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Suppose there are $6$ balls with label $1-6$. There are two boxes, box A and B. Each box can contain exactly $3$ balls. At the beginning, both boxes are empty.

I pick up the ball with label $1$ and throw it into one of the two boxes with probability $\frac{1}{2}$.

Then I pick up the ball with label $2$ and throw it into one of the two boxes with probability $\frac{1}{2}$.

Then I pick up the ball with label $3$ and throw it into one of the two boxes with probability $\frac{1}{2}$.

Then I pick up the ball with label $4$, but there are two cases: (a) if the first $3$ balls lands in the same box, then there is only $1$ box remaining and the probability that label $4$ lands into one of the two boxes is $1$ or is it $0$ since it can't go to either of the two boxes? (b) if the first $3$ balls does NOT land in the same box, then the probability that label $4$ lands into one of the two boxes is $\frac{1}{2}$.

Then I pick up the ball with label $5$ and there are also two cases: (a) if any box is fulfilled with exactly $3$ balls, then there is only $1$ box remaining and the probability that label $5$ lands into one of the two boxes is $1$ (or $0$?); (b) if NO box is fulfilled with exactly $3$ balls, then the probability that label $5$ lands into one of the two boxes is $\frac{1}{2}$.

Then I pick up the ball with label $6$ and there is only $1$ box remaining. The probability that label $6$ lands into one of the two boxes is $1$ (or $0$?).

The $6$ balls will be randomized to the $2$ boxes in this way. For a reason, I will mark the fastest ball of each box with green color.

That is, at the very beginning when both boxes were empty, suppose label $1$ ball lands to box B and there was no ball before. So I will mark it.

If label $2$ lands to the box A where there was no ball before, then I will mark label $2$ and I haven't any concern about the remaining balls. But if label $2$ lands to the same box with label $1$, then I have nothing to do with label $2$. And I will wait for the first landing ball in the empty box A. As soon as I will get the first landing ball in the empty box A, I will mark it and no more concern about the other remaining balls.

Now my question is:

If the balls are thrown in order of their labels, what is the probability that I will mark the ball with label $i$, where $i=1,2,\ldots,6$?

I know here the probability that the label $1$ will be marked is always $1$. The probability that the label $5$, $6$ will be marked is always $0$.

1 Answers 1

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You already did the cases $i=1,5,6$.

The probability of marking ball $2$ is the probability that the second ball goes to the unoccupied box. So it is $1/2$.

In order to mark ball $3$, you need ball $2$ to go into the box that contains ball $1$, and then ball $3$ to go to the other box. The probability is $(1/2)(1/2)=1/4$.

In order to mark ball $4$, you need balls $2$ and $3$ to go into the box that contains ball $1$. Then ball $4$ automatically must go to the other box. The probability is $(1/2)(1/2)=1/4$.

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    I am not understanding what is the first, the second and the last half in $(1/2)(1/2)(1/2)=1/8?$ – 2017-01-22
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    $1/2$ chance that second ball goes in the box with the first ball, $1/2$ chance that third ball goes into same box, and $1/2$ chance that fourth ball goes in other box. – 2017-01-22
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    if ball $1$, $2$, $3$ go to the same box, then why $1/2$ chance that fourth ball goes in other box? – 2017-01-22
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    @ABC You are right, I've corrected my mistake. – 2017-01-22
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    In order to mark ball $3$, $1/2$ chance that second ball goes in the box with the first ball and $1/2$ chance that third ball goes to the other box. That's why the probability to mark ball $3$ is $(1/2)(1/2)= 1/4$. Have I correctly interpreted it? – 2017-01-22
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    @ABC That's correct – 2017-01-22