%parrondo.tex: ``Remarks on the Parrondo Paradox''
%a Plain TeX file by Shalosh B. Ekhad Doron Zeilberger (2 pages)
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\bf
\centerline
{\bf Remarks On The PARRONDO PARADOX}
\rm
\bigskip
\centerline{ {\it
Shalosh B. EKHAD{$\,^1$} and Doron ZEILBERGER
}\footnote{$\,^1$}
{\eightrm \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA.
%\break
{\eighttt [ekhad, zeilberg]@math.temple.edu}
{\eighttt http://www.math.temple.edu/\Tilde [ekhad,zeilberg]/ },
Oct. 12, 2000. The second author is supported in part by the NSF.
This article is exclusive to the Personal Journal of
Ekhad and Zeilberger \hfill\break
{\eighttt http://www.math.temple.edu/\Tilde zeilberg/pj.html }
This article is accompanied by a Maple package PARRONDO
downloadable from Zeilberger's website.
}
}
The celebrated {\it Parrondo Paradox} (Harmer and Abbott,
Nature {\bf 402} (Dec. 23/30 1999), p. 864 and its
references, see also Paulos's gripping account in
www.ABCnews.com) loses some of its shock value
if instead of the `gambling' losing games\footnote{$^2$}
{\eightrm \raggedright
If your current capital is $n$ dollars then in game A
you win or lose one dollar with probabilities
$1/2-e$ and $1/2+e$ respectively, while in game B
you win or lose one dollar with probabilities
$p_1:=1/10-e$ and $1-p_1=9/10+e$ resp. if $n$ is divisible by $3$
otherwise you win or lose one dollar with probabilities
$p_2:=3/4-e$ and $1-p_2=1/4+e$ resp.}
A and B, whose
combinations (e.g. AABB) turned out to be winning, we take the following
completely deterministic `games'.
Suppose that your current capital is $n$ dollars. Then
Game Y (resp. Z) : If $n$ is even (resp. odd)
you WIN $1$ dollar, otherwise you LOSE $3$ dollars.
{\it Only} playing Y, or {\it only} playing Z,
results in a steady loss of one dollar
per move, but playing YZYZYZYZ... (starting with $0$ dollars) results in
a steady win of 1 dollar per move.
So Parrondo's paradox seems to be saying that it is better
to play $(YZ)^n$ rather than $Y^nZ^n$, in other words,
order matters!
But this is hardly a new message.
We were all taught to {\it think} before {\it acting},
to {\it measure} before {\it cutting}, to
{\it study} before {\it playing} and so on. Also that
{\it variety is the spice of life}. Also {\it timing
is everything},
to sell a stock when it is high, and buy it when it is low,
(etc.)$^*$.
Going back to the above `games' Y and Z, they still constitute
a `paradox' if you replace the loss of three dollars by the
loss of one dollar. Now both Y and Z are {\it fair}, i.e.
you neither win nor lose, yet playing YZ in succession
(starting with $0$ dollars) still guarantees a steady win.
Here is a variant of Paulos's very apt spatial analog. Suppose
that we have infinitely many cities labelled by the integers.
The train company Y (resp. Z) has service from city $n$ to city $n+1$
and back to $n$ for $n$ even (resp. odd). If you only buy Y tickets,
or only buy Z tickets, you'll
be going back and forth between your starting city and its immediate
neighbor, but if you buy both kinds of tickets, and
alternate between them, then you can go as far as you want.
But most of us use this wisdom every day when we alternate
between {\it walking}, {\it driving}, and {\it flying}.
The mathematical reason for the general Parrondo paradox is
the fact that matrices (usually) do not commute.
This brings to mind Quantum Mechanics with its own chock-full of
`paradoxes'.
\vfill\eject
{\bf The Maple Package PARRONDO Tells You the BEST Periodic Strategy}
The Maple package {\tt PARRONDO}
far extends the mathematical analysis of Harmer, Abbott and Taylor
(Proc. Royal Soc., to appear).
With A and B of the original Parrondo
paradox, and $e=0$ (i.e. both A and B are
fair), the best strategy is to play $(ABBAB)^*$ with
a steady win of $7.5646$ cents, about three times
the amount one obtains by playing $(AABB)^*$, that only yields
$2.4539$ cents. This is still true
when $p_2=3/4$ is replaced by any other $p_2>1/2$.
The best pay-off is when
game B becomes deterministic with $p_2=1$, resulting
(still with playing $(ABBAB)^*$) with a steady gain of
$36$ cents per move. When $p_2 \leq 1/2$ then any
non-trivial combination of
$A$ and $B$ is losing, so in this case {\it mixing is bad},
and one should stick to {\it only} A , or {\it only} B.
For specific $p_2$, period-length $n$,
`modolus' $M$, and bias $e$,
the function call is {\tt GBestWord(n,A,B,M,$p_2$,e);}.
For example, in the original fair case one should type
{\tt GBestWord(5,A,B,3,3/4, 0);} in order to get the
best period-5 strategy. If the bias is $e=1/100$ then
one should type {\tt GBestWord(5,A,B,3,3/4, 1/100);}.
In either case Maple responds with
the set consisting of {\tt [A,B,B,A,B]} and
its cyclic shifts, followed by the (asymptotic) gain per move.
Not as good as $(ABBAB)^*$, but far better than
$(AABB)^*$, is the period-$3$ strategy $(ABB)^*$, that yields
$6.786$ cents. Finally picking $A$ or $B$ at random
with probabilities $p$ and $1-p$ resp. is optimized
when $p=.4145$ but even with that optimal value it only
yields $2.62$ cents (do {\tt Bestp(0);} in {\tt PARRONDO}).
Finally, if you replace $M=3$ by $M=4$
(i.e. instead of having the `bad luck' amounts
be multiples of $3$, now they are multiples of
$4$), then {\tt PARROND}
(type e.g. {\tt GBestWord(12,A,B,4,3/4, 0);})
tells us that the best strategy is $(AB)^*$.
We hope that other Parrondoholics will not be paranoid, and
use our Maple package to find further interesting properties of
this, not quite paradoxical, but yet very interesting, phenomenon
called the Parrondo paradox.
\bye