3 uses of balance- odd one out of 13 MARBLES

Ok let's say we divide the marbles in 4 groups 13, 13, 13, and 1
We now weigh two of the 13 groups. We have two cases (marked A) and B) below:
A) Both groups are equal and therefore we have 26 GOOD stones and 14 in which is the odd. We take 9 of the 14 and weigh it with 9 of the GOOD . A1) they are equal. We have 5 undecided stones and two weighings left. We divide the 5 U stones in groups of 3 U and 2U. We measure the group of 3 with 3 good stones. If they are equal we have 2 U stones. We measure 1 with 1 good, if equal the final stone is the odd A2) the 9U from the 14U are not equal to the GOOD, so without loss of generality let's assume they are heavier. We divide the Heavier group in three groups of 3. We know that the odd stone is heavier and we have two weighings left. We divide the heavy group in three groups of 3 and we find the stone with two measures.
B) The two groups are not equal. We now have 13 LIGHT, 13 HEAVY, and 14 GOOD stones. Since we don't know if the odd stone is L or H we are faced with a dillema.We now form (without mixing) group C: 9H + 9 L and measure it with group D: 4 H + 4 L + 10 G (again without mixing them). We have without the loss of generality two cases: C is lighter/heavier than D, or vice versa. Let C is heavier (for lighter anallogicaly), then we have the 9H stones and 2 weighings left. We divide them in groups of 3 and we measure two of them. And with the final weighing we find the stone 🙂
Cheers!

EDITED: Just to add it is clear that in the case of 39 objects we can even determine if the odd is heavier or lighter.

......We now form (without mixing) group C: 9H + 9 L and measure it with group D: 4 H + 4 L + 10 G (again without mixing them). We have without the loss of generality two cases: C is lighter/heavier than D, or vice versa. Let C is heavier (for lighter anallogicaly), then we have the 9H stones and 2 weighings left.......

No.NOT SO.

You are left , (at this stage of your metod,) with not just 9H , but 9H+4L , to locate the unknown ghost from. For , group C(LHS) could be heavier not only because of the 9H on the LHS, but, equally because of the 4L on the RHS.

But this has been taken care of in my method. This part was not missing in my method.

What was missing in my method , has been successfully solved by you. You have rightly plugged the missing part in my method. Thanx.

You're totally right. Hehehe, and yes your method seems to cover it 🙂
Well spotted.

Originally posted by ranjan sinha
Let me put forth my steps. Am I missing something? Where ?
Let us have 4 boxes named U, E, H & L , for keeping the marbles for the following different categories.
(i)U-box - undecided; the odd marble may be in thi ...[text shortened]... ter above(TheMaster37).

Am I missing something?
😕🙂😕🙂😠😵

Yes, your method works for 39 marbles case.
And alongwith, the supplementary by Ilywrin , it works even for 40 marbles.
However , I must point out that at the stage of case C212, the content of the L-box is 3 and not 4 making the contents U-0, E-36(E-37 in the case of 40 marbles), H-0 &.L-3 . LHS.NE.RHS will mean that one of the 3L marbles on the lighter side only contains the odd one and that is lighter. Otherwise with L-4 you cannot locate the odd one in 1 additional weighing.

Originally posted by CoolPlayer
However , I must point out that at the stage of case C212, the content of the L-box is 3 and not 4 making the contents U-0, E-36(E-37 in the case of 40 marbles), H-0 &.L-3 . LHS.NE.RHS will mean that is lighter. Othe ...[text shortened]... L-4 you cannot locate the odd one in 1 additional weighing.

Spotted correctly. hmm hmm hmm... That was a minor typo- error .Which is clear from the correct inferences thereafter.

Sarathian ! why dont you give the mathematical basis of the conjecture of Plumber , which you seem to endorse in your last post above. Or Plumber should give the proof of his conjecture.. v

Okay, here is what I found on the web (concerning this problem): http://home.att.net/~numericana/answer/weighing.htm#counterfeit

Double post 🙂