Re: 3-Way Ties (fwd)

[Martin "J." Duerst <mduerst@ifi.unizh.ch>]

Recently, there were questions about the chances in a 3-way tie and some
answers, by Hiroki Kimata, Richard L. Webb, and -Lee (sorry, don't know
the full name).

>   >    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?
>   > 

The numbers above are correct. Another long analysis that was posted to
our list recently seems to have some flaws (as it doesn't agree).
I will not discuss it, but just try to explain the above results,
which are not that difficult to calculate.

First, we of course assume that the probability of winning an individual
bout is 1/2. The next thing to realize is that after the second bout
and ever after, until someone wins two in a row, the situation is allways
the same:

One rikshi, call him A, that has won the last one and just needs one.
One rikshi, call him B, that has still has to win two.
One rikshi, call him C, that can only hope for his next chance.

Now, the rikshi change roles after each bout. There are two other roles,
W for winner, and L for looser, for the end of the whole thing.
Now lets see how the probabilities of winning, denoted wa, wb, and wc,
depend on each other:

wa = 1/2 + 1/2 wb
wb = 1/2 wa
wc = 1/2 wb

This, for example, means that C has a 50% chance of being B in the next
bout, and therefore 50% of the chance that B has of winning (plus a 50%
chance of loosing, which doesn't count towards the chance of winning).

The above simultaneous (linear) equations can be solved easily, giving:
wa = 4/7
wb = 2/7
wc = 1/7
Please check it or solve it yourself.

Now let's go back to the first bout. The tow rikshi that clash in the first
bout obviously each have a 50% chance of being in position A and in
position C after the bout. This ends up with 1/2 (4/7 + 1/7) = 5/14 of
a chance of winning. The one rikshi not participating in the first bout
will in any case play the role of B in the second bout, so his chances
are 2/7 or 4/14, and so 20% lower than those of the two other ones.

----
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
----