What am I doing wrong in applying the principle of inclusion-exclusion to probability? I am challenging myself to apply the two theorems we had time to explore in my class's unit on probability. The problem I am exploring is this: Given a six-sided die, rolled six times, what is the probability that a 1 will occur? I first applied this theorem:
$$Theorem: P(E) = 1 - P(E')$$
Thus, we have $1-(5/6)^6 \approx ~.6651 = 66.51\%$
This matches the value I have approximated via 100,000 or so simulations, but not the value I get when trying the problem with the inclusion-exclusion principle as given in our textbook:
$$Theorem: P(E_1\cup E_2 ) = P(E_1)+P(E_2)-P(E_1\cap E_2)$$
Using this, I have:
$$P(E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6)=1/6+1/6+1/6+1/6+1/6+1/6 - (1/6)^6 \approx .9999 = 99.99\%$$
where $E_n$ is the occurence of  a 1 on the $n^{th}$ roll. What have I done wrong?
 A: I think your understanding of the Inclusion-Exclusion Principle may be a little off. And in my opinion, this concept cannot be better explained than through Venn diagrams.
Below are two series of diagrams decomposed by operations - the first corresponding to two events, the second to three. In each case, regard the desired probability as the area of the appropriate figures. 
In the case of two events, the derivation of the principle is obvious without a diagram, but the picture helps to construct a pattern. To find the area of the red ($A\cup B$), we add each blue ($A,B$), but we've added the intersection ($A\cap B$) twice, so we take away the green. From this, we can derive: $\color{red}{P(A\cup B)} = \color{blue}{P(A) + P(B)} - \color{green}{P(A\cap B)}$.
In the case of three events, to find the area of the red ($A\cup B \cup C$), we add each blue ($A, B, C$), and, like before, we've added too much, so we take away each green ($A \cap C, B \cap C, A \cap B$). However, notice that now we've taken away too much - namely, the intersection of all three ($A\cap B \cap C$). So we add the orange back in. This yields the three-event version: $\color{red}{P(A\cup B \cup C)} = \color{blue}{P(A) + P(B) + P(C)} - ( \color{green}{P(A\cap B) + P(A\cap C) + P(B\cap C)}) + \color{orange}{P(A\cap B \cap C) }$
Extending this to four events, $A, B, C, D$, we have:
$P(A\cup B \cup C \cup D) = [P(A) + ...+ P(D)] -[P(A\cap B) + ... +P(C\cap D)] + [P(A\cap B \cap C) +...+P(B \cap C \cap D)]-[P(A \cap B \cap C \cap D)] $ 
Inductively, this principle becomes evident for any number of events:
$P(\text{the union of all events}) = [\text{the sum of all }P(\text{each distinct event})] - [\text{the sum of all }P(\text{the intersection of any two distinct events})]+ [\text{the sum of all }P(\text{the intersection of any three distinct events})] - ... \pm [P(\text{the intersection of all distinct events})]$
(Clearly, the last operation "$\pm$" depends on how many events there are.)
Hope this helps (and apologies if it seems overly rudimentary).


