TITLE: what is the probability that a scissor became the champion?
QUESTION [20 upvotes]: Here is a question from one of my students:
suppose 8 players are in an elimination match. The players are marked with marked with either R (for rock), P (for paper) or S (for scissors). If two players are marked with the same letter, then one is picked as winner of this round. An example of this sort of matches is:
First round：R-P S-S R-S P-S
Second round： P-S R-S
Third round： S-R
Champion：    R
The question is：Given r+p+s=8, for the match that there are r many R, p many P and s many S, if the table for the elimination match is assigned randomly, what is the probability that a S became the champion?
A more general question is: for a match with 2^n players, and given r+p+s=2^n, how much chance  can some S wins the champion?
My answer did not meet his satisfaction. What I know is, firstly for small n, one can list all  possibilities (there are (2^n)! many, if we make a distinction between two rocks etc.)and find the answer; secondly, one may use a computer program to solve the general question by inputting r,p and s; thirdly, one can also do some induction to find a pattern for the general question; and lastly, if the number r,p and s are assigned randomly and we let n goes to infinity, and the answer should be a third.
So can we have some smart ways to figure out these questions? Or can we reduce them to other known questions? I guess graph theory may help but my knowledge is very limited there.

REPLY [16 votes]: Here is a generating function approach. A $P$-tree $T$ of length $n$
is a complete binary tree of length $n$ with vertices labelled $p,r,s$
such that the root is $p$, the children of $p$ are either $p,p$ or
$p,r$, the children of $r$ are either $r,r$ or $r,s$, and the children
of $s$ are either $s,s$ or $s,p$. Similarly define $R$-trees and
$S$-trees. The weight $w(T)$ is the product of all leaf
vertices. Let $P(n)=\sum_T w(T)$, summed over all $P$-trees of length
$n$, and similarly define $R(n)$ and $S(n)$. Thus
  $$ P(0)=p,\ \ \ P(1)=p^2+2pr, $$
  $$ P(2)=p^4+4p^3r+6p^2r^2+4sp^2r+4pr^3+8pr^2s. $$
Clearly
   $$ P(n+1) = P(n)^2+2P(n)R(n), $$
and similarly for $R(n+1)$ and $S(n+1)$. We can eliminate two of the
functions from these three recurrences and obtain a recurrence for
each function alone. For $P(n)$ the recurrence is the following,
writing e.g. $p3$ for $P(n+3)$:
$$ 0 = 4 p2^4 p0^4+p2^2 p1^6-p2 p1^8-8 p1^5 p0^2 p3+16 p1^4 p0^4 p3 $$
$$ -8 p1^3 p0^6 p3+4 p1^2 p0^4 p2^3-16 p1 p0^6 p2^3+12 p1^7 p0^2 p2 $$
$$ -18 p1^6 p0^4 p2+12 p1^5 p0^6 p2-4 p2^2 p1^5 p0^2+12 p0^6 p2^2 p1^3$$
$$ -6 p2^2 p1^4 p0^4+9 p1^2 p0^8 p2^2-9 p2 p1^4 p0^8. $$
Is there a simpler recurrence? The highest term is $p3=P(n+3)$, so the
initial conditions are $P(0),P(1),P(2)$. The coefficient of $p3$ is
 $$ -8 p1^5 p0^2+16 p1^4 p0^4-8 p1^3 p0^6. $$
Thus there is a "Laurent phenomenon" behavior, since it is not at all a
priori clear why we get an answer that is a polynomial (with
nonnegative coefficients). For more on the Laurent phenomenon, see
e.g. http://arxiv.org/abs/math/0104241.