Does the likelihood of an event increase with the number of times it does not occur? I would seem logical that the more times an event does not happen, the more likely it is to happen, for example: If a coin is flipped and it lands on tails 10 times in a row it would seam more likely that the next flip will result in heads.
The Infinite Monkey Theorem is one such idea that suggests this is true,
http://en.wikipedia.org/wiki/Infinite_monkey_theorem
It states that if some number of monkeys are left in a room with typewriters for an infinite amount of time then they will eventually compose all written texts ever produced.  This seems to suggest that since the chance of the monkeys writing a work, say Shakespeare's Romeo and Juliet, is very low. The more times they do not write it, the more likely they are to write it, until the chance becomes significant and it, the writing of the play, happens.
However another idea, Gambler's Fallacy states quite the opposite.
http://en.wikipedia.org/wiki/Gambler%27s_fallacy
It states that the chance of an event does not increase with the number of times it does not occur.
So what is the answer? Does the likelihood of an event go up the more times it does not happen, or does it stay the same? And if it does stay the same then how does one explain the Infinite Monkey Theorem?
 A: The Infinite Monkey Theorem does not suggest that the "more times they do not write it, the more likely they are to write it, until the chance becomes significant and it, the writing of the play, happens."  Rather, what it says informally is that the longer they have been writing, the more likely they are to have written a given string.  The monkey is just as likely to start with the complete works of Shakespeare from keystroke 1 as from keystroke $10^{400,000}$.  However, the longer the string of successive keystrokes, the more likely any given substring can be found there.  Thus, for example, the complete works of Shakespeare are much more likely to be found in the string of the first $10^{400,000}$ keystrokes than in the string of the first $10^{300,000}$ keystrokes.  That's because the former is $10^{100,000}$ times as long.
A: The Infinite Monkey Theorem (I didn't know it was a theorem!) basically says that a given finite string of text will appear with probability 1 in an infinite truly random string of text. What it means to be "truly random" is the delicate point.
Anyway, in practice, you cannot produce an infinite string of text "at once" but what you can do (employing monkeys, tossing dice, or instructing your laptop) is to print out a random sequence of letters of increasing, albeit finite, length. As the length increases, so does the probability to find a precise string embedded in that sequence, and this may give the false impression that the chance of producing it improves, because of previous "failures".
In fact it is not so, as the Gambler's Fallacy says. If a perfect coin ("perfect" meaning that "head" has exactly a 50% chance) is tossed ten times and you get "head" ten times, "head" has still a 50% chance at the eleventh toss. Believing the opposite, namely that the chance of getting another "head" is lower than 50%, would be equivalent to believing that the coin has some sort of "internal mechanism" that remembers the past flippings, which is rather absurd. 
A: There is another example that may help you. Suppose you are waiting a bus. Let us make a simple mathematical model of this situation. Consider that the waiting time is a random variable with an exponential distribution (say of parameter 1). This is a very common model for waiting times of all kinds. However with this model you have the following surprising property:
Suppose you have already been waiting the bus for 5 minutes. Then the probability to wait more than 15 minutes knowing you have already been waiting for 5 minutes is the same as the probability of waiting for 10 minutes. So it is just as if you had just arrived.
This property is called "memoryless" and is a common feature of only two distributions : the exponential and the geometric one. The geometric distribution appears in you coin problem as the law of the time of the first head (first success in Bernoulli trials). 
However, if you take another model for the bus waiting time distribution, say for instance a Gamma distribution, then you do not have the memoryless property anymore, meaning that the probability to wait more than 15 minutes knowing you have already been waiting for 5 minutes is not the same as the probability of waiting for another 10 minutes, which might be more in accordance with our intuition somehow.
So, it this example, you see that according to the choice of the randomness model, you can have both answers to you initial question : yes and no, it depends !
A: Try this example: You buy a lottery ticket. There are calculated odds of you winning the jackpot for that particular lottery drawing. Let's say you lose that day. If you only played the lottery that one time, your chances of winning in your life time are much lower than if you had played again the following week, increasing the number of times in your life you have played the lottery to two. The more times you play the lottery, even with the same odds of winning each time, the more likely you are to win. Saying you are more likely to win is the same as saying you are less likely to lose. It seems that although there are given odds to winning the lottery, there are overlapping personal odds of winning. You can only lose so many times before you'll win in an infinite scenario. You can't live infinitely, so although the odds of you winning are increasing each time you play, you may not live long enough to get to the winning numbers. So long as the event you are determining the odds for is actually possible, the more times it doesn't occur will bring you closer to it actually occurring. You can't win if you don't play; and the more you play the better your chances of winning, or the better your chances of not losing. 
A: 
If a coin is flipped and it lands on tails 10 times in a row it would seam [sic] more likely that the next flip will result in heads.

Really? I'd say that the coin is not fair, and that it's more likely for tails to come up again.

But really, this is an example of the Gambler's Fallacy. If the coin is fair (and so is the flip), each outcome doesn't depend on the previous. Yes, it's unlikely that heads comes up 10 times in a row, but given the fact that it just did, it's equally likely for tails to come up as it is for heads to come up yet again and make it 11 times.
It seems you're confused here about the a priori probability. Yes, if you calculate the odds of heads coming up 11 times in a row before you flip for the first time, it's indeed a very small chance ($\tfrac{1}{2^{11}} = 0.00048828125$ to be exact) that this will happen. But the coin has no memory. It does not think "Oh, I've done heads 10 times already, I guess it's time for tails." 
If you draw the probability tree (I can't do that in here, or I would've done that for you), you'll see that at each node, the probability of taking either path is $0.5$. The combined chance of taking the path that ends up at the single node where you have 11 heads and no tails, is just as infinitesimal as it is for any other path — it's just that most of those paths have numbers of heads and tails that are roughly equal. But if you already are at the "10 heads"-node, taking the path to tails or to yet another heads has the same probability associated: $0.5$.

But it depends. There are events that are not independent of each other. For instance, if (in a perfect world, where buses run exactly on time) a bus stops at this bus stop every 6 minutes, and you've been waiting for 5 already, you know that the bus will arrive in the next minute.
And even in a not-so-perfect world, you know a bus will come eventually. So yes, there are events that are not independent of each other. Another example would be drawing marbles from a bag without putting them back. 
So the first thing you need to do, is to figure out if you're dealing with independent events, or not.
