Probability I have another pack of sweetener I drove my motorcycle to a fast food restaurant the other day.  As I was waiting for my lunch, I noticed they still had their coffee condiments out.  Not having any at home, I decided I'd grab a small handful and toss them into my motorcycle bag for later.
I always put two sweeteners in my coffee, but one day I pulled a sweetener out of my motorcycle bag and I didn't see a second.
I thought to myself, "I should keep looking; there's a 50/50 chance there's another one in here."
I'm a math novice so I asked some of my friends if they thought I was right.  They weren't willing to assert either way, so I thought I'd ask here.  Was I statistically sound in my conclusion?
EDIT: I haven't accepted an answer on this question yet, but I will be reviewing the answers soon.  Part of the reason I haven't is because I feel like a lot of the answers are a smidgen pedantic (and I don't mean that in a negative way at all).
While I recognize that it's impossible to predict the distribution of a handful of sweeter packets because the universe is or is not random (whatever the case), I feel like a lot of people would fail a statistics course if they were to tell the professor a coin flip isn't 50/50 because, "nobody shuffles cards completely randomly, no die is unbiased, and you impart bias in a coin flip."
I was really hoping to get answers to this question based on the same simple model of the universe that allows statistics professors to teach and allows casinos to make fortunes based on the knowledge that a six sided die has, ceteris paribus, 6 equally likely outcomes.
I'm by no means saying that my original conclusion was correct, however, I'm not prepared to accept that it was wrong because of the "orientation of the hairs on my skin, the amount of blood distending my vessels, or the imperfections in my skin."
 A: Let
  \begin{align*}
  A &= \text{"I pulled a sweetener out of my motorcycle bag."}\\
  B &= \text{"I didn't see a second sweetener."}\\
  C &= \text{"There's another sweetener in there."}
  \end{align*}
You want $P(C\mid A \cap B)$. Applying Bayes' theorem to the probability measure $P(\;\cdot\mid A)$ gives
  $$
  P(C\mid A\cap B) = \frac{P(C\mid A)P(B\mid A\cap C)}
    {P(C\mid A)P(B\mid A\cap C) + P(C^c\mid A)P(B\mid A\cap C^c)}.
  $$
Now, $P(C^c\mid A)$ is the probability that your bag is empty after taking the first sweetener that day, computed without regard for the fact that you did not immediately see another one. It could, in principle, be anything. For example, if you have any rough idea about how many sweeteners you originally took (about 5? about 50?), and if you have a rough idea about how many days ago it was, then this could influence your estimate of $P(C^c\mid A)$.
But let us assume that you are happy to model this as $P(C^c\mid A)=1/2$ by considering no other information but the fact that you use two sweeteners per day, and by assuming that you were as likely to have originally grabbed an odd number of sweeteners as to have grabbed an even number. Then the above probability reduces to
  $$
  P(C\mid A\cap B) = \frac{P(B\mid A\cap C)}
    {P(B\mid A\cap C) + P(B\mid A\cap C^c)}.
  $$
Now notice that if $C^c$ holds, then the bag is empty and you cannot see a second sweetener. Thus, $P(B\mid A\cap C^c)=1$. If we let $p=P(B\mid A\cap C)$, then we now have
  $$
  P(C\mid A\cap B) = \frac p{p + 1}.
  $$
The only way for the above probability to be 1/2 is if $p=1$. Verbally, $p$ is the probability that you would fail to see another sweetener, despite the fact that there is one in there.
So even if you assume that an odd and even number of sweeteners were equally likely to have been originally taken, and even if you ignore whatever information you may have about how many you took and how long ago it was, the probability that you have another sweetener, given that you did not immediately see one, can still only be narrowed down to somewhere between 0 and 1/2. The messier and more cluttered your bag is, the closer it is to 1/2. The cleaner and more organized your bag is, the closer it is to 0.
A: Is there a 50/50 chance that you put an even/odd number of sweeteners in your bag? 
Yes.   (That statement requires too many disclaimers at best, and is just wrong at worst!) 
But as to the probability that, even after not seeing one during an initial glance, you still had one left, ... can't say!
A: This depends on the distribution of the number of sweeteners you grab.  It is not possible to have a "randomly chosen integer" in which all integers are equiprobable.  
A: Suppose the number of packets follows a Poisson distribution, that is, the probability of grabbing $k$ packets is $\lambda^ke^{-\lambda}/k!$, where $\lambda$ is the average number of packets grabbed. Then the probability of grabbing an even number of packets is $e^{-\lambda}\cosh\lambda$, and the probability of grabbing an odd number is $e^{-\lambda}\sinh\lambda$. So, no matter what $\lambda$ is, you're more likely to grab an even number, since $\cosh\lambda\gt\sinh\lambda$ for all $\lambda$. BUT 


*

*The difference is very small: $e^{-\lambda}\cosh\lambda-e^{-\lambda}\sinh\lambda=e^{-2\lambda}$ is miniscule for, say, $\lambda=6$, and 

*I haven't the faintest idea whether it's reasonable to assume a Poisson distribution. 
So all I'm really doing here is pointing out that if you hypothesize a distribution, you can get an answer, and there is one well-known and highly useful distribution that has a (tiny) bias toward even numbers. 
A: There is a famous problem with a related flavour called the Banach matchbox problem. The original problem goes like this. You have two matchboxes, each originally with $n$ matches, one in each pocket. Every time you smoke (Banach was a heavy smoker) you pick a pocket at random, and remove a match.  What is the probability that when you first find a matchbox to be empty, the other box is empty?
A: I think you are wrong. You are more likely to have initially put an odd number of candy in your bag. This is because the positive natural numbers start with an odd number. For any number there can be more odd numbers then even numbers less then it but not the other way around.
A: I'll interpret the question as: is there a mathematical model of this situation for which the probabilities are not equal?  The answer to that is yes and an example is a binomial distribution: if you grab on average six packets, then you will usually grab an even number, and if you grab on average seven packets, then the number will usually be odd.  If you grab on average exactly six and a half packets then the probabilities will be closer to equal, as can be seen from the symmetry of the normal distribution which the binomial approaches.  This simple model is not really of any use if you want to know the truth: to do that, you must calculate the distribution by performing repeated experiments.  But the model does suggest that the answer depends to first order on whether the sample mean is closer to an even or an odd.
