What is an intuitive way to think about the probability of simultaneous vs. sequential independent events? Consider a biased coin with probability $p$ of landing heads up. Then the probability of getting $N$ heads in a row is $p^N$.
The way I think about this intuitively is that the slice of reality where the first coin lands heads up is $p$. Or in other words, the subarea of the unit square corresponding to "heads up" equals $p$ square units. Now if I throw the coin once more, then given that I am in the world where the first coin landed heads up, the slice of reality where heads shows up again is $p^2$. Or in other words, the unit square is now partitioned into 4 portions with subareas $p^2$, 2 times $p(1-p)$ and $(1-p)^2$. And after seeing two heads we are located in the subarea of size $p^2$.
You could also imagine to randomly throw a dart on a unit square that is partitioned into subareas of size $p$ and $(1-p)$. Then, if you hit the subarea of size $p$, which corresponds to heads up, you randomly throw another dart. But this time the range is limited to the subarea you hit previously, which is partitioned into subareas of size $p^2$ and $p(1-p)$.
My problem is that this intuition breaks down when dealing with simultaneous flips of biased coins with equal probabilities. If instead the coins were fair, then I could reason that each outcome of throwing $N$ coins is equally likely (each outcome demands its slice of reality, but the demands are indistinguishable) and thus has a probability of $2^{-N}$. But given biased coins, all I can do is apply the same formula that I'd apply in the sequential case, $p^{N-K}(1-p)^K$, without a good intuition of why it is true.
 A: Here is an extension of your "equally likely outcome" example to any rational probability $p$: 
Suppose the probability $p$ is a rational number $p=r/z$ (with $r$ and $z$ positive integers).  You can imagine throwing a single dart at a region with finite area that is divided into $z$ equal sub-regions, $r$ of which are colored red (the remaining $z-r$ sub-regions are blue).  The number of outcomes of the single dart throw is $z$ (one for each sub-region), and $r$ of these outcomes correspond to a dart landing in a red sub-region.  So, the probability that this single throw lands the dart in a red sub-region is $r/z=p$.
For $N$ experiments we can imagine simultaneously throwing $N$ darts at $N$ separate but identical regions (with each region divided into $z$ equal sub-regions).  The result can be represented by a vector: $$(\omega_1, \omega_2, \ldots, \omega_N)$$ 
where $\omega_i \in \{1,2, \ldots, z\}$ for each $i \in \{1, \ldots, N\}$. There are $z^N$ possible outcomes, and there are $r^N$ outcomes that have all $N$ darts landing on a red sub-region.  If we assume that all outcomes are equally likely, then they all must have probability $1/z^N$, and we get:
\begin{align}
Pr[\mbox{all $N$ darts land on a red sub-region}] &= \frac{\mbox{number of such outcomes}}{\mbox{total number of outcomes}}\\
&= \frac{r^N}{z^N} \\
&= p^N
\end{align}
Notice that I have not anywhere used the word "independent." However, the above calculation gives the same probability as $N$ independent (and "sequential," if you like) dart throws--where "independence means multiply."  So, in this way, you can intuitively think of the "independence means multiply" rule as being defined according to equally likely outcomes. 
We can approximate any real number in $[0,1]$ by a rational number with any desired accuracy, so the above thought experiment about rational probabilities gives good intuition about why the "independence means multiply" rule should even hold for irrational probabilities. 

A special case is when $p=1/2$, in which case we have $z=2$ and $r=1$, and the probability that all $N$ darts land red is $1^N/2^N=(1/2)^N$, which is the same as the example in your original question.  So, the above generalizes that example to any rational probability $p$.
