Party Time!!

There will always be the same number of black b***s in the first bag as red b***s in the second, regardless of how many black b***s you happen to transfer from the second to the first bag! Perhaps the best way to see this is by means of a simple example. If you removed 3 red and 9 black b***s from the second bag and placed them in the first bag, there will be 9 black b***s in the first bag. But, by taking only 3 red b***s across, you have left 9 red b***s behind in the second bag!.

I'll give this a shot

Afer taking 12 red b***s from the first bag and putting them into the second :

The first bag conains 18 b***s  [all red ]

And the second bag now contains :

 72 b***s..... 60 black,  and 12 red  =   5/6 black and 1/6 red

Assuming the second bag is well−mixed and taking 12 b***s from it and putting hem back into the  first bag.:

We should select 10 black b***s  (5/6)*12 and 2 red ones (1/6)*12 to be put back into the first bag

So the second bag [ theoreically] now contains   50 black and 10 red

And the first bag should now contain 10 black and 20 red

So.......the first bag should contain as many black b***s as the second bag contains red ones !!!