Say we roll 2 die. What is the probability of getting two 6s?

Why is $1-(1/6)^2 \ne (5/6)^2$? The prob of 1-5 happening twice and the prob of 6s happening twice should be the compliment of eachother shouldnt it?

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    $\begingroup$ Have you considered the possibility of getting exactly one six? $\endgroup$ – whuber Feb 28 '16 at 19:59
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    $\begingroup$ Ahh ok. So 1-(1/6)^2 -2(1/6)(5/6)=(5/6)^2. Is there a rigorous way you can check to see if you overlooked some events when doing these calculations? $\endgroup$ – Qwertford Feb 28 '16 at 20:07
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    $\begingroup$ Your check was one of the quickest and easiest: when the probabilities add up to less than $1$, then either you have overlooked some outcomes or you have miscalculated a probability. When they add up to more than $1$, then either you have miscalculated a probability or you have defined some overlapping events. $\endgroup$ – whuber Feb 28 '16 at 20:42

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