19 April 2021

Answer to the marbles in Brownian motion puzzle…

The easiest way to think about this problem is not to think of number of marbles in each color but the differences in number between colors. Let’s say “a”, “b” and “c” are the pairwise differences in colors. Their values are 10, 10 and 20 to begin with.

Let’s see what happens whenever there is a “hit”.
If the marbles are of same colors, of course no change in count happens – so a, b and c remain the same.
If the marbles are of different colors – then those two colors go down by 1 each and the third color goes up by 2. Which means the pairwise differences either remain the same (the two colors that hit) or they change by 3 (one color goes down by 1 and the resulting color goes up by 2).

So, the first concept to realize is that a, b and c always change in steps of 0 or steps of 3. There are no other step changes they can have.

Now, let’s say what if the marbles actually did become all of one color. What as the step right before it? We would have had to have all marbles of the same final color excepting two – one each of the other two colors. Thus if finally everything is green, in the penultimate step, there has to be one red, one blue and rest green. The red hits the blue to become green.

Note that in that penultimate step, the pairwise difference between two colors (that have one marble left each) is 0. And that is the second concept to realize.

Now putting these two concepts together…

Well, we started with a, b and c being 10, 10 and 20. Going in steps of 0 and 3, can any of these numbers ever get to 0? No. Therefore it will never reach the penultimate step.

Which means the marbles will never be of the same color !!

Posted April 19, 2021 by Rajib Roy in category "Puzzles

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