Solving this problem again involves reading it very
carefully and knowing that it means just what it
says. We are told exactly what is in each box and
also that all three labels are wrong (not just possibly
wrong). That is a lot of information, so let's draw a
grid as in problem 1, crossing off the possibilities
that any box is label correctly:
Labeled B-
B
Labeled B-
W
Labeled W-
W
Contains B-
B X
Contains B-
W X
Contains W-
W X
This problem now differs from the first in that we
must choose judiciously to fill in the rest of the
table in the fewest number of looks inside the
boxes. The best way I know to discover the answer
is to look for a choice where you are guaranteed to
learn something new. If you look first in the box
labeled B-B and you pull out a white marble, you
still haven't learned what that box contains because
it might be B-W or W-W. Similarly if you look in
the W-W box, a choice of B does not allow you even
to fill in one more cell of the table. But if you
choose the B-W box, then you are guaranteed to be
able to fill in one more cell, because if you draw out
a white marble, then that box must be the W-W. If
you draw out a black, then it must be the B-B. Let's
suppose it is white, and fill in the table:
Labeled B-
B
Labeled B-
W
Labeled W-
W
Contains B-
B X
Contains B-
W X
Contains W-
W O X
Now we can fill in the rest of column two and row
three because only one cell is left:
Labeled B-
B
Labeled B-
W
Labeled W-
W
Contains B-
B X X
Contains B-
W X
Contains W-
W X O X
And now, wonder of wonders, the first column only
has one empty cell, as does the first row, so we can
fill them in too:
Labeled B-
B
Labeled B-
W
Labeled W-
W
Contains B-
B X X O
Contains B-
W O X
Contains W-
W X O X
So the amazing thing is that by drawing only one
marble out of the box labeled B-W, we can deduce
what is in all three boxes!
If we had been bright enough to solve it without
the table we could have said that the B-W box must
be the W-W because we drew a white marble out of
it. That would mean the box labeled B-B must
contain the B-W because it is mislabeled and
cannot contain two black marbles. That leaves the
box labeled W-W to contain two black marbles.
Thus, the puzzle can be solved without using a grid,
but to me the grid method is more convincing
because the logic is so clear at each step.

The summary of my epistle is that by drawing out a single marble from the box labelled B-W, u can deduce the content of the other boxes



Google answereth all things. my 2 kobo

MoyoGENERAL:
one has two white, one has two black and one has one white and one black marble. .

What is the fewest number of marble you could remove from the boxes and look at in order to definitely determine the content of all the three boxes?

the answer is ONE! Cos there is ONE BLACK MARBLE.

MoyoGENERAL:
hmmm...you are way close bro...but you are still wrong
but you missed one important information.. the question said the boxes were labelled incorrectly!!!!!...

if the boxes were named incorrectly it summifies that there will be no correct answer to the incorrect question.

Wait oooooo let me break my fasting first so that I can be able to answer ur question.

5 marbles. I hope my logic is still working fine. Am I right?

currently chopping rice and beans,would be back when i'm done.

This is called Bertrand's box paradox in Mathematics (Simple probability theory)

More ideas before I help out

useless puzzle. it looks incorrect. or just useless

Let's define the boxes as follows
Box A = 2 kind each.
Box B = 2 kind each
Box C = 1 of each kind.

Minimum number is 3, and this is only valid if you picked the first 2 from Box C.

Explanation:
When Box C is eliminated (by picking the two tiles from there), you can can be sure after the third pick (since they contain similar tiles). Otherwise, if you picked the first two from Box A or Box B, and then the third tile from either of the remaining, then you cannot be certain.

Since you said cheating is allowed. Remove one marble and peep the contents of the remaining boxes to determine the contents.

50calibre: 3 boxes each containing 2 marbles

Box A = WW

Box B = BB

Box C = BW

Assuming they were incorrectly labelled, the only way of ascertaining the contents of all three boxes will be to pick all the contents of two boxes, so that is a minimum of

4 Marbles

Am I right?

no u r left.

As somebody has already pointed out, you only need to pick one marble from the box labelled 1 black 1 white.

Since you know that box is incorrectly labelled, if you pick out a black marble, then it's the box having 2 blacks, if you pick out white then you know it's the box having 2 whites.

Let's assume you pick a black marble from that box, that means the remaining two must be the boxes containing 2 whites and 1 white 1 black.

How do you know which is which? Since the boxes are all incorrectly labelled, that simply means the one labelled 2 whites contains 1 white 1 black, and the box labelled 2 blacks contains 2 whites.

Get it?

You only have to pick ONE marble from the box labelled 1 white 1 black and apply simple logic to figure it out.

McWhillion:

Broda, its only 1 marble you need to pick and that's from the bag wrongly labelled as having 1black and 1 white marble.

(I didn't pay attention to the wrongly labelled part.)

I was assuming it had no labels at all.

Puzzle has been discussed on Nigerian Info 95.1fm Abuja. Its just one pick from the wrongly labelled bag having 2colors . Then swap d labels on d other 2bags

Hmmmm

4 marbles.

You pick the first two marbles to know the content of the first box.
Then another pair of marbles reveal what the second box contains.

With these two selections (4 marbles all together), there wouldn't be any need to open the last box.

it is 6ways I will explain later

50calibre: 3 boxes each containing 2 marbles

Box A = WW

Box B = BB

Box C = BW

Assuming they were incorrectly labelled, the only way of ascertaining the contents of all three boxes will be to pick all the contents of two boxes, so that is a minimum of

4 Marbles

Am I right?

No!

Your three picks will be (B,W,B or W)
On the for pick you will have trouble guessing the box with BW.

Third box=BWB or W. Suppose
You pick a B from BW,

BWW

On your forth pick:

You get B (determined)
-------W (you pick again).

So minimum pick is 4 or 5 times.

deadie: Let's define the boxes as follows
Box A = 2 kind each.
Box B = 2 kind each
Box C = 1 of each kind.

Minimum number is 3, and this is only valid if you picked the first 2 from Box C.

Explanation:
When Box C is eliminated (by picking the two tiles from there), you can can be sure after the third pick (since they contain similar tiles). Otherwise, if you picked the first two from Box A or Box B, and then the third tile from either of the remaining, then you cannot be certain.

There is no way to determine wch box is box C since they are labeled incorrectly. You can't be sure you're going to b picking from box C

You need to pick out the two marbles in the box in order to confirm if it is black black, white white or black white.

Ichiato:
There is no way to determine wch box is box C since they are labeled incorrectly. You can't be sure you're going to b picking from box C

Well, the question is what is the minimum number of times that you can pick in order to correctly identify each box, and that is your answer.

Toh! If una don sweat finish make una laff with these videos



https://www.youtube.com/watch?v=4z8XZLqIMmw


https://www.youtube.com/watch?v=5vdC0i9uqRw

Who know were i feet buy cond0m abeg

Seems obvious how to do this with 4 draws (empty any of the two boxes fully determining the contents, and so by elimination you know what the third one contains).

But not obvious to me how to do this in less than 4 draws.

Even 3 seems challenging, no matter how I do it. Here are essentially the ways I can choose three draws:

1) Draw two balls from first box, one from second => If I'm unlucky in draws, still ambiguity about what box #2 and #3 contain
2) Draw one ball from each => Again, same deal. I could draw white, black, white. Does this correspond to {white, white}, {black, black}, {white, black}, or {white, black}, {black, black}, {white, white}? I'd need more draws to figure it out with this choices.

I seem to be missing something that allows one to do less than 4.

cannonnier:

No!

Your three picks will be (B,W,B or W)
On the for pick you will have trouble guessing the box with BW.

Third box=BWB or W. Suppose
You pick a B from BW,

BWW

On your forth pick:

You get B (determined)
-------W (you pick again).

So minimum pick is 4 or 5 times.

You didn't understand me, the rules didn't state pick only 1 marble from each box or pick from all 3 boxes.

Therefore I'm saying since the order is this way

BB WW BW

I'm just going to empty two boxes, 2 from each and based on what I pick out of these 2 boxes, I can say for sure what's in the 3rd regardless of it's label.

On second thought, since they are all labelled wrongly, picking from the box which is labelled as having both colours, you can determine what's in the other boxes.
So my answer is once

G12: Toh! If una don sweat finish make una laff with these videos



https://www.youtube.com/watch?v=4z8XZLqIMmw


https://www.youtube.com/watch?v=5vdC0i9uqRw

Lovely son you got there, what's his name?

Since the box labels are potentially wrong, you may as well assume that they are unlabeled (the labels don't help you at all).

AdeMoss: 4

na four be your answer? olodo.

Are we assuming that each incorrect label is one of "WW, BB, and BW" or that a completely incorrectly label may be used?

deadie: Let's define the boxes as follows
Box A = 2 kind each.
Box B = 2 kind each
Box C = 1 of each kind.

Minimum number is 3, and this is only valid if you picked the first 2 from Box C.

Explanation:
When Box C is eliminated (by picking the two tiles from there), you can can be sure after the third pick (since they contain similar tiles). Otherwise, if you picked the first two from Box A or Box B, and then the third tile from either of the remaining, then you cannot be certain.

Come OP, I don give you the answer wey you want. Better prove me wrong or give me my price

The answer is one marble.
EXPLANATION
Pick one marble from the box labelled one white one black, but which we know to contain either two white or two black marbles. Then take the correct label for it from one of the remaining two boxes. Now you have one box correctly labelled, you have one label in your hand, one box without a label and one box with a wrong label. Take label from the incorrectly labelled box and put on the box without a label then put the label in your hands on the last box.QED

ektbear: Since the box labels are potentially wrong, you may as well assume that they are unlabeled (the labels don't help you at all).

Gbam!

@ OP

Your problem statement is confusing. You can make it clearer by saying that the boxes are unlabeled. Then the question will be thus: what is the minimum number of picks that will enable a person to ascertain the content of the boxes?