Probability without replacement - given label chosen third

Question

What is the probability for each of the four labels that it is chosen third?

There are 10 balls in total A: 4, B: 3, C: 2, D: 1 and the object is to pick out 4 but no label can be reused. If an A ball is picked, all balls labelled A are removed. Complete sequences are thus all permutations of: A B C D with that also being the most likely sequence at $$\frac{4}{10}\cdot\frac{3}{6}\cdot\frac{2}{3} = \frac{2}{15}$$ and all 4! = 24 possible permutations can be calculated in a similar way. Sum the probabilities grouped by the third selection to get the answers.

$$A = \frac{26}{126} \\ B = \frac{33}{126} \\ C = \frac{40}{126} \\ D = \frac{27}{126}$$

However this brute force permutation approach falls apart when there are large numbers of labels (eg A-Z) each with varying numbers of balls. Is there a better approach to this which doesn't involve n! calculations?

\begin{array}{|c|c|c|c|c|c|c|} \hline & & & & 1st & 2nd & 3rd & probability \\ \hline A & B & C & D & \frac{4}{10} & \frac{3}{6} & \frac{2}{3} & \frac{2}{15} \\ \hline A & B & D & C & \frac{4}{10} & \frac{3}{6} & \frac{1}{3} & \frac{1}{15} \\ \hline A & C & B & D & \frac{4}{10} & \frac{2}{6} & \frac{3}{4} & \frac{1}{10} \\ \hline A & C & D & B & \frac{4}{10} & \frac{2}{6} & \frac{1}{4} & \frac{1}{30} \\ \hline A & D & B & C & \frac{4}{10} & \frac{1}{6} & \frac{3}{5} & \frac{1}{25} \\ \hline A & D & C & B & \frac{4}{10} & \frac{1}{6} & \frac{2}{5} & \frac{2}{75} \\ \hline B & A & C & D & \frac{3}{10} & \frac{4}{7} & \frac{2}{3} & \frac{4}{35} \\ \hline B & A & D & C & \frac{3}{10} & \frac{4}{7} & \frac{1}{3} & \frac{2}{35} \\ \hline B & C & A & D & \frac{3}{10} & \frac{2}{7} & \frac{4}{5} & \frac{12}{175} \\ \hline B & C & D & A & \frac{3}{10} & \frac{2}{7} & \frac{1}{5} & \frac{3}{175} \\ \hline B & D & A & C & \frac{3}{10} & \frac{1}{7} & \frac{4}{6} & \frac{1}{35} \\ \hline B & D & C & A & \frac{3}{10} & \frac{1}{7} & \frac{2}{6} & \frac{1}{70} \\ \hline C & A & B & D & \frac{2}{10} & \frac{4}{8} & \frac{3}{4} & \frac{3}{40} \\ \hline C & A & D & B & \frac{2}{10} & \frac{4}{8} & \frac{1}{4} & \frac{1}{40} \\ \hline C & B & A & D & \frac{2}{10} & \frac{3}{8} & \frac{4}{5} & \frac{3}{50} \\ \hline C & B & D & A & \frac{2}{10} & \frac{3}{8} & \frac{1}{5} & \frac{3}{200} \\ \hline C & D & A & B & \frac{2}{10} & \frac{1}{8} & \frac{4}{7} & \frac{1}{70} \\ \hline C & D & B & A & \frac{2}{10} & \frac{1}{8} & \frac{3}{7} & \frac{3}{280} \\ \hline D & A & B & C & \frac{1}{10} & \frac{4}{9} & \frac{3}{5} & \frac{2}{75} \\ \hline D & A & C & B & \frac{1}{10} & \frac{4}{9} & \frac{2}{5} & \frac{4}{225} \\ \hline D & B & A & C & \frac{1}{10} & \frac{3}{9} & \frac{4}{6} & \frac{1}{45} \\ \hline D & B & C & A & \frac{1}{10} & \frac{3}{9} & \frac{2}{6} & \frac{1}{90} \\ \hline D & C & A & B & \frac{1}{10} & \frac{2}{9} & \frac{4}{7} & \frac{4}{315} \\ \hline D & C & B & A & \frac{1}{10} & \frac{2}{9} & \frac{3}{7} & \frac{1}{105} \\ \hline \end{array}

Answer 1

It can be written in a short hand form, but the calculation remains the same. Let us say there are n ball types, and you want the kth type to be chosen at the mth choice, then the probability $P=\sum_{j_1\ne k}\sum_{j_2{\ne k,\ne j_1}}.....$$\sum_{j_{m-1}{\ne k\ne j_1,..,\ne j_{m-2}}}p_{j_1}\frac{p_{j_2}}{1-p_j{_1}}....\frac{p_{j_{m-1}}}{1-\sum_{n=1,m-2}p_{j_n}}$$\frac{p_k}{1-\sum_{n=1,m-1}p_{j_n}}$.

Here $p_j$ is the probability of picking ball type j.

Answer 2

Comment: Reality check using simulation. Labeling the balls 1, 2, 3, 4, with frequencies in the box as 4, 3, 2, 1, respectively, a simulation in R statistical software with a million draws of three balls confirms your results to two or three place accuracy.

ball.3 = replicate( 10^6, sample(1:4, 3, prob=c(.4,.3,.2,.1))[3] )
table(ball.3)/10^6
ball.3
       1        2        3        4 
0.206875 0.262178 0.317036 0.213911          # simulated probabilities
c(26,33,40,27)/126
[1] 0.2063492 0.2619048 0.3174603 0.2142857  # your computations

If you draw four balls, the probability that each color is last is approximated below:

ball.4 = replicate(10^6, sample(1:4, 4, prob=c(.4,.3,.2,.1))[4])
table(ball.4)/10^6
ball.4
       1        2        3        4 
0.078033 0.129446 0.241200 0.551321