What is the probability that candidate A was never behind at any point in the counting?

1. Take a face of the regular tetrahedron and 3 orthogonal projections onto that face of the 3 centers of the 3 balls, touching the same face - they form a smaller homothetical equilateral triangle (with sides, parallel to the sides of the larger). The dihedral angle in the regular tetrahedron is arccos(1/3) ≈ 70°, hence the distance d between the sides of the inner and outer triangle is

d = cot(arccos(1/3)/2) = √((1 + cos(arccos(1/3))/(1 - cos(arccos(1/3))) =

= √((4/3)/(2/3)) = √2, then q = 2 + 2d * cot(30°) = 2 + 2√6

2. This is cited as 'Ballot Theorem' in 'An Introduction to Probability Theory and its Applications' by William Feller, John Wiley & Sons, 1970, Vol. 1, Chapter 3, #1:

"Let candidate A has p votes, candidate B - q votes, and p > q. Then the probability A will be strictly before B in the counting is (p - q)/(p + q)".

Let us consider p + q addends, each of them ±1, p of them 1s, q of them -1s, then we can construct a path, consisting of line segments, connecting (0, 0) with (p + q, p - q), each line segment with slope ±1. The number of all such paths is

C[p + q, p] = C[p + q, q]

The number of paths with vertexes strictly above the x-axis (except the origin) is

C[p + q - 1, p - 1] - C[p + q - 1, p] - they represent the favorable outcomes of the votes counting. Assuming all paths (or sequences of 1s and -1s) equally probable the required probability is

/***/ (C[p + q - 1, p - 1] - C[p + q - 1, p]) / C[p + q, p] = (p - q)/(p + q).

Since the required probability is A never to be behind B (the latter includes a sequence like ABABAB... - the corresponding path can have some vertexes on the x-axis also) the similar considerations show the number of the favorable paths is equal to the number of strictly "positive" paths with 1 more segment, connecting (0, 0) with (1, 1). That leads to a probability

(p - q + 1)/(p + 1) instead of /***/ and the required answer is 62/1033.

3. The Art of Problem Solving Forum can wait some 30 hours until 2010 - I think that's enough for me for the time being. Happy New Year to all Y!Answers Mathematical Community!

For number 2:

There are 971 + 1032 = 2003 total votes tallied.

So for the very first random vote counted there is a 1032 / 2003 chance that candidate A was voted for.

But right away that gives about a 48% chance that B will be the first vote, instantly taking the lead.

I tried to thing about it more but there is great unforeseen complexity in this problem. I really can't tell if the chance of always staying ahead gets smaller and smaller or stays around 50%.

1. 2+2√6

http://i299.photobucket.com/albums/mm286/rozeta53/...