20 balloons are distributed amongst 6 children: Probability that one child gets no balloon? 20 balloons are randomly distributed amongst 6 children. What is the probability, that at least one child gets no balloon? 
What's the mistake in the following reasoning (I know there has to be a mistake; by simulation I know, that the actual probability has to be appr. 0.15, which is not what the following formula gives): 
I start to think about the opposite case: What is the probability that every child gets at least one balloon. There are all together ${20+6-1\choose 20} = {25\choose 20}$ ways to distribute the balloons amongst the children. The number of the desired ways (i.e. distribute balloons so that every child gets at least one balloon) is ${14+6-1\choose 14} = {19\choose 14}$.
So, the probability that every child gets at least one balloon, when the balloons are randomly distributed amongst the children should be $$ \frac{19\choose 14}{25\choose 20}$$ 
For the opposite  case, i.e. the probability that at least one child gets no balloon is: 
$$ 1 - \frac{19\choose 14}{25\choose 20} = 0.78114...$$
At which point did I get wrong??
BTW: I used the following R-Code to simulate: 
  v <- vector()
  for (i in 1:100000){
     t <- table(sample(1:6, 20, replace=T))
     v[i] <- length(t)<6
  }
  print mean(v)

One Remark:
The answer from mlu is in my opinion correct; thank you very much for it! However: My questions was, where my mistake is in the above reasoning?
The number of different ways to distribute k indistinguishable balls (=balloons) into n distinguishable boxes (=children) is ${n+k-1\choose k}$. So: where did I actually got wrong, because the denominator as specified above is correct, right? So whats wrong about the counter? 
Solution
Thank you very much, again, mlu, for the answer as a commentary below. Now I got it: I counted the number of partitions and tried to calculate the probability with the Laplace-Technique (the nominator for the total number of cases, and the counter for the number of cases we are interested in) but I missed, that not every partition is equally probable. For instance the partition where one child gets all balloons is much more improbable than the partition, that child1 to child4 gets 3 balloons and child5 and child6 get 4 balloons is much more probable, which is clear even by intuition: In the first case, there is always just one possibility to put the balloon whereas in the second case there are (at least at the beginning) many possibilities to put balloons.
 A: Lets assume  both children and balloons are distinguisable, labeled. Then the number of distributions corresponds to selecting a 20 digit sequence of numbers 1 to 6, giving $6^{20}$ possibilities. Let $E_k$ be the event that child k does not receive a ballon, this event corresponds to selecting a 20 digit sequence not containing the number k giving $5^{20}$ possibilities.
$$P(\cup_k E_k) = \sum_k P(E_k) - \sum_{k,l} P(E_k \cap E_l) + \sum_{k,l,m} P(E_k \cap E_l \cap E_m) \dots$$
$$ P(\cup_k E_k) = \sum_{n=1}^5 (-1)^{n+1}\frac{\left(\begin{matrix} 6 \\ n \end{matrix}\right)(6-n)^{20}}{6^{20}} = $$
$$ 6 \left(\frac{5}{6} \right)^{20} - 15 \left(\frac{4}{6}\right)^{20} + 20 \left(\frac{3}{6} \right)^{20} - 15 \left( \frac {2}{6} \right) ^{20} + 6 \left( \frac{1}{6} \right)^{20} $$
A: Comment: To me the kids are distinguishable and the balloons are not. (Seems, a kid that doesn't get a balloon will soon 
distinguish himself or herself by making a fuss.) 
Consider this problem. Roll a fair die 20 times. What is the
probability not all six faces are seen? Is this problem fundamentally
the same as yours?
My R code (and result) for this problem:
 m = 10^6;  n = 6;  b = 20;  t = numeric(m)
 for(i in 1:m) {
   s = sample(1:n, b, rep = T)
   t[i] = length(unique(s))  }
 mean(t < 6)
 ## 0.151852

My 95% margin of simulation error is about  $\pm .00075$.
Additional runs gave 0.15175, 0.151887
 and  0.152094. These results are not
quite in agreement with @Andrey's. (I think your code and
mine are equivalent, except mine runs very much faster on
an archival Windows XP machine.)
Moreover, if there are 6 balloons and 6 kids it seems the
answer is $1 - 6!/6^6$, and results agree with my code,
with constant b suitably changed. 
A: I like thinking in terms of balls and boxes. Here we have $20$ balls, and $6$ boxes to put our $20$ balls into. In addition, we allow for empty boxes to exist. The question is now, "How many ways can we put our balls into our boxes such that there is at least one box with no balls?" To find the probability, we would simply divide this by the total number of ways we can put our balls into our boxes.
Great, but wait a second. Are our boxes labeled? Are our balls labeled? Who knows? The discrepancy between your two calculations stems from this.
In the R code, the balls are distinguishable, and the kids are indistinguishable.${}^\text{1}$ This kind of problem deals with Stirling numbers of the second kind. The aforementioned probability here would be
$$ \frac{\sum\limits_{k=1}^5\begin{Bmatrix} 20\\k \end{Bmatrix}}{\sum\limits_{k=1}^6\begin{Bmatrix} 20\\k \end{Bmatrix}} \approx 0.155853. $$
In the calculation you performed by hand, you assumed the balls are indistinguishable, and the kids are distinguishable.
If you want to assume that the neither the kids nor the balls are distinguishable, then you would be dealing with partitions of a number. Specifically, we would find number of $k$-partitions of $20$ for $k\leq 5$, and divide this by the number of $k$-partitions of $20$ for $k\leq 6$.

${}^\text{1}$As a disclaimer, I can't read that R code. I just found the probability under the mentioned assumptions, and it matched the probability included in the problem!
