[Math] 20 balloons are distributed amongst 6 children: Probability that one child gets no balloon

combinatoricsdiscrete mathematicsprobability

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.

Best Answer

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.

$$P(\cup_k E_k) = \sum_k P(E_k) - \sum_{k,l} P(E_k \cap E_l) + \sum_{k,l,m} P(E_k \cap E_l \cap E_m) \dots$$

$$ P(\cup_k E_k) = \sum_{n=1}^5 (-1)^{n+1}\frac{\left(\begin{matrix} 6 \\ n \end{matrix}\right)(6-n)^{20}}{6^{20}} = $$ $$ 6 \left(\frac{5}{6} \right)^{20} - 15 \left(\frac{4}{6}\right)^{20} + 20 \left(\frac{3}{6} \right)^{20} - 15 \left( \frac {2}{6} \right) ^{20} + 6 \left( \frac{1}{6} \right)^{20} $$

Related Question