## Ball bag

Posted by Chris on October 15, 2013 – 3:50 pm

A bag contains either a white ball or a black ball, and each case is equally likely. A white ball is added. You now randomly take a ball from the bag. If it is white, what is the probability that the remaining ball is white?

I’m not sure if this has been posted before.

October 16th, 2013 at 1:39 am

2/3

October 16th, 2013 at 7:47 am

Agreed. This is similar to the “black or white?” logic.

October 16th, 2013 at 2:53 pm

I can’t see the similarity to any of the “Black or White” problems.

Lewis Carroll invented this problem. He gave two solutions. The first (and simplest) is that upon removing the white ball, the bag must then be in its original state, and so the probability of the remaining ball being white is 1/2.

October 17th, 2013 at 1:02 pm

I guess my thought is that if you were to do this 100 times you would get the following results (on average):

a) 25 times you start with a BLACK ball in the bag, you add a WHITE ball and pull the BLACK ball out.

b) 25 times you start with a BLACK ball in the bag, you add a WHITE ball and pull the WHITE ball out.

c) 25 times you start with a WHITE ball in the bag, you add a WHITE ball and pull it directly back out.

d) 25 times you start with a WHITE ball in the bag, you add a WHITE ball and pull the original WHITE ball out.

Note: Since you are equally likely to start with either a WHITE of BLACK ball, I’ve set it up so you start with each 50 times out of 100.

Since the problem statement says we pull a WHTIE ball first we can ignore case (a). Now (b), (c), and (d) are equally likely since they each happened 25 out of 100 times, and 2 of them [(c) and (d)] have a remaining WHITE ball in the bag. [(c)+(d)]/[(b)+(c)+(d)] = [25+25]/[25+25+25] = 50/75 = 2/3

October 17th, 2013 at 3:05 pm

Hi DP. Are you disagreeing with Lewis Carroll’s analysis? Where is the fault in it?

October 17th, 2013 at 5:38 pm

… I was trying to wind you up a bit. 2/3 is the correct answer and Lewis Carroll deliberately gave a fallacious argument. You have covered it in your analysis, but can you say where Lewis Carroll’s argument was in error?

October 18th, 2013 at 9:15 am

The phrase “upon removing THE white ball” is ambiguous. If indeed we had removed THE ADDED white ball, the argument would be correct. But if we had removed the unknown ball, and it happened to be white, then the bag would not be in its original state (with a 50-50 chance of the ball being black or white) and the remaining ball would be white (with a probability of 1).

Lewis Carroll’s second (and correct) explanation was along the lines: let b or w represent the original ball and W represent the added ball. Then after drawing a ball we must have one of the following equally likely cases

and each case is equally likely. But we are told that the drawn ball is white, so we can eliminate the first possibility. Of the three remaining possibilities, two have the remaining ball being white, so the probability of the remaining ball being white is 2/3.

October 18th, 2013 at 10:38 am

I’m glad I didn’t fall for it. I see now that you alluded to having a second solution. Little did I realize that you meant a solution that provided a different answer. Nice ToM.

When I first saw the problem, I thought of 2 coins, which then reminded me of the “black or white?” card problems.

In my coin example you would have a bag with a normal heads-tails coin, then add a double-sided heads coin. Now you blindfold yourself, reach in and grab a coin, then flip. Now when you take off your blindfold and see that it is heads, what is the probability that the other side is heads?

Almost identical to the “black or white?” problem, only I left out the double-sided tails coin.

By the way, I like the addition of spell-check within the reply window through the blog site. Much better than having to go to the main site, or worse, having to copy-paste into a word editor and back again. I am a terrible speller. In fact, I’ve already edited this post…

October 18th, 2013 at 3:35 pm

Hi DP. I’m also glad you didn’t waver (publicly at least).

It turns out I had misread the source site. Carroll didn’t give the argument I gave above. His argument for 2/3 wasn’t so neat.

Of course he was fully aware that 1/2 was wrong and 2/3 was right.

October 22nd, 2013 at 6:35 am

Well….it looks like no one else is going to comment. Jan typically has some pretty sound solutions, but gave none here. I’m guessing to give others (like me) a chance.

So Chris, what IS the reason Carroll gave for 2/3?

October 22nd, 2013 at 1:02 pm

Hi DP. I hadn’t intended to show Carroll’s inelegant workings. Sadly I haven’t found the site that definitely shows it. However, the following is the sort of thing he did:

The chances, before the addition, that the bag contains (a) 1 white (b) 1 black, are (a) 1/2 (b) 1/2. Hence the chances after the addition, that it contains (a) 2 white (b) 1 white, 1 black, are the same, viz. (a) 1/2 (b) 1/2. Now the probabilities, which these 2 states give to the observed event, of drawing a white counter, are (a) certainty (b) 1/2. Hence the chances, after drawing the white counter, that the bag, before drawing, contained (a) 2 white, (b) 1 white, 1 black, are proportional to (a) 1/2· 1 (b) 1/2 · 1/2; i.e. (a) 1/2 (b) 1/4; i.e. (a) 2 (b) 1. Hence the chances are (a) 2/3 (b) 1/3. Hence, after the removal of a white counter, the chances, that the bag now contains (a) 1 white (b) 1 black, are for (a) 2/3 and for (b) 1/3.

Thus the chance, of now drawing a white counter, is 2/3.