Given a distribution find the probability. There are 4 elevators. So far...
Elevator 1 opened 45.455% of the time (5/11).
Elevator 2 opened 27.273% of the time (3/11).
Elevator 3 opened 18.182% of the time (2/11).
Elevator 4 opened 9.091% of the time (1/11).
Given this information, what is the probability of getting each elevator the next time you're waiting for one. Assume that the elevators are independent of each other.
The goal is to guess, accurately as possible, which elevator door will open the next time.
 A: Two approaches:
1)  Your lifts could be appearing with equal probability and you are just seeing this pattern because of your small sample.  It slightly depends on how you measure distance from what might be expected 
If you take a sum of squares approach, I think you may find that the probability of getting as extreme an observation as you did or a more extreme one is about $44.3\%$ so you cannot reasonably reject such a hypothesis of equal probability.
2) You could try to use Bayesian methods and start off thinking that certain elevators may be more likely than others, but you do not know which or by how much.  First you need a prior distribution for the probabilities of the different elevators arriving (constrained so the probabilities add up to $1$ which makes this less than easy).  Then you need to combine these with the likelihood of seeing what you observed to give a posterior distribution.   
If you assuming all combinations of probabilities summing to $1$ are equally likely, then I think you will find a $95\%$ central credible interval for the first elevator's posterior probability is between about $18\%$ and $65\%$, for the second elevator's probability is between about $8\%$ and $51\%$,  for the third elevator's probability is between about $5\%$ and $43\%$,  and for the fourth elevator's probability is between about $2\%$ and $34\%$.  
With this prior, the posterior means for the elevator probabilities are $\frac{6}{15}, \frac{4}{15}, \frac{3}{15}, \frac{2}{15}$ respectively, i.e. about $40\%, 27\%, 20\%, 13\%$, but given the wide range of uncertainty resulting from your small sample, this does not tell you much.  [It is not a coincidence that $6=5+1, $ $4=3+1, $ $3=2+1,$ and $ 2=1+1$.] 
Whichever approach you take, the conclusion is really that your observation size is too small to draw a helpful conclusion.
Some of the results above are empirical, but I expect them to be reasonably accurate.    
