Faculty of Economics and Business
Thesis
Master of Science
Finite Games, the Textbook Model is a
Special Case
Abstract
This thesis finds deviations from all considered standard games for which the fixation probability as estimated by the textbook model produces wrong conclusions. This may lead to concluding that a strategy or individual is favored by selection, while in fact it is not. Also, a relatively straight forward formal model is presented and advice is given on how to use different methods in the models and their pitfalls.
By: Birk Jonker, BSc
6165931
Supervisor: Prof. Dr. Matthijs van Veelen Second Reader: Prof. Dr. Joep Sonnemans
Date: 13-03-2015
Specialization: Behavioural Economics and Game Theory Number of Credits: 20
Page 2 of 38
Index
1. Introduction ... 3
2. Method... 5
2.1 Population and game ... 5
2.2 Textbook model ... 5 2.3 Formal model ... 7 2.4 Scenario ... 10 2.5 Games ... 11 3. Results ... 12 3.1. Wrong conclusions ... 12
3.1.1. Summary for each game ... 12
3.1.2. Over or underestimation by the textbook model ... 13
3.1.3. Fitness: exponent vs. 1-w ... 13
3.1.4. Conclusion ... 14
3.2. Equal results ... 14
3.2.1. Patterns ... 14
3.2.2. Odd versus even i ... 21
3.2.3. General case N = i + x ... 25
3.2.4. Proof of concept ... 29
3.2.5. Conclusion ... 31
4. Discussion ... 32
References ... 33
Page 3 of 38
1. Introduction
Evolutionary game theory is a mathematical tool for the biological processes of evolution. Recently it has also been adopted in economics. As behavioural and experimental economics cannot duplicate game theoretic results, evolutionary game theory might give a rational explanation to –according to Nash– irrational behaviour. In order for a group to survive, it could be very useful to also have co-operators in the group, alongside the rational acting defectors to support the group as a whole. When anomalies from game theoretical predictions can be rationalised, microeconomic models can be adjusted to incorporate more advanced rational expectations. Thus, give a better prediction of economic behaviour.
Another use of evolutionary game theory is to analyse so called evolutionary dynamics. These dynamics describe how treats, ideas, cells, etc. evolve in a given population. Infinite populations can be described in deterministic differential equations (Weibull, 1995) while a finite population requires a stochastic analysis. Nowak (2006) provides a textbook model for the stochastic analysis of a finite population. He compares the probability of a mutant taking over the entire population to the probability of that by cause of a neutral drift or 1/𝑁. The model is based on the expected payoffs of either strategy, where each individual always plays one and the same strategy. The strategy is their treat.
The textbook model takes one source of randomness into account. This randomness is because of the update step when the payoffs are realized. This, however, ignores a second source of randomness when the payoffs are gathered because of the matching pattern. Certain matching patterns can occur with a higher probability than others. And because those patterns have different effects on the average expected payoffs, the second source of randomness could have an important effect on the dynamics. The question is whether this matters.
This thesis will present a more formal model, which incorporates both sources of randomness. This will allow to both find situations in which the textbook model should be used with caution and situations in which the textbook model is accurate.
The formal method is expected to produce different results from the textbook method. More importantly, it is expected to –in some cases– produce different conclusions. In this thesis emphasis will be on the latter case. While it is important to realise that the model will produce inexact values, a model will always produce stylised results. A model comes to fault however, when it produces the wrong conclusions. This expected value model produces the wrong conclusions if the fixation probability is on the wrong side of 1/𝑁 because that is the neutral drift and on opposite sides selection is in favour of one or the other individual (strategy).
On the other hand, it is to be expected there are special cases in which the two methods produce the same result. These special cases should not be dependent on 𝑁. Finding a general proof for these cases might be a challenge as matchings occur in many discrete variations of scenarios.
Page 4 of 38 In the following chapter, I will set out the game, two methods and the scenario. Thereafter, I will present the results from the comparison and a special case. Finally, I will conclude with a discussion.
Page 5 of 38
2. Method
The textbook model will be compared to a more formal method, which takes into account an extra source of randomness in the matching stage of the process. The most interesting value to compare is the fixation probability, the probability that one mutant takes over the entire population. This value is assessed by comparing it to 1/𝑁. If they are equal, the mutant is said to be neutral as the probability for it to take over the population is equal to the random drift. If it is larger, selection favours the mutant, if it is smaller it favours the incumbent individuals. If the two models results in values on different sides of 1/𝑁, the textbook method would prove inadequate to correctly judge the dynamics of evolution.
Below I will describe the population and game that are the basis for this study, followed by a description of the textbook model, the formal model and finally a description of the chosen scenario.
2.1 Population and game
The population consists of 𝑁 individuals, of whom 𝑖 are Red individuals and 𝑁 − 𝑖 are Blue individuals. The states 𝑖 = 0 and 𝑖 = 𝑁 are considered absorption states, from this point there will be no changes in 𝑖. The birth-death process is determined by the Moran process. One individual is chosen for reproduction and one individual is chosen to die. The state of 𝑖 can therefore be changed by 1 maximally. Reproduction is (partly) determined by a standard, symmetric game with the strategies Red and Blue and the payoff matrix:
𝑅𝑒𝑑
𝐵𝑙𝑢𝑒(𝑎 𝑏𝑐 𝑑)
Selection for death is random. Red is chosen with probability 𝑖/𝑁 and Blue with probability (𝑁 − 𝑖)/ 𝑁.
2.2 Textbook model
The basic textbook model is taken from Nowak (2006). This model is an expected payoff model and is defined by the expected payoff for Red and Blue respectively:
𝐹𝑖 =𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖) 𝑁 − 1 𝐺𝑖=𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)
Page 6 of 38 With 0 ≤ 𝑤 ≤ 1 as a measure for the intensity of selection, the amount to which the birth process is determined by the game. 𝑤 = 0 indicates that the game has no influence on selection at all, also known as ‘weak selection’ and 𝑤 = 1, ‘strong selection’ indicates that selection is entirely determined by the game. The fitness of Red and Blue are given by:
𝑓𝑖𝑤= 1 − 𝑤 + 𝑤𝐹 𝑖
𝑔𝑖𝑤= 1 − 𝑤 + 𝑤𝐺𝑖
More recent studies (Traulsen et al., 2008) often use a definition for the fitness that makes use of an exponent:
𝑓𝑖𝑒= 𝑒𝑤𝐹𝑖
𝑔𝑖𝑒= 𝑒𝑤𝐺𝑖
Both definitions will be separate methods in both the textbook as the formal model. Next, the probability to move from 𝑖 to 𝑖 + 1or 𝑖 − 1 is given by:
𝑇𝑖+= 𝑖𝑓𝑖 𝑖𝑓𝑖+ (𝑁 − 𝑖)𝑔𝑖 𝑁 − 𝑖 𝑁 𝑇𝑖−= (𝑁 − 𝑖)𝑔𝑖 𝑖𝑓𝑖+ (𝑁 − 𝑖)𝑔𝑖 𝑖 𝑁
The probability to move up is given by the relative fitness of Red times the probability that a Blue individual is selected to die. The probability to move down is the relative fitness of Blue times the probability that a Red player is selected to die. The down to up transition ratio is then given by:
𝑇𝑖−
𝑇𝑖+=
𝑔𝑖 𝑓𝑖
Page 7 of 38 Which leads to a fixation probability of Red, the probability that Red will go from 𝑖 = 1 to 𝑖 = 𝑁:
𝜌𝑅 = 1 1 + ∑ ∏ 𝑇𝑖− 𝑇𝑖+ 𝑘 𝑖=1 𝑁−1 𝑘=1 = 1 1 + ∑ ∏ 𝑔𝑖 𝑓𝑖 𝑘 𝑖=1 𝑁−1 𝑘=1
Red is a neutral if 𝜌𝑅= 1/𝑁. If 𝜌𝑅> 1/𝑁, selection favors Red replacing Blue and vice-versa.
2.3 Formal model
The formal model is a more extensive model that follows all steps in the process more precisely. It differs most notably from the textbook model where it imposes a second source of randomness: the matching stage. Where the textbook expected value model just looks at if one is Red, what is the probability that one meets another Red or a Blue individual and multiplies this with the correspondent payoff. The formal method takes into account the various possible matching situation, given any 𝑁, 𝑖. If, e.g. 𝑁 = 8 and 𝑖 = 2 there are two possible situations: one where the two Red individuals are matched together and one where the both Red individuals are matched to a Blue individual. Such matching occurs with a certain probability, the second form of randomness. This method first calculates all down and up probabilities for any situation, then weight them with the probability of the situation occurring and sums them to come to the general down and up probability, given 𝑖.
Situation RBC, given 𝑁, 𝑖, is one matching in which there are R all Red individual pairs, B all Blue individual pairs and C pairs that consist of a combination of a Red and a Blue individual. I will use the notation 𝑘, 𝑙, 𝑚 for the number of Red, Blue and Combined pairs. Number of pairs, in situation RBC, consisting of:
Two Red individuals: 𝑅𝑘,𝑙,𝑚= 𝑘
Two Blue individuals: 𝐵𝑘,𝑙,𝑚 = 𝑙 One Red and one Blue individual: 𝐶𝑘,𝑙,𝑚 = 𝑚
Page 8 of 38 The probability of being in situation RBC, given 𝑁, 𝑖 is determined by two parts. First by the probability of it being drawn in one order for it to form the situation RBC, this is the same for each possible combination. The second part counts the total number of possible combinations to form the situation RBC.
𝑝𝑘,𝑙,𝑚=𝑖! (𝑁 − 𝑖)! 𝑁!
(𝑁2) ! 2𝑚
𝑘! 𝑙! 𝑚!
The payoff, in situation RBC, of Red and Blue players respectively are:
Π𝑘,𝑙,𝑚 𝑅 =2𝑎𝑘 + 𝑏𝑚
𝑖 Π𝑘,𝑙,𝑚 𝐵 =𝑐𝑚 + 2𝑑𝑙
𝑛 − 𝑖
The total payoff in situation R,B,C for Red individuals is the number of Red pairs times the payoff For Red meeting Red times two (since both Red individuals earn this payoff) plus the number of Combination pairs times the payoff of Red meeting Blue.
For the intensity of selection I use the two methods, analogue to the textbook method. Fitness is now given by:
π𝑘,𝑙,𝑚 𝑅,𝑤 = 1 − 𝑤 + 𝑤Π𝑘,𝑙,𝑚 𝑅 π𝑘,𝑙,𝑚 𝐵,𝑤 = 1 − 𝑤 + 𝑤Π𝑘,𝑙,𝑚 𝐵
And for the exponent method by:
π𝑘,𝑙,𝑚 𝑅,𝑒 = 𝑒𝑤Π𝑘,𝑙,𝑚 𝑅
Page 9 of 38 The probability that a Red or a Blue individual is being selected for reproduction in a situation RBC is the payoff of the specific strategy divided by the sum of both payoffs. For red players e.g., this is
𝑖𝜋𝑘,𝑙,𝑚 𝐴
𝑖𝜋𝑘,𝑙,𝑚 𝐴 + (𝑛 − 𝑖)𝜋 𝑘,𝑙,𝑚 𝐵
The probability that a Red or Blue player is selected to die is random: 𝑖 𝑁⁄ and (𝑁 − 𝑖) 𝑁⁄ respectively. The probability that 𝑖 will move down to 𝑖 − 1, given RBC is therefore:
𝑆𝑘,𝑙,𝑚 − = (𝑛 − 𝑖)𝜋𝑘,𝑙,𝑚 𝐵 𝑖𝜋𝑘,𝑙,𝑚 𝑅 + (𝑛 − 𝑖)𝜋 𝑘,𝑙,𝑚 𝐵 ∙ 𝑖 𝑁
And the probability that 𝑖 will move up to 𝑖 + 1, given RBC is:
𝑆𝑘,𝑙,𝑚 + = 𝑖𝜋𝑘,𝑙,𝑚 𝑅
𝑖𝜋𝑘,𝑙,𝑚 𝑅 + (𝑛 − 𝑖)𝜋 𝑘,𝑙,𝑚 𝐵
∙𝑁 − 𝑖 𝑁
To account for all possible matching situations, we sum the product of the probability of being in situation RBC, given 𝑁, 𝑖 and their respective probability of moving up or down over each possible situation RBC. 𝑆𝑖−= ∑ [𝑝 𝑘,𝑙,𝑚 ∙ 𝑆𝑘,𝑙,𝑚 − ] 𝑘,𝑙,𝑚 𝑆𝑖+= ∑ [𝑝 𝑘,𝑙,𝑚 ∙ 𝑆𝑘,𝑙,𝑚 + ] 𝑘,𝑙,𝑚
Page 10 of 38 𝜚𝑅 = 1 1 + ∑ ∏ 𝑆𝑖− 𝑆𝑖+ 𝑘 𝑖=1 𝑁−1 𝑘=1
2.4 Scenario
To proof the textbook method is imprecise, only one example would suffice. Therefore, one scenario is presented: 𝑁 = 8. This is the smallest 𝑁 for which there are more than two possible matching situations for at least one 𝑖. For each 𝑖 the possible situations RBC and their probability are given in Table 1. The table gives a representation of each population and matching situation in a visual representation. For all 𝑖 = 𝑦, the situation in 𝑖 = 𝑁 − 𝑦 gives the reversed outcome and can therefore be depicted in the same way.
The presented parameters for the intensity of selection will be 𝑤 = 1, 𝑤 = 0,5 and 𝑤 = 0,1. This will show results for a spectrum of strong selection to weak selection. True weak selection of 𝑤 = 0 would by definition result in the random drift fixation probability of 1/𝑁.
i p x o o o x o o o 1 | | | | o o o o o o o o x x o o 6/7 | | | | x x o o o o o o o o o o x - x o o 1/7 | | o - o o o x x x o 4/7 | | | | x x x o o o o o o o o o x - x x - o 3/7 o - o o - o x x x x 8/35 | | | | o o o o x x x x x - x x x 24/35 | | o o o o o - o o o x - x x - x 3/35 o - o o - o Population Matching
Table 1: Populations, matchings, situations and probabilities i=1 or i=7 i=2 or i=6 i=3 or i=5 i=4
Page 11 of 38
2.5 Games
Five different types of standard games will be considered. Each has standard values for the payoffs 𝑎, 𝑏, 𝑐, 𝑑. For each element in the matrix the ceteris paribus values for which the textbook and the formal method result in fixation probabilities of Red on opposing sides of 1/𝑁. If there are such values, the upper and lower bound of the element will be given for which this is the case. The considered games are: coordination, anti-coordination, prisoners dilemma, chicken game and the hawk-dove game. For calculation purposes, in the case of a standard payoff matrix that consists of one or more zero values or negative, a number will be added to each entry in the matrix such that the lowest value in the matrix is 1. The standard game for the coordination game will be considered:
𝑅𝑒𝑑
𝐵𝑙𝑢𝑒(𝑎 𝑏𝑐 𝑑) = (2 11 2)
The standard game for the anti-coordination game will be considered:
𝑅𝑒𝑑
𝐵𝑙𝑢𝑒(𝑎 𝑏𝑐 𝑑) = (1 22 1)
The standard game for the prisoners dilemma game will be considered:
𝑅𝑒𝑑
𝐵𝑙𝑢𝑒(𝑎 𝑏𝑐 𝑑) = (3 14 2)
The standard game for the chicken game will be considered:
𝑅𝑒𝑑
𝐵𝑙𝑢𝑒(𝑎 𝑏𝑐 𝑑) = (10 911 0)
The standard game for the hawk-dove game will be considered:
𝑅𝑒𝑑
Page 12 of 38
3. Results
In this chapter, I will present the results for both the question where the textbook method produces wrong conclusion and where it produces the same results as the formal method. These separate parts will be concluded in their respective sections.
3.1.
Wrong conclusions
For each game and different intensity of selection values the textbook method produces different conclusions from the formal method. In Appendix 1 all upper and lower bounds are given, together with information on whether the textbook method over or underestimates the fixation probability of Red, the length of the interval and the distance from the interval to the standard payoff.
In general, for each type of game there are payoff matrices to be found for which the textbook model produces wrong conclusions.
3.1.1. Summary for each game
In the coordination game e.g. 𝑎 has, for an intensity of selection of 1, 0,5 and 0,1 respectively the range < 4,64; 6,61 >, < 3,75; 4,13 > and < 3,14; 3,16 > for the original w-textbook method where the textbook fixation probability of Red is on the opposite side of 1/𝑁 from the formal method. As we move from strong selection to weak selection, the length of the interval of wrong conclusion payoffs and its distance to the standard payoff decrease. Furthermore, the textbook overestimates the fixation probability in the standard game for the w-method while it underestimates it for the exponent method. For the wrong conclusion intervals of 𝑎 (above standard) and 𝑑 (below standard) the standard over or underestimation holds while for 𝑏 (above standard) and 𝑐 (below standard) the signs switches.
For the anti-coordination the latter result also holds. The distance to the interval decreases when selection weakens, while the length of the interval increases.
The prisoners dilemma shows to be a special game. The original textbook method and the formal method produce the exact same fixation probability for the standard game. Small deviations from the standard do not result in opposite conclusions between the two methods. In section 3.2. I will set out a proof of concept to find 𝑎, 𝑏, 𝑐, 𝑑 for which this holds.
If we move from strong to weak selection in the chicken game, the length of the interval decreases. The distance from the standard game to the interval varies for the exponent method and decreases for the w-method. For the latter method, the textbook method underestimates the fixation
Page 13 of 38 probability while the exponent method overestimates for the standard game but estimation varies in the wrong conclusion intervals.
The hawk-dove game shows a decreasing interval length when moving from strong to weak selection. The distance to the interval increases. The w-textbook method overestimates and the exponent method underestimates the fixation probability in both the standard game and the intervals.
3.1.2. Over or underestimation by the textbook model
For the coordination, anti-coordination and the hawk-dove game, the w-textbook method overestimates the fixation probability of Red in the standard games. The overestimation follows intuition. If a Red individual matches with the same type, the other type, the Blue individuals, also pair up. Hence, not only the nominator of the fitness of Red is increased but also the denominator by at least as much, decreasing the fitness of Red. The textbook model fails to incorporate these dynamics and therefore overestimates the fixation probability.
In the wrong intervals of 𝑏 and 𝑐, the sign switches. Now the textbook model underestimates the fixation probability. In the formal model, the positive effect on the fixation probability of the change in 𝑏 and 𝑐 is accounted less strongly compared to the textbook model then a positive effect on the fixation probability by a change in 𝑎 and 𝑑. The latter are accounted for twice in the formal model because if there is one extra Red couple, both are assigned 𝑎 while if there is an extra Combined couple the positive payoff is accounted only once, either 𝑏 or 𝑐 has changed. This effect creates values for 𝑏 and 𝑐 where the sign switches from over to underestimation.
The exponent method mirrors the sign of the w-textbook method in most cases. This unintuitive result finds its origin in the values of 𝑇− and 𝑆− and 𝑇+ and 𝑆+. For 𝑖 = 1, 𝑖 = 𝑁/2 and 𝑖 = 𝑁 − 1, the
𝑇−= 𝑆− and 𝑇+= 𝑆+ for both methods. However, between 𝑖 = 1 and 𝑖 = 𝑁/2 and between 𝑖 = 𝑁/2
and 𝑖 = 𝑁 − 1 the signs are opposite between the two methods. Also the sign switches around 𝑖 = 𝑁/2. The first part weighs heavier in the summed product of the quotient, determining the overall over or underestimation of the textbook model.
3.1.3. Fitness: exponent vs. 1-w
The length of the interval is in general larger for the w-textbook method compared to the exponent method. The distance to the interval does not show any conclusive results. Making the exponent method the more reliable method.
Page 14 of 38
3.1.4. Conclusion
First, the results above show that in general the textbook method should be interpreted with caution. Only the prisoners dilemma seems, in general, to produce reliable conclusion. Second, when using the textbook model with the standard (anti-)coordination games, the w-textbook method tends to overestimate the fixation probability while the exponent method underestimates it. Third, the exponent method seems to produce more reliable results compared to the original textbook method.
3.2.
Equal results
Below I will set out a proof of concept in order to find an expression for 𝑎, 𝑏, 𝑐, 𝑑 that produces the same fixation probability for the formal and the textbook method. First, I will describe certain patterns in the formal model with their effects. Thereafter, I will look at the effects of an even and an odd 𝑖 and finally, to give a proof of concept by showing how the model looks for various 𝑁.
3.2.1. Patterns
Matchings follow a very specific pattern. For even 𝑖, the maximum number of Red pairs are 𝑖/2, of Blue pairs (𝑁 − 𝑖)/2 and 0 Combined pairs. For odd 𝑖, the maximum number of Red and Blue pairs are rounded down to the first integer and there is one Combined pair. From that maximum Red matching, one step down results in one less Red, one less Bleu and two more Combined pairs. This pattern repeats itself until there are a maximum number of Combined pairs. For instance:
𝑅1,1,2= 𝑅2,2,0− 1
𝐵1,1,2= 𝐵2,2,0− 1 𝐶1,1,2= 𝐶2,2,0+ 2
If we define the situation RBC in which there are as many Red pairs as possible ‘1’, each function of 𝑅𝑅,𝐵,𝐶, 𝐵𝑅,𝐵,𝐶 and 𝐶𝑅,𝐵,𝐶 can be reduced to a function of 𝑅1, 𝐵1 and 𝐶1. Each step down is +1 in the
subscript. The probability to be in situation 𝑘, 𝑙, 𝑚:
𝑝𝑘,𝑙,𝑚 =
𝑖! (𝑁 − 1)! 𝑁!
(𝑁2) ! 2𝑚
Page 15 of 38 Is then defined by:
𝑝𝑗=𝑖! (𝑁 − 𝑖)! 𝑁! (𝑁2) ! 2𝐶1+2𝑗−2 (𝑅1− 𝑗 + 1)! (𝐵1− 𝑗 + 1)! (𝐶1+ 2𝑗 − 2)! With, 𝑝1= 𝑖! (𝑁 − 𝑖)! 𝑁! (𝑁2) ! 2𝐶1 𝑅1! 𝐵1! 𝐶1!
The probability to move down as before, is defined by a sum of the probabilities to go down for each situation multiplied by the probability of being in that situation:
𝑆𝑖−= ∑ [𝑝𝑘,𝑙,𝑚 ∙ 𝑆𝑘,𝑙,𝑚 – ] 𝑘,𝑙,𝑚 With 𝑆𝑘,𝑙,𝑚 − = (𝑁 − 𝑖)𝜋𝑘,𝑙,𝑚 𝐵 𝑖𝜋𝑘,𝑙,𝑚 𝑅 + (𝑁 − 𝑖)𝜋 𝑘,𝑙,𝑚 𝐵 ∙ 𝑖 𝑁 = 𝑖 𝑁 [ (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑚 + 2𝑑𝑙𝑁 − 𝑖 )) 𝑖 (1 − 𝑤 + 𝑤 (2𝑎𝑘 + 𝑏𝑚𝑖 )) + (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑚 + 2𝑑𝑙𝑁 − 𝑖 ))] = 𝑖 𝑁[ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝑚 + 2𝑑𝑙) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑘 + 𝑏𝑚) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝑚 + 2𝑑𝑙)]
Page 16 of 38 𝑆𝑖−= 𝑖 𝑁[𝑝1 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) + 𝑝2 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶2+ 2𝑑𝐵2) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅2+ 𝑏𝐶2) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶2+ 2𝑑𝐵2) + 𝑝3 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶3+ 2𝑑𝐵3) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅3+ 𝑏𝐶3) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶3+ 2𝑑𝐵3) + ⋯ ] = 𝑖 𝑁[𝑝1 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝2 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1 + 2𝑑𝐵1+ 2(𝑐 − 𝑑)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1+ 2(𝑏 + 𝑐 − 𝑎 − 𝑑)) + 𝑝3 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1+ 4(𝑐 − 𝑑)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1+ 4(𝑏 + 𝑐 − 𝑎 − 𝑑)) + ⋯ ] If 𝑎 + 𝑑 = 𝑏 + 𝑐, then: 𝑆𝑖−= 𝑖 𝑁[𝑝1 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝2 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1+ 2(𝑐 − 𝑑)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝3 (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1+ 4(𝑐 − 𝑑)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + ⋯ ] = 𝑖 𝑁[ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) + 𝑤(𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) ] When we equate this with the probability to move down from the textbook model:
Page 17 of 38 𝑇𝑖−= (𝑁 − 𝑖)𝑔𝑖 𝑖𝑓𝑖+ (𝑁 − 𝑖)𝑔𝑖 𝑖 𝑁 = 𝑖 𝑁 [ (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑖 + 𝑑𝑁 − 1(𝑁 − 𝑖 − 1))) 𝑖 (1 − 𝑤 + 𝑤 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)𝑁 − 1 )) + (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)𝑁 − 1 )) ] = 𝑖 𝑁 [ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 𝑁 − 𝑖𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 (𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑖 𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)))]𝑁 − 𝑖 We get: 𝑇𝑖−= 𝑆 𝑖− ⇔ 𝑖 𝑁 [ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 𝑁 − 𝑖𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 (𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑖 𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)))]𝑁 − 𝑖 = 𝑖 𝑁[ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) + 𝑤(𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) ]
If the denominators equate:
𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 (𝑁 − 1𝑖 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) ⇔ (𝑁 − 1𝑖 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1
Page 18 of 38 Then 𝑇𝑖−= 𝑆 𝑖− ⇔ (𝑁 − 𝑖)(1 − 𝑤) + 𝑤𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) + 𝑤(𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔ 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 𝑐𝐶1+ 2𝑑𝐵1+ (𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ )
The probability to move up as before, is defined by a sum of the probabilities to go up for each situation multiplied by the probability of being in that situation:
𝑆𝑖+= ∑ [𝑝 𝑘,𝑙,𝑚 ∙ 𝑆𝑘,𝑙,𝑚 + ] 𝑘,𝑙,𝑚 With 𝑆𝑘,𝑙,𝑚 + = 𝑖𝜋𝑘,𝑙,𝑚 𝑅 𝑖𝜋𝑘,𝑙,𝑚 𝑅 + (𝑁 − 𝑖)𝜋 𝑘,𝑙,𝑚 𝐵 ∙𝑁 − 𝑖 𝑁 =𝑁 − 𝑖 𝑁 [ 𝑖 (1 − 𝑤 + 𝑤 (2𝑎𝑘 + 𝑏𝑚𝑖 )) 𝑖 (1 − 𝑤 + 𝑤 (2𝑎𝑘 + 𝑏𝑚𝑖 )) + (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑚 + 2𝑑𝑙𝑁 − 𝑖 ))] = [ 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑘 + 𝑏𝑚) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑘 + 𝑏𝑚) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝑚 + 2𝑑𝑙)]
Page 19 of 38 When defined by the situation of the maximum number of red pairs this becomes:
𝑆𝑖+=𝑁 − 𝑖 𝑁 [𝑝1 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶1+ 2𝑑𝐵1) + 𝑝2 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅2+ 𝑏𝐶2) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅2+ 𝑏𝐶2) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶2+ 2𝑑𝐵2) + 𝑝3 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅2+ 𝑏𝐶3) 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅3+ 𝑏𝐶3) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(𝑐𝐶3+ 2𝑑𝐵3) + ⋯ ] =𝑁 − 𝑖 𝑁 [𝑝1 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝2 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1+ 2(𝑏 − 𝑎)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1+ 2(𝑏 + 𝑐 − 𝑎 − 𝑑)) + 𝑝3 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1+ 4(𝑏 − 𝑎)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1+ 4(𝑏 + 𝑐 − 𝑎 − 𝑑)) + ⋯ ] If 𝑎 + 𝑑 = 𝑏 + 𝑐, then: =𝑁 − 𝑖 𝑁 [𝑝1 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝2 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1+ 2(𝑏 − 𝑎)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + 𝑝3 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1+ 4(𝑏 − 𝑎)) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1) + ⋯ ] =𝑁 − 𝑖 𝑁 [ 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) + 𝑤(𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1)]
Page 20 of 38 When we equate this with the probability to move down from the textbook model:
𝑇𝑖+= 𝑖𝑓𝑖 𝑖𝑓𝑖+ (𝑁 − 𝑖)𝑔𝑖 𝑁 − 𝑖 𝑁 =𝑁 − 𝑖 𝑁 [ 𝑖 (1 − 𝑤 + 𝑤 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)𝑁 − 1 )) 𝑖 (1 − 𝑤 + 𝑤 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)𝑁 − 1 )) + (𝑁 − 𝑖) (1 − 𝑤 + 𝑤 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)𝑁 − 1 )) ] =𝑁 − 𝑖 𝑁 [ 𝑖(1 − 𝑤) + 𝑤𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖))𝑖 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 (𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑖 𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)))]𝑛 − 𝑖 We get: 𝑇𝑖+= 𝑆𝑖+ ⇔𝑁 − 𝑖 𝑁 [ 𝑖(1 − 𝑤) + 𝑤𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖))𝑖 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤 (𝑁 − 1 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑖 𝑁 − 1 (𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)))]𝑁 − 𝑖 =𝑁 − 𝑖 𝑁 [ 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) + 𝑤(𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) 𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1)]
If the denominators equate:
𝑖(1 − 𝑤) + (𝑁 − 𝑖)(1 − 𝑤)
+ 𝑤 (𝑁 − 1𝑖 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑁 − 𝑖
𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 𝑖(1 − 𝑤) + (𝑛 − 𝑖)(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1)
Page 21 of 38 ⇔ 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 2𝑎𝑅1+ 2𝑑𝐵1+ (𝑏 + 𝑐)𝐶1 Then 𝑇𝑖+= 𝑆 𝑖+ ⇔ 𝑖(1 − 𝑤) + 𝑤 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) = 𝑖(1 − 𝑤) + 𝑤(2𝑎𝑅1+ 𝑏𝐶1) + 𝑤(𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔ 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) = 2𝑎𝑅1+ 𝑏𝐶1+ (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ )
3.2.2. Odd versus even i
Because of the impossibility of a general proof when 𝑆𝑖−= 𝑇
𝑖− and 𝑆𝑖+= 𝑇𝑖+, as the formal model has
too many (discrete) variables, I have to resort to a proof of concept. First, one should consider two possible sets of 𝑖: even 𝑖 and odd 𝑖. For odd 𝑖:
𝑅1=𝑖 − 1 2 𝐵1 = 𝑁 − 𝑖 − 1 2 𝐶1= 1
This gives us:
𝑝𝑗=𝑖! (𝑁 − 𝑖)! 𝑛!
(𝑁2) ! 22𝑗−1
Page 22 of 38 And for the probability to move down if,
( 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 2𝑎𝑖 − 1 2 + 2𝑑 𝑁 − 𝑖 − 1 2 + (𝑏 + 𝑐) ⇔ (𝑁 − 1𝑖 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 𝑎(𝑖 − 1) + 𝑑(𝑛 − 𝑖 − 1) + 𝑏 + 𝑐 ⇔ ( 𝑖 𝑁 − 1− 1) 𝑎(𝑖 − 1) + ( 𝑁 − 𝑖 𝑁 − 1− 1) 𝑑(𝑁 − 𝑖 − 1) = 𝑏 − 𝑖 𝑛 − 1𝑏(𝑁 − 𝑖) + 𝑐 − 𝑁 − 𝑖 𝑁 − 1𝑐𝑖 Then 𝑇𝑖−= 𝑆 𝑖− ⇔ 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 𝑐 + 𝑑(𝑁 − 𝑖 − 1) + (𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) And for the probability to move up if,
𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 2𝑎𝑖 − 1 2 + 2𝑑 𝑁 − 𝑖 − 1 2 + (𝑏 + 𝑐) ⇔ 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 𝑎(𝑖 − 1) + 𝑑(𝑛 − 𝑖 − 1) + 𝑏 + 𝑐
Page 23 of 38 ⇔ ( 𝑖 𝑁 − 1− 1) 𝑎(𝑖 − 1) + ( 𝑁 − 𝑖 𝑁 − 1− 1) 𝑑(𝑁 − 𝑖 − 1) = 𝑏 − 𝑖 𝑁 − 1𝑏(𝑁 − 𝑖) + 𝑐 − 𝑁 − 𝑖 𝑁 − 1𝑐𝑖 Then 𝑇𝑖+= 𝑆 𝑖+ ⇔ 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) = 2𝑎 𝑖 − 1 2 + 𝑏 + (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) = 𝑎(𝑖 − 1) + 𝑏 + (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) For an even 𝑖: 𝑅1= 𝑖 2 𝐵1 =𝑁 − 𝑖 2 𝐶1= 0 This gives: 𝑝𝑗= 𝑖! (𝑁 − 𝑖)! 𝑛! (𝑁2) ! 22𝑗−2 (2 − 𝑗 + 1)! (𝑖 𝑁 − 𝑖2 − 𝑗 + 1)! (2𝑗 − 2)!
And for the probability to move down, if
( 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 2𝑎 𝑖 2+ 2𝑑 𝑁 − 𝑖 2
Page 24 of 38 ⇔ ( 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 𝑎𝑖 + 𝑑(𝑛 − 𝑖) ⇔ 𝑖 𝑁 − 1𝑎(𝑖 − 1) − 𝑎𝑖 + 𝑁 − 𝑖 𝑁 − 1𝑑(𝑁 − 𝑖 − 1) − 𝑑(𝑁 − 𝑖) = − 𝑖 𝑁 − 1𝑏(𝑁 − 𝑖) − 𝑁 − 𝑖 𝑁 − 1𝑐𝑖 Then 𝑇𝑖−= 𝑆𝑖− ⇔ 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1)) = 𝑑(𝑁 − 𝑖) + (𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) And for the probability to move up, if
( 𝑖 𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) + 𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 2𝑎 𝑖 2+ 2𝑑 𝑁 − 𝑖 2 ⇔ (𝑁 − 1𝑖 (𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) +𝑁 − 𝑖 𝑁 − 1(𝑐𝑖 + 𝑑(𝑁 − 𝑖 − 1))) = 𝑎𝑖 + 𝑑(𝑁 − 𝑖) ⇔ 𝑖 𝑁 − 1𝑎(𝑖 − 1) − 𝑎𝑖 + 𝑁 − 𝑖 𝑁 − 1𝑑(𝑁 − 𝑖 − 1) − 𝑑(𝑁 − 𝑖) = − 𝑖 𝑁 − 1𝑏(𝑁 − 𝑖) − 𝑁 − 𝑖 𝑁 − 1𝑐𝑖 Then 𝑇𝑖+= 𝑆 𝑖+
Page 25 of 38
⇔ 𝑖
𝑁 − 1(𝑎(𝑖 − 1) + 𝑏(𝑁 − 𝑖)) = 𝑎𝑖 + (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ )
3.2.3. General case N = i + x
For both odd and even 𝑖, 𝑁 can be defined generally by 𝑖 plus any integer 𝑥: 𝑁 = 𝑖 + 𝑥. To maintain the assumption of an even 𝑁, 𝑥 must be odd or even parallel to 𝑖. For the odd case and the probability to move down, if ( 𝑖 𝑖 + 𝑥 − 1− 1) 𝑎(𝑖 − 1) + ( 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1− 1) 𝑑(𝑖 + 𝑥 − 𝑖 − 1) = 𝑏 − 𝑖 𝑖 + 𝑥 − 1𝑏(𝑖 + 𝑥 − 𝑖) + 𝑐 − 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1𝑐𝑖 ⇔ (𝑖𝑥 − 𝑖 − 𝑥 + 1)(𝑏 + 𝑐) = (𝑖𝑥 − 𝑖 − 𝑥 + 1)(𝑎 + 𝑑) ⇔ 𝑎 + 𝑑 = 𝑏 + 𝑐 Then 𝑇𝑖−= 𝑆𝑖− ⇔ 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1(𝑐𝑖 + 𝑑(𝑖 + 𝑥 − 𝑖 − 1)) = 𝑐 + 𝑑(𝑖 + 𝑥 − 𝑖 − 1) + (𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔(𝑥𝑖 − 𝑖 − 𝑥 + 1)(𝑐 − 𝑑) (𝑖 + 𝑥 − 1)(𝑐 − 𝑑) = 2𝑝2 + 4𝑝3+ ⋯ ⇔𝑥𝑖 − 𝑖 − 𝑥 + 1 𝑖 + 𝑥 − 1 = 2𝑝2 + 4𝑝3+ ⋯ And for the probability to move up, if
Page 26 of 38 ( 𝑖 𝑖 + 𝑥 − 1) (𝑎(𝑖 − 1) + 𝑏(𝑖 + 𝑥 − 𝑖)) + ( 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1) (𝑐𝑖 + 𝑑(𝑖 + 𝑥 − 𝑖 − 1)) = 𝑎(𝑖 − 1) + 𝑑(𝑖 + 𝑥 − 𝑖 − 1) + 𝑏 + 𝑐 ⇔ (𝑖𝑥 − 𝑖 − 𝑥 + 1)(𝑏 + 𝑐) = (𝑖𝑥 − 𝑖 − 𝑥 + 1)(𝑎 + 𝑑) ⇔ 𝑎 + 𝑑 = 𝑏 + 𝑐 Then 𝑇𝑖+= 𝑆 𝑖+ ⇔ 𝑖 𝑖 + 𝑥 − 1(𝑎(𝑖 − 1) + 𝑏(𝑖 + 𝑥 − 𝑖)) = 𝑎(𝑖 − 1) + 𝑏 + (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔(𝑥𝑖 − 𝑖 − 𝑥 + 1)(𝑏 − 𝑎) (𝑖 + 𝑥 − 1)(𝑏 − 𝑎) = 2𝑝2 + 4𝑝3+ ⋯ ⇔𝑥𝑖 − 𝑖 − 𝑥 + 1 𝑖 + 𝑥 − 1 = 2𝑝2 + 4𝑝3+ ⋯
This shows that both equations are equivalent, for the odd case:
𝑇𝑖+= 𝑆𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔𝑥𝑖 − 𝑖 − 𝑥 + 1
𝑖 + 𝑥 − 1 = 2𝑝2 + 4𝑝3+ ⋯ And the probability to be in situation 𝑗, for the odd case is given by:
Page 27 of 38 𝑝𝑗= 𝑖! 𝑥!
(𝑖 + 𝑥)!
(𝑖 + 𝑥2 ) ! 22𝑗−1
(𝑖 − 12 − 𝑗 + 1)! (𝑥 − 12 − 𝑗 + 1)! (2𝑗 − 1)!
For the even case and the probability to move down, if
( 𝑖 𝑖 + 𝑥 − 1) 𝑎(𝑖 − 1) − 𝑎𝑖 + ( 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1) 𝑑(𝑖 + 𝑥 − 𝑖 − 1) − 𝑑(𝑖 + 𝑥 − 𝑖) = − 𝑖 𝑖 + 𝑥 − 1𝑏(𝑖 + 𝑥 − 𝑖) − 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1𝑐𝑖 ⇔ 𝑖𝑥(𝑏 + 𝑐) = 𝑖𝑥(𝑎 + 𝑑) ⇔ 𝑎 + 𝑑 = 𝑏 + 𝑐 Then 𝑇𝑖−= 𝑆𝑖− ⇔ 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1(𝑐𝑖 + 𝑑(𝑖 + 𝑥 − 𝑖 − 1)) = 𝑑(𝑖 + 𝑥 − 𝑖) + (𝑐 − 𝑑)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔ 𝑥𝑖(𝑐 − 𝑑) (𝑖 + 𝑥 − 1)(𝑐 − 𝑑)= 2𝑝2 + 4𝑝3+ ⋯ ⇔ 𝑥𝑖 𝑖 + 𝑥 − 1= 2𝑝2 + 4𝑝3+ ⋯ And for the probability to move up, if
Page 28 of 38 ( 𝑖 𝑖 + 𝑥 − 1) (𝑎(𝑖 − 1) + 𝑏(𝑖 + 𝑥 − 𝑖)) + ( 𝑖 + 𝑥 − 𝑖 𝑖 + 𝑥 − 1) (𝑐𝑖 + 𝑑(𝑖 + 𝑥 − 𝑖 − 1)) = 𝑎𝑖 + 𝑑(𝑖 + 𝑥 − 𝑖) ⇔ 𝑥𝑖(𝑏 + 𝑐) = 𝑥𝑖(𝑎 + 𝑑) ⇔ 𝑎 + 𝑑 = 𝑏 + 𝑐 Then 𝑇𝑖+= 𝑆 𝑖+ ⇔ 𝑖 𝑖 + 𝑥 − 1(𝑎(𝑖 − 1) + 𝑏(𝑖 + 𝑥 − 𝑖)) = 𝑎𝑖 + (𝑏 − 𝑎)(2𝑝2 + 4𝑝3+ ⋯ ) ⇔ 𝑥𝑖(𝑏 − 𝑎) (𝑖 + 𝑥 − 1)(𝑏 − 𝑎)= 2𝑝2 + 4𝑝3+ ⋯ ⇔ 𝑥𝑖 𝑖 + 𝑥 − 1= 2𝑝2 + 4𝑝3+ ⋯
This shows that both equations are also equivalent. For the even case:
𝑇𝑖+= 𝑆
𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔
𝑥𝑖
𝑖 + 𝑥 − 1= 2𝑝2 + 4𝑝3+ ⋯ And the probability to be in situation 𝑗, for the even case is given by:
𝑝𝑗= 𝑖! 𝑥! (𝑖 + 𝑥)!
(𝑖 + 𝑥2 ) ! 22𝑗−2
Page 29 of 38
3.2.4. Proof of concept
Now we have an expression for a general 𝑁, I will show a proof of concept by showing that 𝑇𝑖+= 𝑆 𝑖+
and 𝑇𝑖−= 𝑆𝑖− for 𝑥 = 1, 2, 3, 4. Results found for these 𝑥, also hold in general. However, it is not possible to proof this statement in general. For the case of 𝑥 = 1 we have:
𝑇𝑖+= 𝑆 𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔ 𝑥𝑖 − 𝑖 − 𝑥 + 1 𝑖 + 𝑥 − 1 = 2𝑝2 + 4𝑝3+ ⋯ ⇔𝑖 − 𝑖 − 1 + 1 𝑖 + 1 − 1 = 0 For 𝑥 = 2 𝑇𝑖+= 𝑆 𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔ 𝑥𝑖 𝑖 + 𝑥 − 1= 2𝑝2 + 4𝑝3+ ⋯ ⇔ 2𝑖 𝑖 + 2 − 1= 2𝑝2 ⇔ 2𝑖 𝑖 + 1= 2 𝑖! 2! (𝑖 + 2)! (𝑖 + 22 ) ! 22 (2 − 2 + 1) ! (𝑖 22 − 2 + 1) ! 2! ⇔ 2𝑖 𝑖 + 1= 2 𝑖 𝑖 + 1 For 𝑥 = 3 𝑇𝑖+= 𝑆 𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔ 𝑥𝑖 − 𝑖 − 𝑥 + 1 𝑖 + 𝑥 − 1 = 2𝑝2 + 4𝑝3+ ⋯ ⇔3𝑖 − 𝑖 − 3 + 1 𝑖 + 3 − 1 = 2𝑝2
Page 30 of 38 ⇔2𝑖 − 2 𝑖 + 2 = 2 𝑖! 3! (𝑖 + 3)! (𝑖 + 32 ) ! 23 (𝑖 − 12 − 2 + 1) ! (3 − 12 − 2 + 1) ! 3! ⇔2𝑖 − 2 𝑖 + 2 = 2 𝑖 − 1 𝑖 + 2 And, for 𝑥 = 4 𝑇𝑖+= 𝑆 𝑖+⇔ 𝑇𝑖−= 𝑆𝑖−⇔ 𝑥𝑖 𝑖 + 𝑥 − 1= 2𝑝2 + 4𝑝3+ ⋯ ⇔ 4𝑖 𝑖 + 4 − 1= 2𝑝2 + 4𝑝3 ⇔ 4𝑖 𝑖 + 3= 2 𝑖! 4! (𝑖 + 4)! (𝑖 + 42 ) ! 22 (2 − 2 + 1) ! (𝑖 42 − 2 + 1) ! 2! + 4 𝑖! 4! (𝑖 + 4)! (𝑖 + 42 ) ! 24 (2 − 3 + 1) ! (𝑖 42 − 3 + 1) ! 4! ⇔ 4𝑖 𝑖 + 3= 2 6𝑖 𝑖2+ 4𝑖 + 3+ 4 𝑖2− 2𝑖 𝑖2+ 4𝑖 + 3 ⇔ (𝑖 + 3)(𝑖2+ 𝑖) = 𝑖(𝑖2+ 4𝑖 + 3) ⇔ 𝑖3+ 4𝑖2+ 3 = 𝑖3+ 4𝑖2+ 3
The results above show that
𝑇𝑖−= 𝑆
Page 31 of 38 And
𝑇𝑖+= 𝑆𝑖+ ∀𝑤, 𝑁 ∈ 2ℤ, 𝑎 + 𝑑 = 𝑏 + 𝑐
Hence, the fixation probability of Red shows equal results for the textbook and formal model:
𝜌𝑅 = 1 1 + ∑ ∏ 𝑇𝑖− 𝑇𝑖+ 𝑘 𝑖=1 𝑁−1 𝑘=1 ⇔ 1 1 + ∑ ∏ 𝑆𝑖− 𝑆𝑖+ 𝑘 𝑖=1 𝑁−1 𝑘=1 = 𝜚𝑅
3.2.5. Conclusion
The values 𝑎 + 𝑑 = 𝑏 + 𝑐 are known as equal benefits from switching. Van Veelen (2009) has already shown inclusive fitness correctly predicts games in this special case. Above, another property is demonstrated i.e. it is the only case in which the textbook model gives a correct result for the probability of moving up or down and subsequently, for the fixation probability.
Page 32 of 38
4. Discussion
In this thesis, I find that only in the special case of equal gains from switching, the textbook model for finite population evolutionary dynamics produces accurate dynamics. For all other standard games considered, I find that deviations in payoffs may cause the textbook model to produce wrong fixation probabilities. Furthermore, this thesis, develops a formal model which requires relatively little mechanical work compared to its extensive form. The formal model is more appropriate for cases where the textbook method fails to produce the correct conclusions from the fixation probability.
For all games –except equal gains from switching– there are situations for which even a small deviation from the standard payoffs can lead to wrong conclusions from the textbook model. The parameter for the intensity of selection is positively correlated with the length of the wrong-interval for all games and with the distance to the interval for the coordination game and negatively elsewhere.
If one chooses to use the textbook model, most cases show that the exponent method is preferred to the original method for fitness. In general, the intervals are shorter while the distance to the intervals varies between the two methods.
When using exponent method, it may be useful to appreciate its tendency to underestimate the fixation probability for the standard games, while the w-textbook model tends to overestimate. It may be an interesting inquiry to find what determines whether each method over or underestimates for a given payoff.
Page 33 of 38
References
Nowak, M.A., 2006, Evolutionary Dynamics; Exploring the Equations of Life, The Belknap Press of Harvard University Press, Cambridge MA and London England.
Traulsen, A., Shoresh, N., Nowak, M.A., 2008, Analytical Results for Individual and Group Selection
of Any Intensity, Bulletin of Mathematical Biology, 70, pp 1410-1424.
Van Veelen, M., 2009, Group selection, kin selection, altruism and cooperation: When inclusive fitness
is right and when it can be wrong, Journal of Theoretical Biology, Vol 259, 3, pp 589-600.
Page 34 of 38
Appendix 1 – Wrong conclusion intervals
Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 4,64 6,61 > 1,97 2,64 2 e 4,66 5,06 < 0,4 2,66 b w 1,444 1,45 < 0,006 0,444 1 e 1,497 1,498 > 0,001 0,497 c w -0,08 0,01 < 0,09 1,08 1 e -0,375 -0,389 > 0,014 1,389 d w 1,37 1,38 > 0,01 0,62 2 e 1,396 1,406 < 0,01 0,594 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 3,75 4,13 > 0,38 1,75 2 e 3,6 3,72 < 0,12 1,6 b w 1,427 1,431 < 0,004 0,427 1 e 1,4522 1,4531 > 0,0009 0,4522 c w 0,14 0,16 < 0,02 0,84 1 e 0,009 0,015 > 0,006 0,985 d w 1,42 1,43 > 0,01 0,57 2 e 1,444 1,448 < 0,004 0,552 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 3,14 3,16 > 0,02 1,14 2 e 3,095 3,104 < 0,009 1,095 > b w 1,4067 1,4077 < 0,001 0,4067 1 e 1,4109 1,4111 > 0,0002 0,4109 c w 0,295 0,3 < 0,005 0,7 1 e 0,2789 0,2799 > 0,001 0,7201 d w 1,4813 1,4846 > 0,0033 0,5154 2 e 1,4881 1,489 < 0,0009 0,511 w=1 w=0,5 w=0,1 Table A1.1 - Coordination game upper and lower bounds for conflicting fixation probabilities, under or overestimation by textbook model, the length of the interval and distance to standard value of payoff.
Page 35 of 38 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 0,53 0,57 > 0,04 0,43 1 e 0,51 0,52 < 0,01 0,48 b w 1,651 1,657 < 0,006 0,343 2 e 1,7088 1,7098 > 0,001 0,2902 c w 2,49 2,5 < 0,01 0,49 2 e 2,3471 2,3481 > 0,001 0,3471 d w 1,38 1,39 > 0,01 0,38 1 e 1,377 1,383 < 0,006 0,377 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 0,38 0,42 > 0,04 0,58 1 e 0,33 0,34 < 0,01 0,66 b w 1,63 1,634 < 0,004 0,366 2 e 1,6557 1,6565 > 0,0008 0,3435 c w 2,55 2,557 < 0,007 0,55 2 e 2,466 2,468 > 0,002 0,466 d w 1,424 1,436 > 0,012 0,424 1 e 1,438 1,442 < 0,004 0,438 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 0,113 0,133 > 0,02 0,867 1 e 0,084 0,089 < 0,005 0,911 b w 1,6069 1,6079 < 0,001 0,3921 2 e 1,6111 1,6112 > 1E-04 0,3888 c w 2,633 2,636 < 0,003 0,633 2 e 2,6175 2,6183 > 0,0008 0,6175 d w 1,482 1,484 > 0,002 0,482 1 e 1,4879 1,4888 < 0,0009 0,4879 w=1 w=0,5 w=0,1
Table A1.2 - Anti-coordination game upper and lower bounds for conflicting fixation probabilities, under or overestimation by textbook model, the length of the interval and distance to standard value of payoff.
Page 36 of 38 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 3 e x x b w x x 1 e x x c w x x 4 e x x d w x x 2 e x x Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 3 e x x b w x x 1 e x x c w x x 4 e x x d w x x 2 e x x Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 3 e x x b w x x 1 e x x c w x x 4 e x x d w x x 2 e x x w=1 w=0,5 w=0,1 Table A1.3 - Prisoners Dilemma upper and lower bounds for conflicting fixation probabilities, under or overestimation by textbook model, the length of the interval and distance to standard value of payoff.
Page 37 of 38 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 10 e x x b w 1,88 2,12 < 0,24 6,88 9 e 3,07 3,55 > 0,48 5,45 c w 23,32 27 < 3,68 12,32 11 e 16,99 18,09 < 1,1 5,99 d w 8,03 8,05 < 0,02 8,03 0 e 7,76 7,78 > 0,02 7,76 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 10 e x x b w 2 2,21 < 0,21 6,79 9 e 2,88 3,17 > 0,29 5,83 c w 23,01 25,98 < 2,97 12,01 11 e 17,46 17,77 < 0,31 6,46 d w 8,03 8,04 < 0,01 8,03 0 e 7,884 7,889 > 0,005 7,884 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w x x 10 e x x b w 2,33 2,44 < 0,11 6,56 9 e 2,62 2,65 < 0,03 6,35 c w 22,15 23,42 < 1,27 11,15 11 e 19,47 19,56 > 0,09 8,47 d w 8,012 8,026 < 0,014 8,012 0 e 7,97833 7,97838 < 5E-05 7,97833 w=1 w=0,5 w=0,1 Table A1.4 - Chicken game upper and lower bounds for conflicting fixation probabilities, under or
overestimation by textbook model, the length of the interval and distance to standard value of payoff.
Page 38 of 38 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w 0,18 0,39 > 0,21 0,61 1 e -0,53 -0,2 < 0,33 1,2 b w 3,33 3,39 > 0,06 0,61 4 e 3,2 3,26 < 0,06 0,74 c w 2,7 2,85 > 0,15 0,7 2 e 2,89 2,97 < 0,08 0,89 d w 3,76 3,85 > 0,09 0,76 3 e 3,95 4 < 0,05 0,95 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w -0,08 -0,12 > 0,04 1,12 1 e -0,66 -0,45 < 0,21 1,45 b w 3,28 3,32 > 0,04 0,68 4 e 3,2 3,23 < 0,03 0,77 c w 2,87 2,98 > 0,11 0,87 2 e x x d w 3,84 3,89 > 0,05 0,84 3 e 3,98 4 < 0,02 0,98 Variant Lower bound Upper bound TB=Formal Difference Upp-Low Distance to standard a w -0,66 -0,55 > 0,11 1,55 1 e -0,91 -0,84 < 0,07 1,84 b w 3,221 3,231 > 0,01 0,769 4 e 3,2 3,206 < 0,006 0,794 c w x x 2 e x x d w 3,96 3,97 > 0,01 0,96 3 e 3,995 4 < 0,005 0,995 w=1 w=0,5 w=0,1
Table A1.5 - Hawk-dove game upper and lower bounds for conflicting fixation probabilities, under or
overestimation by textbook model, the length of the interval and distance to standard value of payoff.
