I consider the following sceneriou. Three balls enter the box, $B_1, B_2, B_3$, the balls are independent and identical. Next, we draw the balls from the box, one after another, waiting an exponential amount of time, we call them $P_x, P_y, P_z$. Since the time the balls spend in the bin is from exponential distribution, i use the memoryless property, that B_1, B_2, B_3 are equaly probably to leave the bin at any given time (so I tried that uniformly). I'm trying to compute the following probabilities.
$$ Pr[P_x = B_1], Pr[P_x = B_2], Pr[P_x = B_3] Pr[P_y = B_1], Pr[P_y = B_2], Pr[P_y = B_3] Pr[P_z = B_1], Pr[P_z = B_2], Pr[P_z = B_3] $$
My intuition is, that all events should have equal probability 1/3. So I computed the following:
$$ Pr[P_x = B_1] = Pr[P_x = B_2] = Pr[P_x = B_3] = 1/3 $$ $$ Pr[P_y = B_1] = Pr[B_1 ~\text{didn't leave as $P_x$}] \cdot Pr[B_1 ~\text{leaves now}] = (1-1/3) \cdot 1/2 = 1/3 $$ The same for $Pr[P_y = B_2] = Pr[P_y = B_3] = 1/3$ $$ Pr[P_z = B_1] = Pr[B_1 ~\text{didn't leave as $P_x$}] \cdot Pr[B_1 ~\text{didn't leave as $P_y$}]\cdot Pr[B_1 ~\text{leaves now}] = (1-1/3)\cdot(1-1/2)\cdot 1 = 1/3 $$ And the same for $Pr[P_z = B_2] = Pr[P_z = B_3] = 1/3$
However, in those calculations there is one thing which is not consitent. I computed that $$Pr[P_y = B_1] = 1/3$$, so the probability that $B_1$ didn't leave as $P_y$ should be 1 - 1/3, however when i do this step by step my inuition tells me that $$Pr[B_1 ~\text{didn't leave as $P_y$}] = (1-1/2)$$ (see computation for $Pr[P_z = B_1]$). Can someone suggest where am I doing a mistake?
Where you found $\frac12$ as outcome you actually did not calculate $Pr(B_1\text{ didn't leave as }P_y)$ there, but you calculated $Pr(B_1\text{ didn't leave as }P_y\mid B_1\text{ didn't leave as }P_x)$.
Besides. Your intuition concerning the probabilities $P[P_x=B_i]=\frac13$ is also valid for $P_y$ and $P_z$. It is immediate that also $P[P_y=B_i]=\frac13$ and $P[P_z=B_i]=\frac13$ for $i=1,2,3$.
Your calculations using conditional probabilities are in fact redundant.