Ok buddy I'm starting to get a bit irritated. Not because I can't solve it but because you are refusing to understand my solution that I typed up. You have yet to point out an accurate flaw in my solution. Every error that you accuse me of is incorrect.

Lets try again
READ MY SOLUTION BELOW:


I have started off with the 4 on the left 4 on the right 4 on the table position that you mentioned earlier.

3 Measures if at first weigh (with the 4 v 4) the scale is balanced:

For this explanation I start off from the beginning detailing how I put 4 balls on the left and 4 on the right.

Weigh 1: Start off with any four balls on the left side of the scale. Then put any 4 of the other 8 on the scale. Now if the scale is balanced then we know the odd ball is in the 4 that are not on the scale at all.

Weigh 2: To solve from there. You take 2 of the 4 that were left over and put them on the left side of the scale. On the right side of the scale you place 2 that we measured earlier and they balanced as these are normal. If the scale is balanced then it is one of the 2 that we didn't put on the scale and we can take the ones on the left off and the ones on the right off.

Weigh 3: Take the 2 left and put one on the scale with a normal ball. If the scale is balanced then it is the ball in your hand that you didn't use. If it is unbalanced then it is the one you put on the left.


If the scale is not balanced AFTER the 4 v 4 weigh then do this for the 2nd weigh.

Remember this situation has 4 balls on the left and 4 on the right and the scale is unbalanced. So the odd ball out is either on the left or the right of the scale. We still have 4 balls on the side that we can use for accurate measurement.
Before weighing the 2nd time:
take 1 on the left and switch it with one on the right it doesn't matter which you would just have to remember which you did switch.
take 3 on the left (excluding the 1 that you switched from the right) and switch them off the scale with 3 that we know are normal (of the 4 that we ignored in the first measure.)

That was our set up for the 2nd weigh. It is always that set up. Now we weigh the 2nd time.:

If the after 2nd weigh the scale changes which side is up and which side is down:
Then we know it is one of the 2 we switched sides with. (Remember we switched 2 balls. One from the left with one from the right)

Your third weigh in this situation would be:
Take one of the 2 balls that we moved and put it by itself on the left. Weigh it with a normal ball. If the scale is unbalanced then it is the ball on the left. If the scale is balanced it is the other ball of the 2 that we didn't weigh.

If when after 2nd weigh and the scale stays in the same position. (doesn't matter if left is high or right is high just if it stays the same):
Then we know it is one of the 3 that we took off from the left.

We then have to remember when we weighed the first 4 (for the first weigh) whether the left side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we had taken off from the left earlier and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.

If when after 2nd weigh and the scale does not change its position (meaning it is still unbalanced but hasn't moved)
Then we know that the ball is one of the 3 on the right side.

We then have to remember when we weighed the first 4 (for the first weigh) whether the right side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we have on the right and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.


DO NOT REPLY UNLESS YOU HAVE FULLY READ MY SOLUTION.

Moderator Edit (Khethil): All-Caps lines sizes taken down in accordance with forum rules

Calm down calm down..on the first weigh you bring it down from twelve to four then down to three..When you have three you discount one and weigh the other two against two known balls.If they are balanced its the odd ball..Does that make it easier to explain?

no no no. On the first weigh if the scale is balanced. Then you narrow it down to 4. 2nd weigh 2, 3rd weigh 1.

if the scale is not balanced the first weigh then you have narrowed down to 8. 2nd weigh either narrowed down to 2 or 3. 3rd weigh 1.

The more I explain this the more I am sure my solution is right.

balanced on not balanced on the first you still have narrow it down to three on the second weigh and then one on the last.The most difficult would be unbalanced on the first as you say on the second transfer one to the other side with three known.If it is then unbalanced you have three left.Split two up and put other to one side weigh them with two others known.Idont know if we are the same but i like you know its the answer.Its our english thats the hard bit.

the 2nd Setup for the weigh is EXTREMELY important.

this is all done after the first weigh.

Take one from the left and switch it with one from the right.

take the other 3 from the left and put them asside and put 3 normal balls on the left.

then read my explanations for possible scale orientations.

I understand, honest i do but i had to get there myself i just took a bit longer..Ive lost more brain cells than youve had beers remember..

do you understand the solution now?

Yes but i had to get there myself with your help..Thanks xris

Hey click here , why the rage man!! :perplexed:

There are others in this thread!!

I chainged the thread title to the odd ball logic problem, as it usually comes with this title

Oh!! and I solved this peroblem over thirty years ago, by both mathematical formula and the methos we are using now

Iwas away sorry, there are various ways of descibing the solution, the one below is just one of many. :bigsmile:

Your very large fonts make me think you are a little manic :perplexed:

I will compare it to yours. and you must compare it to mine OK

For 12 balls we start out with 24 possible hypotheses since each ball can be either heavy or light.

First, since all the balls look alike let's just number them 1 to 12 to tell them apart.

Let's denote the hypothesis that ball 1 is heavier than the rest by 1+ and that it is lighter by 1-.

Similarly denote 2+,2-, 3+,3- etc.

Of course, only one of these 24 hypotheses can be correct.

Our job is to find out which one in 3 weighings.

Let's call the three weighings W1, W2 and W3.

Also, if we weigh balls 1,2,3 (in the left pan) against 4,5,6 (in the right pan), then we'll denote that weighing as 1,2,3 <> 4,5,6.

In general we can use this notation to describe any weighing.

There are three possible outcomes of any weighing:

the left pan is heavy,

the right pan is heavy

or the two pans are even. Let's just denote these outcomes as L, R and E respectively.

We say O(W1)=L to denote that the outcome of weighing one is that left pan is heavier.

Now to the solution! Most people will struggle with various combinations of balls until they have either given up or stumbled on an answer.

That's no fun! We are going to pick our weighings based on a very simple strategy:
Try to pick each weighing in such a way that it is equally likely for the outcome to be

L =LEFT PAN,

R=RIGHT PAN AND E=PANS ARE EQUAL OR BALANCE PAN =E

so use L H E to denote these possible states.

"O" Denotes outcome
The only weighing that results in equally likely outcomes is when you weigh four balls against four other balls. OOOO<>OOOO my hint

So we choose

W1= 1, 2, 3, 4<>5, 6, 7, 8. Clearly if O (W1) = E

Then the oddball is one of 9,10,11,12 and the probability of that is 4/12=1/3.

That leaves 2/3 which is divided equally between the outcomes L and R.


So this weighing matches our strategy since the probability of each outcome is 1/3. Now depending on what happens in W1:

1. O (W1) =E: Then the oddball is one of 9,10,11,12, i.e. there are only 8 remaining hypotheses (see how we eliminated 2/3 of the hypotheses?).
   
   Now for W2 we have to shoot for the same thing, i.e. make the outcomes L, R, and E as close to being equally likely as we can.
   
2. Since the number of hypotheses is not a multiple of 3 we won't be able to get the outcomes to exactly equal but we can come close. There are several ways to do this --here's one:
   

W2: 9, 10, 11<>1, 2, 3. Notice that we already know that 1, 2, 3 are not oddballs. So the probability of outcomes L, R and E are 3/8, 3/8 and 1/4 respectively.

If O (W2) =E then the oddball is ball 12. All we have to do in weighing 3 is to just weigh ball 12 against any other ball, e.g. W3=12<>1.

If the outcome is L, then 12 is heavy and if the outcome is R then ball 12 is light.


If O (W2) = L then we know that the oddball is heavy and the remaining hypotheses are just 9+, 10+, and 11+.

Now the third weighing is easy. For example, W3= 9<>10 will work. If L then 9 is the oddball, if R then 10 is oddball and if E then 11 is the oddball.



If O (W2) =R then this case is almost identical to the previous case except that the oddball is lighter.

1. O(W1)=L: We are left with the hypotheses: 1+,2+,3+4+,5-,6-,7-,8-. Now for the second weighing we want the outcomes to be as equally likely as possible. How about:
   
   
   
2. W2=1, 2, 5<>3, 4, 6? The probabilities of the outcomes L, R and E are 3/8,3/8 and 1/4 respectively which is as close to being equiprobable as you can get.
   
3. Now if O(W2)=E then 7- or 8- are true. W(3)=7<>8 will do the job. And if O(W2)= L we are left with 1+,2+ and 6-.
   
4. Then W(3)= 12 for the third weighing identifies the oddball.
   
5. O(W1)=R: This case is very similar to case 2 and can be completed the same way.
   

Click Here,

Your solution is correct, good thinking, if you really want to remain my buddy tone down your rhetoric, I had all the intention of revisiting your solution, and as you are now aware it is hard to put the solution down on print without becoming involved in a cycle of explanations

This is not an easy problem and you are one of the very few that I have come across who has solved it.

But remember I gave you a hint, if you try this out on your friends don't give them a hint as it spoils the whole thing

I put my own way of solving the solution and by doing this was able to see you were correct

Good man

Alan


Ok buddy I'm starting to get a bit irritated. Not because I can't solve it but because you are refusing to understand my solution that I typed up. You have yet to point out an accurate flaw in my solution. Every error that you accuse me of is incorrect.

Lets try again
READ MY SOLUTION BELOW:


I have started off with the 4 on the left 4 on the right 4 on the table position that you mentioned earlier.

3 Measures if at first weigh (with the 4 v 4) the scale is balanced: CORRECT

For this explanation I start off from the beginning detailing how I put 4 balls on the left and 4 on the right.

Weigh 1: Start off with any four balls on the left side of the scale. Then put any 4 of the other 8 on the scale. Now if the scale is balanced then we know the odd ball is in the 4 that are not on the scale at all. CORRECT

Weigh 2: To solve from there. You take 2 of the 4 that were left over and put them on the left side of the scale. On the right side of the scale you place 2 that we measured earlier and they balanced as these are normal. If the scale is balanced then it is one of the 2 that we didn't put on the scale and we can take the ones on the left off and the ones on the right off.


Weigh 3: Take the 2 left and put one on the scale with a normal ball. If the scale is balanced then it is the ball in your hand that you didn't use. If it is unbalanced then it is the one you put on the left. If the scale is not balanced AFTER the 4 v 4 weigh then do this for the 2nd weigh.

Remember this situation has 4 balls on the left and 4 on the right and the scale is unbalanced. So the odd ball out is either on the left or the right of the scale. We still have 4 balls on the side that we can use for accurate measurement.



Weigh one: 1 2 3 4 V 5 6 7 8 does not balance (9 10 11 12) keep to one side

Pan went up with 1 2 3 4 down for 5 6 7 8

( Switch balls)

8 2 3 12 V 1 9 10 11 correct (keep 4 and 5 one side)


Weigh two

8 2 3 12 GOES UP V 1 9 10 11 GOES DOWN

We know if 8 is the oddball it must be lighter

We know if the oddball is 1 it must be heavier
So the oddball must be 1 , 2 or 8

or

If 8 2 3 12 V 1 9 10 11 Balances

The oddball must be balls set to one side 4 or 5

Weigh three" option one: weigh ball 5 against any ball, scale goes balances it is the other ball 4 and 4 is must be lighter

Weigh 5 against any ball and the scale goes down on 5 side the oddball is 5 and it is obviously heavier



Weigh three: ball 8 against 9 or any ball except ball 1

Balances oddball is 1 and the oddball is heavier

Does not balance the oddball must be ball 8 and the oddball is lighter



Weigh oddball 1 against any ball except ball 8

Balances oddball is 8 and it is lighter

Does not balance the oddball must be ball 1 and the oddball is heavier



Before weighing the 2nd time:
take 1 on the left and switch it with one on the right it doesn't matter which you would just have to remember which you did switch.
take 3 on the left (excluding the 1 that you switched from the right) and switch them off the scale with 3 that we know are normal (of the 4 that we ignored in the first measure.)


That was our set up for the 2nd weigh. It is always that set up. Now we weigh the 2nd time.:




If the after 2nd weigh the scale changes which side is up and which side is down:
Then we know it is one of the 2 we switched sides with. (Remember we switched 2 balls. One from the left with one from the right)

Your third weigh in this situation would be:
Take one of the 2 balls that we moved and put it by itself on the left. Weigh it with a normal ball. If the scale is unbalanced then it is the ball on the left. If the scale is balanced it is the other ball of the 2 that we didn't weigh.

If when after 2nd weigh and the scale stays in the same position. (doesn't matter if left is high or right is high just if it stays the same):
Then we know it is one of the 3 that we took off from the left.

We then have to remember when we weighed the first 4 (for the first weigh) whether the left side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we had taken off from the left earlier and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.

If when after 2nd weigh and the scale does not change its position (meaning it is still unbalanced but hasn't moved)
Then we know that the ball is one of the 3 on the right side.

We then have to remember when we weighed the first 4 (for the first weigh) whether the right side was higher or lower. If it was lower. Then we know the odd ball is heavier. If higher then lighter.

Third weigh:
We then pick 2 of the 3 that we have on the right and put 2 (doesn't matter which) on opposite sides. If balanced we know its the one we didn't weigh.
Now we determined right above whether or not the ball is lighter or heavier.
So if the scale is unbalanced then we just pick the side that has the lighter/heavier ball in it. It is that simple.


DO NOT REPLY UNLESS YOU HAVE FULLY READ MY SOLUTION.

Moderator Edit (Khethil): All-Caps lines sizes taken down in accordance with forum rules

Yes I did get a bit carried away. It had seemed in your replies as though you weren't even reading my response. Nonetheless a good riddle.

Yes you did give me a hint though I think it wouldn't have taken much longer to figure out 4v4 is the only way to go at start. 6v6 is worthless, 5v5 doesn't give you much as you have 10 to go from if it doesn't balance. 3v3 is the one you'd have to think about and figure out why it doesn't work. I'm quite tired right now and I don't remember why it doesn't as I was explaining earlier to my dad why it won't work. I gave it to him without a hint though he gave up pretty quick.

Click here,
You solved a really difficult problem without using mathematics. I got this problem over thirty years ago from a work colleague and believe it or not in all that time only you and a friend has been able to solve it.

Good man!!


I did, however, solve it many years ago