*Originally posted by tomtom232*

**I think 1 is wrong.
**

A) there is a 50% chance that the black ball you pull out is the original ball.

B) if it is the original ball it is 100% chance that the next ball is black.

C) if it is not the original ball then it is an even chance that the next ball is black or white.

D) this means that only 25% percent of the time the second ball will b ...[text shortened]... e W is the probability of drawing a white ball and B is the probability of drawing a black ball.

For clarification: the "original ball" is the one that can be either black or white.

I disagree with your A).

Try this:

A*) There is a bag. Inside are two balls. One is black. The other is either black or white.

B*) The unknown ball has equal chances to be white or black.

C*) We remove a ball from the bag.

D*) When we pulled the ball from the bag, there were two possibilities:

Da*) We pull the guaranteed black ball.

Db*) We pull the unknown ball.

E*) Da*) and Db*) are equally likely.

F*) Now, because there are two possible colors for the second ball, there are four equal possibilities (2x2) before we look at the color of our first pick:

Fa*) Black/White, pull the unknown. It is white.

Fb*) Black/White, pull the known. It is black.

Fc*) Black/Black, pull the unknown. It is black.

Fd*) Black/Black, pull the known. It is black.

G*) We look at our ball. It is black. Therefore Fa*) is not true, leaving us with three possibilities.

H*) If...

Fb*), the second will be white.

Fc*), the second will be black.

Fd*), the second will be black.

Wait, this seems to give me a 67% chance. Hmmm.

I think my mistake was not realizing black/black has two out of two possible ways to pick black first, while white/black has one out of two possible ways to pick black first. If I did not know the ball I picked was black my 50% would be correct I think.