Balls

A bag contains a ball which is known to be either black or white (even chance). A black ball is added. The balls are then drawn from the bag one at a time.

What is the probability of the second ball being black
1. if the first is black?
2. if the first is white?
3. if the colour of the first is unknown?

If the ball originally in the bag is white, there are two possibilities for drawing order:

W B
B W

if the original ball is black:

B B
B B

Therefore, if the first ball draw is white: 100% that next ball is black
if first ball drawn is black: 66% that next ball is black
if first ball drawn is unknown: 75% that next ball is black

Agreed. So what is wrong with this?

I pull a black ball first.

I know it is either the original unknown ball or the black one I put in. Two balls in the bag so either is just as likely. If it was the original ball then the remaining ball MUST BE black. If it was the ball I put in then its 50/50 if its black.

Therefore the chance of second ball being black is
(0.5*1) + (0.5*0.5) = 0.75

... not 0.66

Originally posted by wolfgang59
A bag contains a ball which is known to be either black or white (even chance). A black ball is added. The balls are then drawn from the bag one at a time.

What is the probability of the second ball being black
1. if the first is black?
2. if the first is white?
3. if the colour of the first is unknown?

1. a) 50% chance of black/black, b) 50% chance of black/white. In a), it's 100%. If b), it's 0%. Thus the answer is 50%.

2. 100%

3. You can pick ball A or ball B first. Ball A is black. Ball B can be either black or white.

If you pick A, then it's 50%. If you pick B, it's 100%. 50% x 50% = 25%. 50% X 100% = 50%. 25% + 50% = 75%.

Originally posted by AThousandYoung
1. a) 50% chance of black/black, b) 50% chance of black/white. In a), it's 100%. If b), it's 0%. Thus the answer is 50%.

No, it's a 25% chance of Black1/Black2, 25% of Black2/Black1, and 25% of Black/White, therefore, in 2/3 cases the next ball will be black. You can't count both black balls as being the same ball.

You can think of it this way, instead of having 1 ball with a 50/50 chance of being black/white, use two balls, one black and one white.

Then, instead of adding a single black ball to the bag, add two black balls, you can think of each ball being one "side" of the original two balls.

Now calculate your probabilities.

Originally posted by forkedknight
No, it's a 25% chance of Black1/Black2, 25% of Black2/Black1, and 25% of Black/White, therefore, in 2/3 cases the next ball will be black. You can't count both black balls as being the same ball.

You are mistaken. We know one of them is black. If you want to use your sort of analysis:

Case A) Black1/Black2
Case B) Black1/White2
Case C) White1/Black2
Case D) White1/White2

Which one do we want to label as ball #1? It doesn't matter, but let's try both ways.

If Ball1 is the guaranteed black one, and Ball2 is the one that can be white or black, then Cases C) and D) are 0%.

If Ball2 is the guaranteed black one, then Cases B) and D) are 0%.

In both, the other two Cases are equally probable.

I believe your mistake is labelling the balls as 1 and 2 when they're both black, but failing to do so when one is white.

Originally posted by forkedknight
You can think of it this way, instead of having 1 ball with a 50/50 chance of being black/white, use two balls, one black and one white.

Then, instead of adding a single black ball to the bag, add two black balls, you can think of each ball being one "side" of the original two balls.

Now calculate your probabilities.

Why complicate things like that?

Originally posted by AThousandYoung
1. a) 50% chance of black/black, b) 50% chance of black/white. In a), it's 100%. If b), it's 0%. Thus the answer is [b]50%.

2. 100%

3. You can pick ball A or ball B first. Ball A is black. Ball B can be either black or white.

If you pick A, then it's 50%. If you pick B, it's 100%. 50% x 50% = 25%. 50% X 100% = 50%. 25% + 50% = 75%.[/b]

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 be white and 75% black because

W=.5*.5 = .25 and B= (.5*1.0)+W= .75

Where W is the probability of drawing a white ball and B is the probability of drawing a black ball.

Edit2: I believe what your are calculating in 1 is the probability of black/white or white/ black vs black/black since 100% of the time one ball will be black and the other ball is 50% so you have 50% chance of drawing a white IF you draw both BUT you don't take into account the chances if the first one is already black. This means that the scenario of white/black is impossible lowering the % of drawing a white ball.

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.

Originally posted by wolfgang59
Agreed. So what is wrong with this?

I pull a black ball first.

I know it is either the original unknown ball or the black one I put in. Two balls in the bag so either is just as likely. If it was the original ball then the remaining ball MUST BE black. If it was the ball I put in then its 50/50 if its black.

Therefore the chance of second ball being black is
(0.5*1) + (0.5*0.5) = 0.75

... not 0.66

The problem is that you would not be factoring in the conditional probability correctly. The event of drawing an initial black ball is more likely in the case of black/black than in the case of white/black. The event of drawing a black ball will then provide us with information about which balls are in the bag.

P(2nd = B|1st=B) = (2nd = B and 1st = B)/P(1st = B)

We need:
P(1st = B) = P(1st = B|BB)*P(BB) + P(1st = B|WB)*P(WB) = 0.5 + 0.5*0.5 = 0.75

P(2nd = B and 1st = B) = P(2nd = B and 1st = B|WB)*P(WB) + P(2nd = B and 1st = B|BB)*P(BB) = 0 + 1*0.5 + 0 = 0.5

=> P(2nd = B|1st=B) = 0.5/0.75 = 2/3

Originally posted by AThousandYoung
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 ...[text shortened]... k black first. If I did not know the ball I picked was black my 50% would be correct I think.

Except in scenario 1 the first ball you pull out IS black AND has a 50% chance of being the original ball in the bag.

Originally posted by tomtom232
Except in scenario 1 the first ball you pull out IS black AND has a 50% chance of being the original ball in the bag.

I don't think you're taking into account what Palynka just expressed very well:

The event of drawing a black ball will then provide us with information about which balls are in the bag.

Once you know the ball is black, the probability that you're holding the original ball changes and is no longer 50%, since you now know that you can rule out the possibility that you're holding a white ball.

Originally posted by AThousandYoung
I don't think you're taking into account what Palynka just expressed very well:

The event of drawing a black ball will then provide us with information about which balls are in the bag.

Once you know the ball is black, the probability that you're holding the original ball changes and is no longer 50%, since you now know that you can rule out the possibility that you're holding a white ball.

Ah. Well then #3 would be .75*.66 or 50% to draw black then black

25% chance to draw white then black

And 25% chance to draw black then white so 75% is the chance that the second will be black when the first ball is unknown.