Re: On the mathematics of three way ties

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

|> --- Begin redistributed message ---
|>
|> Subject: Re: On the mathematics of three way ties
|> Date: Tue, 3 Aug 1993 20:55:54 +0200 (MET DST)
|> From: Martin "J." Duerst <mduerst@ifi.unizh.ch>
|> 
|> >
|> >|> --- 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.
|> 
|> Well, thanks for being on my side (I expected that :-). But just for the
|> record, I wouldn't call my analysis a semi-derivation. What was missing
|> was an exact statement of the rules, which I assumed everyone knew, and
|> an explicit mentioning of the independence of the outcome of the fights.
|> 
|> Otherwise, the analysis was full-fledged, just working in another way
|> than you probably expected. From the formula-handling part, it was
|> actually simpler, as I didn't have to use exponentiation or summs with
|> an infinite number of terms. May be it was conceptually more difficult,
|> and so I will try to explain the difference.
|> 
|> May be the difference is somewhat similar to that between a loop and a
|> recursion in computer science or between an oshidashi (push out) and
|> a shitadashinage (kind of underarm throw) is sumo. With the former, you
|> work straightforward but hard, with the later, you need more technique
|> but once you know how to do it, it is much easier.
|> 
|> Now, Richard tried to view the whole situation with all combinations
|> of fights, and Olaf in fact did it; I really like his drawing:
|> 
|> >|>   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
|> 
|> But actually, there is no need to view all these bouts (and potentially,
|> their number is really big) at the same time. Apart from the special situation
|> in the first bout, it suffices to view exactly one bout, and to think
|> about what happens to the rikshi depending on it's outcome. After all,
|> the rikshi just change roles after each bout that doesn't end the whole
|> thing. You can then just express the probabilities of winning of each
|> role in terms of the probabilities of the other roles, because you know
|> with what probability each role will change to another depending on the
|> outcome of the bout.
|> Once you have written the equations down, you have three simple equations
|> with three unknowns, which are prety easily solved. I will not repeat
|> the formulas here. Mathematically, the view of things changing states
|> (or rikshi changing roles) with certain probabilities is called a
|> Markof chain, but that should not bother anyone.
|> 
|> Now that I have tried to explain again my approach, I would like to
|> come back to Olaf's. It actually contains a serious flaw (or two),
|> which, by chance or by trial-and-error, didn't affect the result:
|> 
|> >|> 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.
|> 
|> This would lead to the following probabilities:
|> basho being decided by 1 bout		(0)
|> basho being decided by 2 bouts		1/4
|> basho being decided by 3 bouts		1/8
|> basho being decided by 4 bouts		1/16
|> basho being decided by 5 bouts		1/32
|> and so on.
|> 
|> Obviously, this never adds up to 1, and is too low by a factor of two.
|> The correct term for what Olaf maybe means, and actually uses later,
|> is something like "basho being won by east (west) rikshi in bout n".
|> (east and west are the two starting positions in the dohyou).
|> This is quite a different, and not so obvious, concept.
|> 
|> In conclusion: It is possible to calculate the probabilities for each
|> rikshi both by looking at everything at the same time and adding it all
|> together, as Olaf did, or by just using the symmetries of the situation,
|> looking at one situation and solving the resulting equations. The result
|> is the same:	5/14 - 5/14 - 4/14.
|> 
|> ----
|> Dr.sc.  Martin J. Du"rst			    ' , . p y f g c R l / =
|> Institut fu"r Informatik			     a o e U i D h T n S -
|> der Universita"t Zu"rich			      ; q j k x b m w v z
|> Winterthurerstrasse  190			     (the Dvorak keyboard)
|> CH-8057   Zu"rich-Irchel   Tel: +41 1 257 43 16
|>  S w i t z e r l a n d	   Fax: +41 1 363 00 35   Email: mduerst@ifi.unizh.ch
|> $@%F%e!<%k%9%H!&%^!<%F%#%s!&%d%3%V!J%A%e!<%j%C%RBg3X>pJs2J3X2J!K(J
|> ----
|>
|> --- End of redistributed message ---

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