There are four balls in a hat: a blue one, a white one, and two red ones. Now I draw simultaneously two balls, look at them, and announce that at least one of them is red.

What is the chance that the other is red as well?

(In reply to Who's right? by Ady TZIDON)

Is it so simple?

One method includes numbering the red balls 1 and 2. I'm not sure if that is right, if they are identical. But assume it is, and they are red 1 and red 2:

permute 2 from {1,2,3,4} {1, 2} | {1, 3} | {1, 4} | {2, 1} | {2, 3} | {2, 4} | {3, 1} | {3, 2} | {3, 4} | {4, 1} | {4, 2} | {4, 3} (total: 12)

Order doesn't matter, so we obtain:{1, 2} {1, 3}{1, 4}{2, 3}{2, 4}  {3, 4} (in some order). This results in the 1/5 answer.

But since the reds are identical:

permute 2 from {1,1,3,4} {1, 1} | {1, 3} | {1, 4} | {3, 1} | {3, 4} | {4, 1} | {4, 3} (total: 7)

Order doesn't matter, so we obtain: {1,1,3,4} {1, 1} {1, 3} {1, 4} {4, 3} (in some order). This gives the 1/3 answer.

But if we consider 'red and non-red', it is different again:

permute 2 from {1,1,2,2} {1, 1} | {1, 2} | {2, 1} | {2, 2} (total: 4)

Order doesn't matter, so we obtain: {1,1,2,2} {1, 1}{1, 2}  {2, 2}. This gives an answer of 1/2.

All that being said, I did experiment with 2 blue balls (B) and one yellow one (Y) and the results of 100 draws were 63BY and 37BB. By parity of reasoning, Nature does seem to distinguish between ball Blue 1 and Blue 2, so the 1/5 answer would seem to be preferable.

Edited on May 20, 2017, 3:00 am

Posted by broll on 2017-05-20 02:56:51