How to approach analyzing probability problems? (Specific question included) I've recently become very interested by the concept of probability. After doing some studying, I believe I've become fairly familiar with the terms: probability, random variables, probability distributions, probability density functions, expected value of a random variable, variance etc.'.
However, I came across one particular problem that appears fairly easy to solve, but when I tried to actually solve it, I didn't get a particular result.
The question is: 

'Suppose you have an infinite amount of money. Let's say the currency
  is called Zed. You only have bills of 50 Zeds each. People arrive with
  bags with Zeds demanding that you exchange them for one of your bills.
  The bags contain:
a) 49 Zeds, 30% of the time;
b) 50 Zeds, 60% of the time; c) 51 Zeds, 10% of the time.
The number of Zeds in each bag is independent of any other bag of
  Zeds.
You decide to trust the people and not count the money to see if they
  actually count up to 50 Zeds. By doing this, how many bags of Zeds
  must you obtain to be sure that you're at a loss exactly 99% of the
  time?'

I am stuck. My notebook has similar problems worked out by using Chebishev's inequality, so I tried working with that, but I didn't seem to get anywhere. I assume I ought to express the relation between the expected value of the number of Zeds and the actual value each one is supposed to contain, but after that I can't think of much more.
Any help or piece of advice would be greatly appreciated.
Thanks in advance.
Edit: I am sorry for not being specific enough. The hypothetical 'loss' that's implied in the question occurs whenever you receive a bag with 49 Zeds in it. 
 A: The answer is at least $54$ bags to give a probability of loss of at least $0.99$.
You can calculate this using the recursion $$p(n,x)=0.1p(n-1,x-1)+0.6p(n-1,x)+0.3p(n-1,x+1)$$ where $p(n,x)$ is the probability of a profit of $x$ after you have exchanged $n$ bags.  You start with $p(0,x)=0$ except that $p(0,0)=1$ and you need to extend your calculations to $x$ in the range $[-n,+n]$ and then add up the results to see the probabilities of being in profit or loss after $n$ bags.  You will get for example the following rounded values:
bags    Profit  Even    Loss
48      0.0077  0.0068  0.9856
49      0.0072  0.0063  0.9865
50      0.0068  0.0059  0.9873
51      0.0063  0.0056  0.9881
52      0.0060  0.0052  0.9888
53      0.0056  0.0049  0.9895
54      0.0053  0.0046  0.9901

André Nicolas suggested a suitable normal approximation.   
Each transaction has a expected profit of $-0.02$, with a variance of $0.36$, i.e. a standard deviation of $0.6$.  So the expected net position after $n$ bags is $-0.2n$ with a standard deviation of $0.6\sqrt{n}$.  If you assume a normal distribution, then the point with a cumulative probability of $0.99$ is about $2.32635$ standard deviations above the mean, suggesting a value of about $48.707$ bags is needed, which rounds up to $49$. 
You can do better than this: the normal approximation is continuous. It assumes you can have fractions of a Zed and that the probability of being exactly even is $0$.  To take account of this, you could instead be trying to find $n$ such that the probability of losing at least $0.5$ Zeds is $0.99$: that would involve solving $$\Phi\left(\frac{0.2n-0.5}{0.6 \sqrt{n}}\right)\ge 0.99$$ essentially a quadratic, to give a figure for $n$ of about $53.590$, which rounds up to $54$. So, in this case, using a normal approximation with a continuity correction gives the correct answer.
