Probability question about changing balls between baskets In Basket $A$ we have $2$ Blue and $1$ Red Balls and in Basket $B$ we have $3$ Blue and $3$ Red Balls. we randomly choose 2 balls from each basket (2 from $A$ and 2 from $B$). Then we put the balls from $A$ into $B$ and the balls from $B$ into $A$. Then, We chose one of the baskets randomly and pick a random ball from it. What is the probability that the chosen ball is Blue?
I know a method for solving this but it takes a lot of time. It is examining each condition when 2 blue is selected from $A$, 1 blue and 1 red is selected from $A$ (in order), 1 red 1 blue is selected from $A$ and like this for $B$. At all it is about 12 different conditions.


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*I want a more intelligent solution for this. Not examining all the different conditions. A solution which is applicable on other type of this question.

*The final answer is $\frac{5}{9}$. It is like that we completely mixed the two baskets into one basket and then pick one ball from it. Why this happens? I mean why these too many different conditions lead to this simple answer? the question must have a straightforward answer, mustn't it?

 A: Suppose there are $n_1$ balls in bucket 1, and $n_2$ balls in bucket 2.  If we pick a bucket uniformly at random, then pick a ball from that bucket uniformly at random, what is the probability that a particular ball is the one picked?
It depends on which bucket the ball is in.  If the ball is in bucket 1, the probability is $\frac{1}{2n_1},$ but if the ball is in bucket 2, the probability is $\frac{1}{2n_2}.$
So in the given problem, the probability that a ball is picked depends on which bucket it ends up in, an that depends on which bucket it starts out in.  If the ball starts out in bucket 1, then it ends up in bucket 1 with probability $\frac{n_1-2}{n_1}$ and in bucket 2 with probability $\frac{2}{n_1}.$  The probability that the ball is picked is therefore $$
\frac{1}{2n_1}\frac{n_1-2}{n_1}+\frac{1}{2n_2}\frac{2}{n_1}=
\frac{1}{n_1}\left(\frac{1}{2}+\frac{1}{n_2}-\frac{1}{n_1}\right)\tag 1
$$  Symmetrically, the probability that a ball that starts out in bucket 2 is picked is $$
\frac{1}{n_2}\left(\frac{1}{2}+\frac{1}{n_1}-\frac{1}{n_2}\right)\tag 2
$$
In this problem, we have have $n_1=3, n_2=6.$  You can verify that both the probabilities in (1) and (2) work out to be $\frac{1}{9},$ so each ball has an equal probability of being chosen.
This is fortuitous, however.  If $n_1=4, n_2 = 5,$ then the probability in $(1)$ comes out to $\frac{9}{80}$ and that in $(2)$ comes out to $\frac{11}{100}.$
The formulas in $(1)$ and $(2)$ would allow you to work out the probability that a blue ball is chosen without having to go through every possible scenario, however.  
It's an interesting question whether there are any other $n_1,n_2\ge 2$ that exhibit the phenomenon you noticed (with $n_1\ne n_2,$ of course.)   
EDIT I just worked it out.  The only time this happens with buckets of unequal sizes is when one bucket has $3$ balls and the other $6.$
A: This post hopefully sheds some additional light on WHY the "coincidence" happens that the result is the same as if the two baskets are mixed together.  Let $n_A, n_B$ be the no. of balls originally in A and B, and let $n_C$ be the no. of balls transferred from each basket to the other.  (In the OP, $n_C=2$.) 
This problem is easier if you imagine the original balls from basket A are somehow different, e.g. smaller.  Let FBOA be the event that "the Final Ball ORIGINALLY came from basket A".  So:
Prob(final ball is Blue) = Prob(final ball is Blue | FBOA) * Prob(FBOA) + Prob(final ball is Blue | not FBOA) * Prob(not FBOA).
Note: Prob(final ball is Blue | FBOA) = 2/3, and Prob(final ball is Blue | not FBOA) = 3/6, and both depend only on the initial proportions of the two baskets (before any exchange).
Now, consider two experiments:
If you mix the two baskets together into a giant basket then pick a ball, then Prob(FBOA) = $\frac{n_A}{n_A+n_B} = \frac{3}{3+6} = \frac{1}{3}$.
Meanwhile, in the OP exchange-some-balls experiment: 
Prob(FBOA) = Prob(choose baseket A) * Prob(FBOA | choose basket A) + Prob(choose basket B) * Prob(FOBA | choose basket) = $\frac{1}{2} \frac{n_A-n_C}{n_A} + \frac{1}{2} \frac{n_C}{n_B} = \frac{1}{2} \frac{3-2}{3} + \frac{1}{2} \frac{2}{6} = \frac{1}{3}.$
So you see the "coincidence" happens because in both experiments, Prob(FBOA) = 1/3.  (Once this happens, the two experiments will have equal Prob(final ball is blue), regardless of the initial proportions in each basket.)
More generally this "coincidence" will happen whenever:
$$\frac{1}{2} \frac{n_A-n_C}{n_A} + \frac{1}{2} \frac{n_C}{n_B} = \frac{n_A}{n_A+n_B}$$
For any given $n_A,n_B$, this is a linear equation in $n_C$ with one solution: $n_C = \frac{n_A n_B}{n_A + n_B}$, or equivalently, $\frac{1}{n_C} = \frac{1}{n_A} + \frac{1}{n_B}$.  There are many integer solutions to this, e.g. $(n_C, n_A, n_B) =$ (11, 12, 132) in addition to the original (2, 3, 6) and scaled versions like (200, 300, 600).
Update: It turns out that the solution $n_C = \frac{n_A n_B}{n_A + n_B}$ makes all 3 ratios are equal: $\frac{n_A - n_C}{n_A} = \frac{n_C}{n_B} = \frac{n_A}{n_A + n_B}$.  This leads to a surprising consequence: the "coincidence" will happen even if you pick the final basket using a biased coin flip!  In that scenario, the 2nd experiment's Prob(FBOA) becomes $p_A \frac{n_A-n_C}{n_A} + (1-p_A) \frac{n_C}{n_B}$, which still equals $\frac{n_A}{n_A + n_B}$.
A: When you randomly pick $2$ balls from B, consider like you pick $2$ balls of a $50\%$ bluish tone,
while those from A will have a $66\%$ of blue.
At the end of the picking and exchange process you have
 - in A : 1ball $2/3$+2ball $1/2$ = 3balls with $(2/3+1)/3=5/9$ of tone
 - in B : 2ball $2/3$+4ball $1/2$ = 6balls with $(4/3+2)/6=5/9$ of tone.
So no difference in which basket you are going to pick.
Of course, the whole can be written rigorously by use of the expected value,
but I think you are asking for the intuition behind that.
