All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Probability of constant leading (Posted on 2015-10-05)
Imagine an election between two candidates.

If the ballots are cast one at a time, what is the probability that A will lead all the way throughout the voting process?

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Formula with partial proof | Comment 1 of 5
I'm guessing the formula is P(m,n) = (m-n)/(m+n)

Obviously P(m,0)=1
and
P(m,1)=(m-1)/(m+1) since B's lone vote cannot be first or second.

For P(m,2)
the denominator is C(m+2,2)=(m+1)(m+2)/2
the numerator requires the first two votes to be A = C(m,n)  except the first four votes cannot go AABB which is only a single possibility so C(m,n) - 1 = m(m-1)/2 - 1 = (m+1)(m-2)/2
and the fraction reduces to (m-2)/(m+2)

From there things get complicated
Unallowed starts would be B, AB, AABB, AABABB, AAABBB.  I didn't tabulate them by formula but by hand counting P(4,3)=5/35=1/7 which fits the pattern (4-3)/(4+3)

Edited on October 6, 2015, 12:59 pm
 Posted by Jer on 2015-10-06 08:15:19

 Search: Search body:
Forums (0)