# How can you pick the odd marble by 3 steps in this case?

Imagine i have 12 marbles, all identical in every aspect except that 11 of them have exactly the same weight but you do not know the weight of the 12th one (it may be bigger ot smaller than the standard weight). You are provided with a physical balance, or a balance with two sides or trays, and your task is to find out the defective marble and its relation by weight to the others (whether its weight was more or less than the conventional weight) by just using the physical balance 3 times.

I have worked out a little bit:

I am going to divide 12 marbles into three subsets of 4 each and put a set on one side each. If they come out to be equal, the defective one lies in the 4 i kept. Now i have to deduce from 4 to 1 in two steps. I keep one out, and put the rest three in one side. On the second side, i put any three marbles which i verified to be correct (the ones i tested). If they come out to be equal, the one i left is the one and i can use my third step to determine either it was heavier or not. If they come out to be unequal, the incorrect one must be in the 3 left. I compare only two of them. If they are equal, the one i left is mine and if unequal, the trend shown by the 3 in the previous step determines my answer.

Now i am not able to solve for the unequal part. Any help would be appreciated.

Note that it is possible to distinguish the odd one out of thirteen marbles with three weighings, as follows -

First - weigh $ABCD -v- EFGH$

$(i)$ If these are level, then $ABCDEFGH$ are all fine. Weigh $ABC-v-IJK$

$(ia)$ If these are again level, the odd one is $L$ or $M$ - weigh $L-v-A$ to find out [this does $LM$, but if it is $M$ we have not used $M$ in any weighing, so we cannot tell whether it is heavy or light]

$(ib)$ If $IJK$ are heavy [or light, crucially you know which] weigh $I -v- J$ to determine which is the odd one from $IJK$

$(ii)$ If $ABCD$ is heavier than $EFGH$ [for lighter make the obvious adjustment] we know that $IJKLM$ are all fine. Weigh $ABCE-v-DIJK$.

$(iia)$ If $DIJK$ is heavy then $D$ must be heavy, or $E$ must be light - so weigh $D-v-I$ to find out (this does $DE$)

$(iib)$ If $DIJK$ is light then $ABC$ must be heavy, and weighing $A-v-B$ will tell which of $ABC$ it is.

$(iic)$ If these balance then one of $FGH$ must be light, and weighing $F-v-G$ will tell which it is.

• Note that in only one case do we not know whether the odd marble is heavy or light. With twelve marbles we can know heavy or light in all cases (do the same weighings, but $M$ doesn't exist). – Mark Bennet Aug 26 '13 at 18:14
• In the 2nd step, I recommend that you weigh ABE-v-CFI. Actually I got my answer now by this. Think it up. – Rohinb97 Aug 28 '13 at 16:31
• And also, note that you can deduce the odd one out of 3 in just one step. With the above I wrote, 3,3,2 can be deduced. – Rohinb97 Aug 28 '13 at 16:33
• @Rohinb97 Your second stage reduces the number of balls weighed in total. See also math.stackexchange.com/questions/476780/… for a reference to some general results. – Mark Bennet Aug 28 '13 at 16:47

If the first four are heavier than the second four, say, you are left with 8 possible outcomes: One of $A,B,C,D$ is heavy or one of $E,F,G,H$ is light. Make your second weighing so that equaloty occurs in oly two cases, for example compare $ABE$ vs. $CDF$ (so equality means $G$ or $H$ is light; left heavy means $A$ heavy or $B$ heavy or $F$ light; left ligt means $E$ light or $C$ heavy or $D$ heavy). The last weighing is easily set up.

For generalizations see e.g. http://arxiv.org/pdf/0906.0693