Identify heavier ball in group of $14$ We have $14$ identical looking balls, one of them is heavier than the rest of $13$ (all the others weigh exactly the same), how many of weighings (minimum) to find the heavier one?
And I want to know if there is mathematical formula for this problem?

(Sorry for my English)
You need at most 3 weightings.

One possible solution algorithm is:


  Divide the balls in 2 groups of 7 balls.
  Compare the 2 groups.
  Divide the heavier group of 7 balls into 3 groups (3-3-1)
  Compare the 2 groups containing 3 balls. If they are equal, then the 1 remaining is the heaviest.
  Else: Divide the heavier group of 3 balls into 3 groups containing 1 ball.
  Compare 2 of those balls. If they are equal, the remaining ball is the heaviest.


One can also start with 3 groups (5-5-4) and terminate in 3 weighting steps (and I bet there are more different solutions). It is obvious that you can't terminate with 2 weighting steps at most, but I can't prove this formally.
You need just three weighings.  

Number the balls in odd numbers, and the remaining number is 20, except 13.  Then use numbers 0 and 26

So the balls are numbered, 0,1,3,5,7,9,11,15,17,19,20,21,23,25,26 or in base 3, 

000, 001, 010, 012, 021, 100, 102, 120, 122, 201, 210, 212, 221,222

454 v 454 v 454  for the odd numbers, and 5,4,5 for this set.

Label the pans 0 and 2, and leave 1 off.  Weigh the three digits, and record which pan goes down, or 1 for equal.  The number written is the base 3 value of the heavy ball.  

This is actually a three-state system, and the trinary numbers give the weighings,  The solution is good for 26 balls, numbered in trinary numbers 0 to 27 (222) except 13 (111).

0 = put ball in left pan   heavy 0 makes this pan go down

1 = leave ball off  heavy 1 makes pans equal

2 = put ball in right pan.  heavy 2 makes right pan go down.

Since the heavy ball is comprised of three trinary numbers, the weighings will give up to 26 unique numbers.

The solution as given is a slightly modified version of the problem where one coin is known good, and one of thirteen coins is defective, can be either heavy or light.  This has a solution in

M = left pan;   0 = left off;  1 = right pan.

The 13 coins are numbered from 1 to 13, in trinary, with the even numbers in reverse sign, viz 1, -2, 3, -4, ..., -13.  In trinary, they look like

001, 0m1, 010, 0mm, 1mm, m10, 1m1, m01, 100, m0m, 11m, mm0, 111.

The known good coin is numbered -7, ie m1m.

You weigh the coins according to first, second, third digit, and get the numbers as m = left pan down, 0 = balance, 1 = right pan down.

If the number is even (ie an odd number of 0's), swap the m and 1's, eg m10 = 1m0,

You have then determined the number of the coin (as the weightings are 9,3,1), and whether it's heavy or light.  So 1m0 gives +9-3.1 = +6, coin 6 is defective, and heavy.

But you can't do it with 14 coins in three weighings unless you know the defect is +.  In that case, you can do as many as 26 weighings.