probability that something will occur given infinite time thanks in advance for the help.  Let r be a round.  Also let N be a set of coins.  Suppose that every round I flip all of the coins.  Is it correct to say given infinite rounds the probability that there will be at least one round where all N coins are heads or tails is 1?  If so how would I express this conclusion mathematically?  I believe that I could express this conclusion (as true) using a limit, but I'm not completely convinced that using a limit approach is logically correct.
 A: This is a situation that considers the convergence of random variables, and so I believe the strongest statement you can make is that it converges "with probability 1" or "almost surely".
To clarify, let's define $X$ to be a random variable that represents the number of heads that turn up when you flip the $N$ coins in a single trial.  Then let's say that you repeat the experiment $k$ times and represent the number of heads in each trial by $X_i$ so that you have a sequence of independent random variables $\{X_1,X_2,...,X_k\}$. Finally, let's define $$Y_k=\min\{X_1,X_2,...,X_k\}$$
So if $Y_k=0$ we had a round that had zero heads/all tails.  We can state that $Y_k$ will converge to this result "with probability 1" if we can show that 
$$P\left(\lim_{k \to 0}Y_k=0\right)=1$$
So let's do that.
Clearly $Y_k$ is nonincreasing since it's defined as the minimum of the $X_i$'s. Furthermore, the lowest that it could conceivably be is $0$.  So let's take $\epsilon\gt0$.  Now we can get $Y_k\ge\epsilon$ if and only if $X_i\ge\epsilon$ for all $i$, which implies that
$$P(Y_k\ge\epsilon)=P(X_1\ge\epsilon,X_2\ge\epsilon,...,X_k\ge\epsilon)=\left(\sum_{j=\epsilon}^Np_X(j)\right)^k$$
where $p_X(x)$ is the PDF of the binomial distribution
$$p_X(x)={N\choose x}p^x(1-p)^{N-x}$$
$p$ being the probability of tossing a head with one of the coins.  So we know that
$$\sum_{j=0}^Np_X(j)=1$$
which means that
$$\sum_{j=\epsilon}^Np_X(j)\lt1$$
for $\epsilon$ greater than $0$.  (Note that this is a strict less than due to the definition of $p_X(x)$ and my assumption here that $p\neq0$ and $p\neq1$ since that would be trivial.)  This, in turn, implies
$$\lim_{k\to\infty}P(Y_k\ge\epsilon)\le\lim_{k\to\infty}\left(\sum_{j=\epsilon}^Np_X(j)\right)^k=0$$
so that $P(Y_k\ge\epsilon)=0$ for any $\epsilon\gt0$ as $k\to \infty$.  We can then conclude that $P(Y_k\gt0)=0$ as $k\to \infty$, which implies that $P(Y_k=0)=1$ as $k\to \infty$.  So now we have that $Y_k$ converges to $0$ with probability $1$, which is to say that it converges to the result of getting all tails on one of the tosses with probability $1$.  In a virtually identical manner, the same can be said about getting all heads.
In conclusion I repeat that the strongest statement I can make about this situation is that it converges to the result you're asking about with probability 1, which actually isn't to say that it's certain to happen.  It's basically as close to certain as you can get without being certain, though.
A bit more info if you're interested in convergence of random variables.
