Conditional probability in case of balls on succesive addition in urns

Question

Four balls are placed successively in four urns, all arrangements being equally probable. Given that the first two balls are in different urns, what is the probability that one urn contains exactly three balls?

Answer 1

I'm assuming balls and urns are not labeled. One can trace the probability, starting with $u=4$ empty urns and probability 1 if no balls are thrown, with a state representation. A state is a list of $i^{u_i}$ strings, meaning $u_i$ urns contain $i$ balls, $i\ge 0$. These frequqencies $u_i$ fulfil $\sum_i u_i=u$, the number of urns, and $\sum_i iu_i=b$, the number of balls. Adding a ball means that with a probability of $u_i/u$ a state can be deployed to other states when a ball is inserted into urns that already contain $u_i$ balls. The initial state is $0^2 1^2$, meaning 2 urns have no balls and 2 urns have 1 ball. The sequence of states and their probabilities (after the bar) after adding the next 0 to 4 balls are

# 0 balls 1 states
 0^2 1^2 | 1 1.0

# 1 balls 2 states
 0^1 1^3 | 1/2 0.5
 0^2 1^1 2^1 | 1/2 0.5

# 2 balls 4 states
 0^0 1^4 | 1/8 0.125
 0^1 1^2 2^1 | 5/8 0.625
 0^2 1^0 2^2 | 1/8 0.125
 0^2 1^1 2^0 3^1 | 1/8 0.125

# 3 balls 5 states
 0^0 1^3 2^1 | 9/32 0.28125
 0^1 1^1 2^2 | 3/8 0.375
 0^1 1^2 2^0 3^1 | 7/32 0.21875
 0^2 1^0 2^1 3^1 | 3/32 0.09375
 0^2 1^1 2^0 3^0 4^1 | 1/32 0.03125

# 4 balls 7 states
 0^0 1^2 2^2 | 39/128 0.3046875
 0^0 1^3 2^0 3^1 | 1/8 0.125
 0^1 1^0 2^3 | 3/32 0.09375
 0^1 1^1 2^1 3^1 | 11/32 0.34375
 0^1 1^2 2^0 3^0 4^1 | 29/384 0.07552083333333333
 0^2 1^0 2^0 3^2 | 3/128 0.0234375
 0^2 1^0 2^1 3^0 4^1 | 13/384 0.033854166666666664

That means after 2 balls are thrown, 4 in total in the urns, the probability is $1/8$ that one urn contains 3 balls.

That means after 4 balls are thrown, 6 in total in the urns, there is a probability $39/128$ that 2 urns have 1 ball and 2 urns have 2 balls, there is a probability $1/8$ that 3 urns have 1 ball and 1 urn has 3 balls, etc. So the probability that exactly 1 urn has 3 balls is $1/8+11/32=1 5/32$, the probability that 2 urns have 3 balls is $3/128$, and the probability that at least 1 urn has at least 3 balls is $1/8+11/ 32+29/384+3/128+13/384=77/128$.