On the mathematics of three way ties

[osumo3@native.usc.edu (Olaf Meeuwissen)]

|> --- Begin redistributed message ---
|>
|> Date: Tue, 3 Aug 93 10:04:37 PDT
|> From: olaf@rcf.usc.edu (Olaf Meeuwissen)
|> Subject: On the mathematics of three way ties
|> 
|> Konnichi wa.
|> 
|> I couldn't quite get to sleep a few  nights ago and was  slightly disturbed
|> by  the differing results of Richard Webb's and Martin Duerst's analysis so
|> I sat down and did the derivation myself. Richard's was  incorrect, besides
|> being a bit  hard to follow, Martin's got the right numbers, but I couldn't
|> quite follow the semi-derivation.  Below I'll try to explain my findings in
|> laymen's terms.
|> 
|> 
|>   I. The rules
|> 
|> A three way tie is decided by a series of bouts  in  which the winner of  a
|> bout  fights  the  spectating  third rikishi in the next, until one of  the
|> rikishi wins two bouts in a row.
|> 
|>  II. The assumptions
|> 
|> We assume that  the probability of a rikishi winning a bout  is exactly 1/2
|> for every bout he wrestles. This implies that the result  of every  bout is
|> independent   of  all   the  previous   ones,  a  mathematically  important
|> characteristic in probablity theory which we shall use later on.
|> 
|> III. The notation
|> 
|> For simplicity's sake we label the three rikishi A, B and C [mathematicians
|> are not  particularly known for creativity]. The probability of winning the
|> basho will be denoted as P,  so P(A) is the probability riskishi A wins the
|> bout. The probability  that  the basho is decided  after the n-th  bout  is
|> given by:
|> 
|> 	p(n) = 1/2 ^ n	for n > 1 (since two wins in a row are needed)
|> 
|> where "^" denotes exponentiation.  [It is here  that we have  used the fact
|> that the results of previous bouts do not affect the following one.]
|> 
|>  IV. The proof
|> 
|> Suppose that  the tie-breaker starts  of  with A vs.  B. This can always be
|> achieved  by relabelling  the rikishi, so any conclusions reached from this
|> are general. The  following bout is either A vs. C or B vs. C. When writing
|> this out more schematically:
|> 
|>   2    3    4    5    6    7              number of bouts to decide basho
|>   A    C    B    A    C    B              basho winner
|>   |    |    |    |    |    |
|>   AC - CB - BA - AC - CB - BA - (etc.)    following bouts
|>   |
|>   AB                                      first bout
|>    |
|>    BC - CA - AB - BC - CA - AB - (etc.)   following bouts
|>    |    |    |    |    |    |
|>    B    C    A    B    C    A             basho winner
|>    2    3    4    5    6    7             number of bouts to decide basho
|> 
|> Keeping in mind that the probability of deciding the basho in the n-th bout
|> is p(n), we can see that:
|> 
|> 	P(A) = p(2) + p(5) + ... + p(4) + p(7) + ...
|> 	P(B) = p(4) + p(7) + ... + p(2) + p(5) + ... = P(A)
|> 	P(C) = p(3) + p(6) + ... + p(3) + p(6) + ...
|> 
|> Or, in short:
|> 
|> 	P(A) = sum{ p(3i-1) } + sum{ p(3i+1) }	for i = 1, 2, 3, ...
|> 	P(B) = P(A)
|> 	P(C) = 2 * sum{ p(3i) }			for i = 1, 2, 3, ...
|> 
|> Let us look at p(3i-1). This can be written as:
|> 
|> 	p(3i-1) = (1/2) ^ (3i-1)
|> 	        = { (1/2) ^ (3i) } * (1/2) ^ -1
|> 	        = 2 * (1/2) ^ (3i)
|> 	        = 2 * p(3i)
|> 
|> Similarly, one finds: p(3i+1) = (1/2) * p(3i)
|> 
|> Substituting this back in our last expression for P(A), we get:
|> 
|> 	P(A) = sum{ 2 * p(3i) } + sum{ (1/2) * p(3i) }
|> 	     = 2 * sum{ p(3i) } + (1/2) * sum{ p(3i) }
|> 
|> If we write S for the sum part, we have:
|> 
|> 	P(A) = (5/2) * S
|> 	P(B) = P(A)
|> 	P(C) = 2 * S
|> 
|> From the fact that P(A) + P(B) + P(C) = 1 (this is necessary, since we know
|> for a fact that one  of the three _will_ win the basho) it is quite easy to
|> solve for S and one finds:
|> 
|> 	S = 1/7
|> 
|> And hence:
|> 
|> 	P(A) = 5/14
|> 	P(B) = 5/14
|> 	P(C) = 4/14
|> 
|> NOTE: It  is possible  to  calculate  the summation explicitly.  This  also
|>       yields 1/7, showing independently that P(A) + P(B) + P(C) = 1 and re-
|>       assuring us that we didn't make any mistakes.
|>       For those who want to know:
|> 
|> 	sum{ (1/n) ^ i } = 1/(n-1)	for i = 1, 2, 3, ...
|> 
|>       And in our case n = 2 ^ 3 = 8, since (1/2) ^ 3i = { (1/2) ^ 3 } ^ i.
|> 
|> 
|>   V. The conclusion
|> 
|> The rikishi who gets to sit out the first bout  in  a  three way tie  has a
|> smaller chance of winning the basho. Bummer!
|> 
|> 
|> I hope this was understandable. Matane,
|> 
|>     ____
|>    / __ \ /\          ___  Olaf Meeuwissen           (e-mail: olaf@usc.edu)
|>   / / / // /____     / __\ Dept. of Chemistry, Univ. of Southern California
|>  / / / // // __ \  _/ /    ------------------------------------------------
|> / /_/ // // /_/ / /  _/    These days  are precious,  and I'd  rather spend
|> \____//_/ \_____\ / /      them goofing around than studying.
|>                  /_/                                              -- Calvin
|>
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