very nice question... got me thinking

here is my answer and train of thought:

I refrased the question to make it more clear. It should be the same as asking the following:

There is a box with 4 balls inside. 1 BLUE, 1 GREEN, and 2 RED.

If someone randomly says one of those three colours what is the chance that you will pick that exact colour (assuming you cant see throught the box and you are picking at random..)

There are 3 colours, so the person who is specifying what colour you need to pick out has a 1/3 chance to choose each colour, in other words, after many repeats each of the colours will be chosen the same amount.

If you repeat this process A LOT of times, you should get the follwing success probability.

-BLUE = 1/4 = 25%

-GREEN = 1/4 = 25%

-RED = 1/2 = 50%

Now simply take the average of these probabilities:

((1/4) + (1/4) + (1/2)) / 3 = 1/3 = 33.33...%

So the ''chance you'll be correct'' by randomly choosing A,B,C,or D is 1/3

CASE CLOSED!

... unless:

- what if there is more then one correct answer (colour)

- 33.33..% is not one of the proposed answers, so should the final answer be 0%?

- my brain hurts.