Parrondo’s paradox

Valentin Cornaciu

rcor.ro

2023-05-12

Parrondo’s paradox, a paradox in game theory, has been described as:
A combination of losing strategies becomes a winning strategy.

Consider playing three games, Game A, Game B and Game A+B with the following rules:

  1. Winning a game earns us 1 and losing requires us to surrender 1

  2. In Game A, we toss a biased coin, Coin 1 🟠 , with probability of winning \(P_1=1/2−\alpha\)

  3. In Game B, we first determine if our capital is a multiple of 3. If it is, we toss a biased coin, Coin 2 🟤, with probability of winning \(P_2=1/10−\alpha\). If it is not, we toss another biased coin, Coin 3 🟣, with probability of winning \(P_{3}=3/4-\alpha\).

  4. In Game A+B, we first toss a balanced Coin 0 🪙, and if it comes up head, you play Game A or play Game B otherwise.

Clarifications

  • It is named after its creator, Juan Parrondo, who discovered the paradox in 1996
  • It is clear that by playing Game A, we will almost surely lose in the long run
  • Game B posses a real problem: is a Markov chain. Is this a losing game? We will see that it is.
  • Can Game A+B be a winning game? A combination of the two games can, paradoxically, become a winning game.
  • Even if the combination is determined by a random process!

Summary

A fair coin, but we will see that it is not the same as tossing alternatively, starting with Game A, followed by Game B, (ABAB…)

(ABAB …) even if it is predictive, it has not the same predictability as (AABBAABB…)

A fair coin has maximum entropy

In Game A+B it helps when it replaces the Coin 2 🟤, which raises the winning chances.

One loses for sure, but Game B uses it only when you have a balance multiple of three. Lowest entropy from all the coins.

Starting coin in Game B, can be in a loop at at he beginning.

Playing alternatively (ABAB…), not only we do not start with this Coin, but we might not toss it when the balance is multiple of three.

In Game B you toss this coin more often in a row (two times same coin).

Move to different tabs to get all the info

  • The strong form is that even if we play the two games randomly, we can win.
  • It means that the evolution could happen by pure luck.
  • Synergy can be present even with no correlation at all.

  • In Game B, our balance is of the form \(3k+i\), with \(i=0:2\) which means we have a Markov chain.

  • It is important to find how many time one rolls Coin 2 or Coin 3

  • With this in mind, in the steady state, the final distribution will approach the Poisson Binomial Distribution.

  • With \(\alpha=0.005\), the transition matrix of Game B will be: \[B=\begin{bmatrix} 0 & 0.255 & 0.745 \\ 0.095 & 0 & 0.255 \\ 0.905 & 0.745 & 0 \end{bmatrix}\]

library(markovchain)
alpha<-0.005
Q<-matrix(c(0,.5-alpha,1-.5+alpha,1-.5+alpha,0,.5-alpha,.5-alpha,1-.5+alpha,0),3,3)

P<-matrix(c(0,.1-alpha,1-.1+alpha,1-.75+alpha,0,.75-alpha,.75-alpha,1-.75+alpha,0),3,3)

new("markovchain",states=as.character(0:2),byrow=FALSE,P)->mk
new("markovchain",states=as.character(0:2),byrow=FALSE,Q)->m0
new("markovchain",states=as.character(0:2),byrow=FALSE,.5*Q+.5*P)->mf

In both plots, one can see the transition probabilities of Game B for any initial state, but since we start from 0, only that inital state is highlighted.

Evolution of the game’s profit from growing the process 16 times (left) to 36 times.

Game A+B and Game (ABAB…)

  • Because of independence, the transition matrix of Game A+B, will become: \[B=\begin{bmatrix} 0 & 0.255 & 0.745 \\ 0.095 & 0 & 0.255 \\ 0.905 & 0.745 & 0 \end{bmatrix} \implies A+B=\begin{bmatrix} 0 & 0.38 & 0.62 \\ 0.295 & 0 & 0.38 \\ 0.705 & 0.62 & 0 \end{bmatrix}\]
  • Playing 50% of one of the games randomly 🪙, is not the same as playing them alternately (ABAB…), which is the crucial component of Parrondo’s Paradox.
  • The paradox arises precisely because of the alternation between the two games
  • In Game (ABAB…) we first play Game A 🟠, meaning we will play Game B surely after, by tossing the lucky Coin 3 🟣, then again Coin 1 🟠.

Game (ABAB…)

c0<-c(0,0.5-alpha,0,0,0,0.5+alpha)#🟠
c1<-c(0.25+alpha,0,0.75-alpha,0,0,0)#🟣
c2<-c(0,0.5+alpha,0,0.5-alpha,0,0)#🟠
c3<-c(0,0,0.9+alpha,0,0.1-alpha,0)#🟤
c4<-c(0,0,0,.5+alpha,0,0.5-alpha)#🟠
c5<-c(0.75-alpha,0,0,0,0.25+alpha,0)#🟣
AB<-matrix(c(c0,c1,c2,c3,c4,c5),ncol=6)
new("markovchain",states=as.character(0:5),byrow=FALSE,AB,name='(ABAB...)')->mAB
t(steadyStates(mAB))
             0         1         2         3          4          5
[1,] 0.1173357 0.2321729 0.3447361 0.1897981 0.03792821 0.07802898
t(mAB^2*c(1,0,0,0,0,0))
           0 1        2 3        4 5
[1,] 0.50245 0 0.368775 0 0.128775 0
  • We play Game A every time the balance is odd, we will have now a transition matrix of dimension \(6\): \[AB=\begin{bmatrix} 0 & 0.255 & 0 & 0 & 0 & 0.745 \\ 0.495 & 0 & 0.505 & 0 & 0 & 0 \\ 0 & 0.745 & 0 & 0.905 & 0 & 0 \\ 0 & 0 & 0.495 & 0 & 0.505 & 0 \\ 0 & 0 & 0 & 0.095 & 0 & 0.255 \\ 0.505 & 0 & 0 & 0 & 0.495 & 0 \end{bmatrix}\]

Look at the Settings tab for insights. This step generates of lot of mistakes, still they afirm: Parrondo’s Paradox is Not Paradoxical

In both plots, one can see the transition probabilities of Game (ABAB…) for any initial state, but since we start from 0, only that initial state is highlighted.

Evolution of the game’s profit after 100 times (left) and plus a toss of Coin 3 (right)