I would like to know if my partial solution is appropriate:
We have ten urns, in each urn we have 2 balls:
in 1st urn - 2 balls of number 1
in 2nd urn - 2 balls of number 2
.
.
in 10th urn - 2 balls of number 10
In first step, we take one ball from 1st urn (randomly) and throw it to the second one. Then from the second we take one randomly (from 3 that are currently in the urn) and throw it to the third one etc... and in the last (10th) step we take one ball.
What is the probability that the last ball is greater than 6. There is my partial solution:
X - last ball greater than 6
$M_i$ - in the last step we have two balls of number 10 and one of number $i$
$P(X) = \sum_{i = 1}^{n = 9} P(X | M_i)P(M_i)$
For $i > 6$ $P(X | M_i) = 1$.
$\sum_{i = 1}^{n = 9} P(X | M_i)P(M_i) = \sum_{i = 1}^{n=6}P(X | M_i)P(M_i) + \sum_{i = 7}^{n=9} P(M_i)$
$P(X|M_i) = \frac{2}{3}$ for $i <= 6$
$P(M_i) = (\frac{2}{3} * (\frac{1}{3})^{9 - i}$ for i > 1
Is it ok?
After the first six transfers, there will be three balls in the seventh container. Two of them will display a $7$ and the other will be a number that is at most $6$ since it must have come from one of the first six containers.
The only way we could fail to select a number that is greater than $6$ at the end is if the one ball that is left with a number smaller than $7$ is repeatedly transferred to the eighth, ninth, and tenth containers, and then is selected from the tenth container. Let $S$ be the event that the ball labeled with a number smaller than $7$ is selected from the tenth container. Let $S_i$, $i \leq 7 \leq 9$, be the event that the ball selected from the $i$th container is labeled with a ball smaller than $6$. Then \begin{align*} Pr(S) & = \Pr(S_7)\Pr(S_8 \mid S_7)\Pr(S_9 \mid S_8 \cap S_7)\Pr(S \mid S_9 \cap S_8 \cap S_7)\\ & = \left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\\ & = \left(\frac{1}{3}\right)^4 \end{align*}
Note that $\Pr(S_7) = 1/3$ since one of the three balls in the seventh container is labeled with a number with a number smaller than $7$. If that ball is transferred to the next container, then the probability of selecting it from the next container is also $1/3$ since one of the three balls in that container will be labeled with a ball with a number smaller than $7$. If a ball with a number smaller than $7$ is transferred to the tenth container, the probability of drawing it from that container will again be $1/3$ since one of the three balls in the tenth container will be the ball labeled with a number smaller than $7$.
Hence, the probability that a ball with a number larger than $6$ is selected at the end is $$\Pr(S^C) = 1 - \Pr(S) = 1 - \left(\frac{1}{3}\right)^4$$