.

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  • $\begingroup$ thanks so much! $\endgroup$ – spence May 27 '19 at 20:13
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Assume that the first die is showing a $4$ and you are ready to through the second one. The probabity of getting a $4$ For the second one is $1/6$ which is the same as getting any other number from $1$ to $6$

That is probability of $(4,4)$ is the same as any of $(4,1)$, $(4,2)$,...,$(4,6)$

Out of the total $36$ possible outcomes there is only one $(4,4)$ so why should we count it twice?

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