logic puzzle - black, white, and black/white ball boxes

[Komag]

(this is a puzzle that was used in hiring interviews for Cambridge, Mass start-up, Panjiva Inc.)

 

- Imagine you are looking at three cardboard boxes.

- One box contains all white balls, one all black balls, and one a mix of black and white balls.

- Each box is labeled, but the labels are all wrong.

 

How many balls would you need to pull out to determine which box is which?

[Chiron]

Answer:

 

 

It takes one pull.

 

 

 

 

Explanation:

 

 

 

 

The key is the assumption that the labels are all wrong.

 

You draw one ball from the Box labeled "BW"

Since the label is wrong you know it cannot be the mixed Box (BW), and depending on which color you draw you can also exclude the box with all white (WW) or all black balls (BB).

 

You now know the content of the one box you've pulled from, which cannot be the content

of the other boxes, and you also know what the other boxes cannot contain due to the wrong labeling. So there is only one possibility for each of the two remaining boxes left.

 

 

 

Edited September 12, 2010 by Chiron

[Komag]

you got it!

[demagogue]

Not having read the spoiler, I'll take a crack at it:

 

 

You first pull 1 out of the 'Mixed' box, knowing it must be one of the unmixed. Whatever color it is (say white for now) is in that box. Then you have the 'White' & 'Black' boxes, one of which you know is mixed, the other being the opposite-color in 'Mixed'. If it was white in there, then you have black left. It can't be in 'Black', so black must be in 'White', leaving mixed to be in 'Black'. If black had been in 'Mixed', then for the same reason white is in 'Black', leaving mixed to be in 'White'. Either way, you only need to pull out that 1 ball in the mixed bag to solve the whole thing.

 

 

This was a pretty good one. If it's ok, I know a similar riddle like this one I might post. But I have to find it first ... so I'll post it later.

 

Edit: Ok, here is the similar riddle I remember. If it's cool, I'll post it as another challenge along with Komag's for this thread.

 

There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat is it and how can this be?

 

Edit2: I got this riddle from a good collection - here: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml

[Komag]

 

There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat is it and how can this be?

 

 

okay, let me walk through this one writing it out as I think it through...

 

 

 

Okay, so the only way A COULD know what color he was wearing is if B and C both had white on, thus A would know he must be wearing Black. so B and C must not both be white (but one could be). B has heard this, so he knows that he and C must either be both black or one black and one white. The only way B COULD know what color he has on is if C is wearing white, because in that case he knows he CAN'T be wearing white too or else A would known what color A was wearing; thus if C is wearing white B knows that B MUST be wearing black. Since B's NOT sure what color B is wearing, it must be because he sees a Black on C, not white, and thus his own hat could be either white OR black. So, A definitely doesn't see two whites on B and C, and B definitely sees black on C. Thus C knows he's wearing black!

 

 

 

I think that's it!

[Chiron]

you got it!

 

Yay

 

But in a test with time pressure I would have failed miserably. For 20 minutes I drew trees with possibilities, calculating the minimum and maximum number of pulls if you only knew the number of the balls and so on....

[Fidcal]

Yeah I would have failed the time test. This is one of those that in a list of questions you do the quick ones first and come back (unless you immediately see it of course.)

 

I had another alternative:

 

None. Just look in the boxes. If you were at a job interview you might be commended for initiative as there was no rule against it. Or not.

 

These IQ questions raise other interesting ideas such as the way we use words. This question is intended to be literally correct but unless the question page states that who is to know? For instance most people might use the word 'all' in this context in a different way. I suspect the question relies on that assumption to some extent so many people won't even think about the labels. eg, "Damn! Harry just stuck the labels on any box so all the labels are mixed up" (meaning they are random so one might be correct; they might actually all be correct.)

 

Another possibility for error is to find a valid solution but you cannot be certain there is not a better solution. So you would get zero IQ points for what is an intelligent and useful solution and you might arrive at this solution faster:

 

 

Three.

 

Take one from each and you are certain to have two the same and one different; let's say two white and one black.

 

Whichever is single must be its color so the one black must be all blacks.

 

Since the labels are ALL wrong then take the black label from its original and put it near the true black box.

 

The 'remaining' box label must also be wrong so replace it with the only other one available which is the original from the black and put its own label on the box which was originally labelled black.