3-Way Ties (fwd)

[kimata@zeppelin.convex.com (Hiroki Kimata)]

Hi,there.
Here is the great effort made by Richard for my question about 3Way ties.
Thank you,

Hiroki Kimata

---From Here

   From: Richard L. Webb <webb@phys3.physics.wsu.edu>
   Date: Wed, 28 Jul 1993 14:04:00 -0700 (PDT)
   X-Mailer: ELM [version 2.4 PL22]
   Content-Type: text
   Content-Length: 7702      

   Hiroki,

   Here is my reply to your letter (sorry it's so late).  My mailer
   cannot properly reply to a group this large, so can you forward it
   to the entire sumo group for me (I would hate to do all of this and
   then not spread the knowledge)?  Thanks.

   >    Although I cannot confirm or deny any of this, I seem to remember
   >    someone doing a very professional mathematical/statistical job
   >    in s.c.j. when the same subject came up for discussion.  (was
   >    that anyone who is in this group?)  One of the things I think
   >    I remembered was that whoever went first, had a much greater
   >    (statistical) advantage.  Maybe this advantage is given by right
   >    to the highest ranking rikishi, so as to increase the chance of
   >    him winning.  The other 2 then being given an equal "chance"
   >    to get in the 1st match.
   >    Just a thought?
   > 
   > Recently, the same discussion was going on in fj.rec.sports,
   > and the concensus was both two rikishis who have the first match
   > have 5/14 chance of win, while third rikishi has 4/14.
   > This calculation seems a bit complicated for me who is definitely 
   > not math expert.
   > Is there any sumo loving mathematician who can help in this ML?
   > 
   > 
   >    > 
   >    > -Lee
   >    > 
   >    > 
   >    > 
   > 
   > Hiroki
   > 
   > 

   I remembered the odds being something like those quoted above, but
   on doing the calculations myself, I found slightly different odds
   (maybe there is a flaw in my method, huh?)  Anyway, here is what I
   did:

   First, we must assume three exactly even rikishi (named A, B, and
   C) such that the odds of winning each match are exactly 1/2.  Then 
   the probability of the tie-breaker ending in the nth bout is 

   P(n) = 2^(n-1); for n >= 2  (the tie-breaker must go at least 2 bouts)

   To determine the probability for each rikishi being in a position
   to win the nth bout, I made a branching truth table (or whatever
   you call it) as shown below for rikishi A in the first bout (against
   rikishi B)

   #1 - 1 -> #3 - 1 -> #5 - 1 -> #7 - 1 -> #9 --->
    |         |         |         |         |
    1         1         1         1         1
    |         |         |         |         |
   #2 - 1 -> #4 - 2 -> #6 - 3 -> #8 - 4 -> #10--->
	      |         |         |         |
	      1         2         3         4
	      |         |         |         |
	     #5 - 1 -> #7 - 3 -> #9 - 6 -> #11--->
			|         |         |
			1         3         1
			|         |         |
		       #8 - 1 -> #10= 4 -> #12--->
				  |         |
				  1         4
				  |         |
				 #11- 6 -> #13--->
					    |
					    6
					    |
					   #14--->

   with the following key:

   All numbers preceded by a # represent the bout of that number
   (i.e., #5 means the 5th bout).  The numbers on the paths in between
   the bouts represent the number of ways that bout can be reached.
   Lines going down represent by victories, lines going across
   represent by defeat.  From this table, the number of ways to win in
   the nth bout can be determined by summing the numbers pointing down
   to the nth bout.  The numbers for C are shifted down one bout.
   I have tabulated them below for the first 16.

	   Chances to win nth bout by a rikishi A (or B) and C

   n   A(n)           C(n)         a(n) c(n) t(n)

   1    0              0             0    0    0

	1
	-
   2    4              0             1    0    2

	1              1
	-              -
   3    4              4             0    1    1

	5              1
	--             -
   4    16             4             1    0    2

	1              13
	-              --
   5    3              48            1    1    3

	11             9
	--             --
   6    32             32            1    1    3

	7              91
	--             ---
   7    20             320           2    1    5

	677            551
	----           ----
   8    1920           1920          2    2    6

	10877          4423
	-----          -----
   9    30720          15360         3    2    8

	119887         48743
	------         ------
   10   337920         168960        4    3    11

	420017         341531
	-------        -------
   11   1182720        1182720       5    4    14

	15968731       12983953
	--------       --------
   12   44943360       44943360      7    5    19

	159726811      129870253
	---------      ---------
   13   449433600      449433600     9    7    25

	159746761      259770431
	---------      ---------
   14   449433600      898867200     12   9    33

	2496199        519570787
	-------        ----------
   15   7022400        1797734400    16   12   44

	74129429729    15067991723
	------------   -----------
   16   208537190400   52134297600   21   16   58

   The pattern I saw in this (those more astute may see a simpler one)
   is that 

   a(n) = a(n-1) + a(n-5) = number of chances to win in n rounds for A
   c(n) = c(n-1) + c(n-5) = number of chances to win in n rounds for C

   with the first five defined as shown (aren't recursive
   relationships great!?).  Thus, the probability of rikishi A winning
   the tie-breaker in the nth bout is product of his chances to win
   be in a position to win the tie-breaker in the nth bout 

   [ = a(n) / ( 2 * a(n) + b(n) ) ] 

   and the odds of the tie-breaker ending in that bout [P(n)].

   SO, the odds of winning (evaluated after n bouts) are

   A(n) = Sum[ P(m) * a(m) / ( 2 * a(m) + b(m) ) , {m,2,n} ]

   (written in Mathematica format for you computer nerds:-) for all
   others that means summing the probability of the bout ending in
   each round by the probability of A winning in that bout over all
   bouts from the second on)

   This number is given in the table above as well (in fractional
   form) for your viewing pleasure, as well as C(n) for C.

   I had the computer do the tabulation, of course, to get the table I
   have shown above.  I actually went all the way out to the 25th bout
   (I am assuming that this is approaching infinity since the mean
   number of bouts in a tiebreaker of this fashion is only three) and
   I get the following odds:

   Odds of A winning (= odds of B winning) =

   168761566719965063342567/4747367915810043671347200
    = 33.55858%

   Odds of B winning =
   6886068149222782082082358933/2373683957905021835673600
    = 28.9031%

   (I believe the percentages because there is less chance that I
   miss-typed them ;-)

   I believe the 5/14 and 4/14 numbers quoted previously were
   taken from the statistics after the 11th bout (5/14 = 35.71429% and
   4/14 = 28.57143%).  The error in this method can be seen by
   comparing the number after the 11th bout with those after the 12th
   bout (7/19 = 36.84210% and 5/19 = 26.31579%).  Not taking into
   account the probabilities of winning in each round make the series
   not converge properly (although it would still converge if you went
   to infinity).

   Well, that's my mathematical stab at this.


   Richard Webb
   webb@phys3.physics.wsu.edu

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