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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?

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  • $\begingroup$ You should write e.g. $P(P_z=B_1)=P(P_x\neq B_1)P(P_y\neq B_1\mid P_x\neq B_1)P(P_z=B_1\mid B_1\in\{P_x,P_y\})=\frac23\frac121=\frac13$ (the last factor can also be left out). $\endgroup$ – drhab Jul 2 '18 at 10:01
  • $\begingroup$ Typo in former comment: it must be $B_i\notin\{P_x,P_y\}$ instead of $B_i\in\{P_x,P_y\}$. $\endgroup$ – drhab Jul 2 '18 at 10:20
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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.

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  • $\begingroup$ Well, the fact that It is immediate is not so clear for me, i wanted to see the calculation behind it, or an explanaition, that's why i did my computation $\endgroup$ – SugerBoy Jul 3 '18 at 19:59
  • $\begingroup$ That is a good thing, since it is sensible to make calculations in order to confirm intuition. Drawing $3$ balls one after another can be looked at as drawing them at the same moment together with a randomly chosen element from $\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)\}$ determining their order. In this context there is no essential difference between "one by one" and "at the same time and order them randomly". In that light it might become more clear that we can speak of "immediate". $\endgroup$ – drhab Jul 4 '18 at 8:36

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