There are $5$ bags of colours Blue, Green, Pink, Red and yellow. Each of these bags have $5$ balls of same colour as that of the bag. A ball is drawn from blue bag and is transferred to one of the other bags. Then, a ball is drawn from green bag and transferred to one of the other bags. Similarly, $1$ ball is transferred from the pink bag, then from red bag and then from yellow bag (to one of the other bags). Find the probability that at the end each bag has $5$ balls of same colour, given that there are $5$ balls in each bag at the end.
My working: To keep the number of balls same in each bag, none of the bags must receive more than 1 ball i.e., $𝐷_5=44$ cases to do it.
All Favourable Cycles:
(1) $B\to G\to B; P\to R\to Y\to P$
(2) $B\to P\to B; G\to R\to Y\to G$
(3) $B\to R\to B; G\to P\to Y\to G$
(4) $B\to Y\to B; G\to P\to R\to G$
(5) $B\to G\to P\to B; R\to Y\to R$
(6) $B\to G\to R\to B; P\to Y\to P$
(7) $B\to G\to Y\to B; P\to R\to P$
(8) $B\to P\to R\to B; G\to Y\to G$
(9) $B\to P\to Y\to B; G\to R\to G$
(10) $B\to R\to Y\to B; G\to P\to G$
(11) $B\to G\to P\to R\to Y\to B$
Now,
P(Each bags content remain same/Each bag has 5 balls at the end)
$=\frac{\left[10\times 1\times \left(\frac16\right)^3\times \left(\frac14\right)^5+1\times\left(\frac16\right)^4\left(\frac14\right)^5\right]}{44\times\left(\frac14\right)^5}=\frac{61\times \left(\frac16\right)^4\times \left(\frac14\right)^5}{44\times\left(\frac14\right)^5}=\frac{61}{57024}$
Best Answer
As suggested by Daniel Mathias, here is consideration based on the transfer of the blue ball.
However, I am afraid that I may make careless mistakes. Please tell me if I do so. Thanks in advance.
Case (1): the blue ball goes to only one other bag and then come back to the blue bag.
For example, the blue ball goes to red bag and then go back directly to the blue bag.
Since the first bag is arbitray, it has probability $1$. The second draw, going back to blue bag, has probability $\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right) $.
We have to consider the transfer of balls in the remaining $3$ bags, e.g. Green, Pink and Yellow.
If a green ball is transferred to the yellow bag, this green ball has to go back to the green bag. But then we cannot transfer any ball in the pink bag.
So the green ball must go to the pink bag, then the yellow bag and finally come back to the green bag.
This has a probability of $\left( \frac{1}{4} \right) $$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right) $$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right)=\left( \frac{1}{4} \right)^3 $$\left( \frac{1}{6} \right)^2 $.
The overall probability in this case is $\left( \frac{1}{4} \right)^4 $$\left( \frac{1}{6} \right)^3$.
Case (2), the blue ball goes to exactly 2 other bags and then come back to the blue ball.
For example, it goes to pink bag, then yellow bag and back to blue bag. There are $6$ such cases, namely GP, GR, GY, PR, PY and RY. Each has probability $\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right) $$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right) $$\left( \frac{1}{4} \right)=\left( \frac{1}{4} \right)^3 $$\left( \frac{1}{6} \right)^2 $.
(Notice that we can't have cases like RP. This is because once the blue ball is transferred to the red bag, it cannot go back to the pink bag because the withdrawal of ball in the pink bag has been done.)
For the remaining $2$ bags, e.g. Red and Yellow. The red ball must go to the yellow bag and then come back.
This has a probability of $\left( \frac{1}{4} \right) $$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right) =\left( \frac{1}{4} \right)^2 $$\left( \frac{1}{6} \right)$.
The overall probability in case $(2)$ is $6 \times $$\left( \frac{1}{4} \right)^3 $$\left( \frac{1}{6} \right)^2 $$\times $$\left( \frac{1}{4} \right)^2 $$\left( \frac{1}{6} \right)$$=\left( \frac{1}{4} \right)^5 $$\left( \frac{1}{6} \right)^2$
Case $(3)$, the blue ball goes to exactly 3 other bags and come back to the blue bag.
For example, it follows the path Green, Pink, Yellow and then back to Blue.
But this is impossible because, under this condition, the ball in the red bag cannot go to anywhere.
Finally case (4), the blue ball goes to exactly 4 bags and come back.
It must go through Green, Pink, Red, Yellow and back to Blue, in that order. The probability is $\left( \frac{1}{4} \right) $$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right)$$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right)$$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right)$$\left( \frac{1}{4} \right) $$\left( \frac{1}{6} \right)$$=\left( \frac{1}{4} \right)^5 $$\left( \frac{1}{6} \right)^4$
Combining all cases, the final answer is
$\frac{\left( \frac{1}{4} \right)^4 \left( \frac{1}{6} \right)^3+\left( \frac{1}{4} \right)^5 \left( \frac{1}{6} \right)^2+\left( \frac{1}{4} \right)^5 \left( \frac{1}{6} \right)^4}{D_5\times \left( \frac{1}{4} \right)^5}= \frac{61}{44\times 6^4}$