A Game-Theoretic Queueing Exercise From Seinfeld
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A Game-theoretic Queueing Exercise from SeinfeldBismark Singh
AbstractNo abstract for you. Next1!
1. IntroductionIn Season 7, episode 6 of the sitcom Seinfeld, people line up to be served by the Soup
Nazi and must adhere to his strict disciplinary demands. Failure to comply with
these makes a customer abandon the line. Due to the soups delicious taste, which
makes your knees buckle, customers nonetheless brave the hardships and seek to be
served. A customer standing outside the store, facing a decision of whether to enter
or not, wants to know how likely he is to be served and in what time. We quantify some
of these parameters for three standard queueing disciplines. We also analyze two
psychological queueing strategies, from a hawk and dove game perspective, which a
customer may adopt while waiting to attempt to decrease his queue time. Finally we
attempt to see if an evolutionary stable strategy amongst these two is possible or
not.
2. Modeling:We observed from the episode that customers are served in a First-In-First-Out fashion,
and we follow that throughout to highly simplify things. Further, we define thefollowing rewards and costs associated with the queue from the customers
perspective:
Cost to purchase the soup, if served Happiness rewardat achieving the soup (Jambalaya!) if served Discomfort cost arising from standing in a regimented line: $d/min. No soup for you! Dissatisfaction and insult cost of being kicked out of the
restaurant and from the line: $c if not served, $0 if served.
The first two costs can be combined together to a reward, $r if served, and $0 else. We
assume customers arrive to the store with an exponential () process and are served
by the Soup Nazi with an exponential () process. We take that the Soup Nazi becomes
angry at customers with an inter-anger time following an exponential ()
distribution. We attempt to obtain analytical results throughout, but for comparison
purposes whenever we would provide some numerical result, we take the queueing input
parameters,1 1 1
, ,3 2 5
with all values in minutes-1. The motivation for taking
1The inspiration for the abstract comes from (Dixit, 2011)
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for the numerical computation is important and would be clear later. We take costs as1$0.1min , $8, $5d r c , i.e. waiting for 50minutes is monetarily equivalent to being
kicked out. With these as input we can numerically calculate the standard queue
output parameters.
Within FIFO we again make three distinctions:
2.1. Suffering for the SoupFirst we model the system as a simple M/M/1 queue with parameters mentioned above.
This is thus a birth death process- and the model dynamics are just like standard
queuing abandonment problems.
Figure 1 Transition rate diagram for ith state
From the balance equations we can find the steady state probabilities
0 0
1
( )...( ( 1) )( 1)
i i
i i
j
q q qi
j
(1.1)
We also define0
1
, 1
( ( 1) )
i
i i
j
j
and using the normalizing constraint 1iq we
get,
0 0
1, i
o i o i
j j
q q q
(1.2)
The queue is always stable.
We know that the Soup Nazi is an expert at soup making and customers are willing to
bear the costs associated with standing in line upto a certain amount of time with
the anticipation of being served. Next we assume that Newman and Kramer are
preparing a chart to help them decide on whether to join the queue or not. They would
need the probability of not being kicked out while at position i which we call as
Success Probability. Thus success occurs, for customer i, when one of (i-2) customers
ahead of him waiting in line is kicked out, or when the customer currently in service
finishes service -before this ith
customer is himself kicked out. Or, the kick-out time
for customer i must be larger than the minimum of the kick-out time for each of (i-2)
customers ahead of him in line and the service time for customer 1. We note that this
success probability is not the probability that he is served, but only the
probabilities of not being kicked out while at position i. It is also implicitly
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assumed that a customer can count the number already in line thus with this
probability known can decide whether to enter the queue or not. Let the sequence
1 2{ , ... }np p p denote this success probability. Further let iK be the kick-out time for the
ith
customer, andi
Tbe the waiting times at the ith
position alone. The latter definition
should be seen as associated with thei
p sequence as the amount of time a customer has
to wait at position i is the minimum of three quantities- i) service time of the
current customer, ii) the kick-out times of the (i-2) customers ahead of him, and iii) his
own kick-out time.
Thus:
1 1p (1.3)
since the first customer enters service directly.
In general we need the probability 1 2 3 1( min( , , ... ))i iP K S K K K .Since S1 and Ki are
exponentials and , the RHS is a min of (i-1) exponentials which is again anexponential with parameter ( 2)i . The final expression is thus obtained as:
( 2), 2
( 1)i
ip i
i
(1.4)
1 2 3min( , , ... ) ~ exp(( 1) )i iT S K K K i (1.5)
As a sanity check, we can easily verify the case when i.e. the kick out time is
instantaneous.
Since we have the{ }ip sequence, now we can also derive the probability that customer iis not kicked out until reaching the soup service. We define this as his Overall
Success probability. The Overall Success is the real probability of interest whereas
the success probability is only the perceived momentary success probability.
1
( )( 1)
i
i jP Overall Success pi
(1.6)
Similarly, we have a set of recursive equations for the expected waiting time for the
a customer, depending on the position he enters in line, iW . For that we would make use
of the identity,
[ ] [ | ]. [ | ].(1 )i i i i iE W E W not kicked out at i p E W kicked out at i p (1.7)
We note on the RHS of (1.7),the first term can be broken recursively as the time spent
in state i given that the customer is not kicked out and then the process repeating
itself but from state (i-1). The second term is the time spent at state i alone, if the
customer is kicked out. Now, we also have
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1 2 3 1 1 2 3 1[ ] [ | min( , , ... )]. [ | min( , , ... )].(1 )i i i i i i i i iE T E T K S K K K p E T K S K K K p (1.8)
From the memoryless property of exponentials, the conditional distribution of the
time of the first occurrence of a process(say kicked out or not kicked out) given that
this event happened before the first occurrence of the other process (say not kicked
out or kicked out) is the distribution of the first occurrence of the whole process.This gives us
1 2 3 1 1 2 3 1[ ] [ | min( , , ... )] [ | min( , , ... )]i i i i i i iE T E T K S K K K E T K S K K K (1.9)
Substituting (1.5)and (1.9)in (1.7) we get our required recursion as,
1[ ] ( [ ] [ ]) [ ](1 )i i i i i iE W E W E T p E T p (1.10)
which can be simplified again by substituting the expectation of the distribution in
(1.5) to,
1
1[ ] [ ]
( 1)i i iE W E W p
i
(1.11)
This equation has a simple intuitive feel too. The expected waiting time from the ith
stage is the time spent in that state plus the probability of moving forward to the
next stage times the expected waiting time from the (i-1)th
stage. This recursion can be
solved easily by setting [ ]i i iZ E W Y where1
( 1)iY
i
. Then,
1
1
1, 1
1
i iZ Z i
Z
and thus iZ i or
[ ]( 1)
i
iE W
i
(1.12)
Before we proceed, as a sanity check we verify Littles Law on the parameters above. It
is considerably easier to take parameter values and verify that, 10 [ ]i iiL q E W
and
1[ ]i iW q E W
are related as by Littles Law. The indices on L deserve attention as a
customer arriving at the i
th
position sees (i-1) people in the system.
2
We note a few things about the expression for E[Wi]. The expression, given by (1.12)can
be verified to be monotonically decreasing and convex when and monotonically
increasing and concave for .When the kick-out rate equals the service rate i.e.
2A significant amount of time was spent on this verification alone and the author cansafely say that all credit goes to Dr. Hasenbein for this.
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the waiting time is a constant for all entering positions, namely1
. In our
example we take the latter case, i.e. customer farther along the queue are expected to
wait longer in the system.
We present Kramer and Newmans chart in Figure 2, using the same parameter values asabove.
Figure 2 Success and Overall Success probabilities for different entering positions
The expected reward for the ith
customer can be defined by using the costs and
rewards defined above as:
[ ] (1 ) [ ]( )
( 1 )i i i i
E R rdi c
P c P dE Wi r
i
(1.13)
Note that the Pi in (1.13)is the overall success probability as defined in Expanding we
note that,
[ ] 0,
[ ] 0,
[ ] 0,
i
i
i
c rE R if i
d c
c rE R if i
d c
c rE R if id c
(1.14)
This is interesting because there exists a threshold for queue entry positions for
some rewards (and inter-anger time) beyond which the expected reward is always
positive (or negative). This is plotted inFigure 3. We also note from (1.15) that the first
and second derivatives of the expected reward are negative and positive respectively,
for i.e the expected reward is convex and decreasing.
at entry
overall
5 10 15 20Successprobability
0.2
0.4
0.6
0.8
1.0
ntry position
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'
2
''
3
( ) ( )[ ]
(( 1 ) )
2 (( ) ( ))[ ]
(( 1 ) )
i
i
d c rE R
i
c r dE R
i
(1.15)
Another interesting quantity is the expected waiting cost upto position j, whenentered at i. This is particularly of interest when a customer enters at i, and is
kicked out at j, j i , as in(2.2) to calculate his loss. It is plotted inFigure 4
( )( ),
(( 1 ) )([ ] ( [ ] [ ]
1 ))ij i j
d i jiE C d W E W j
jE
i
(2.1)
[ ] ( [( )( )
,(( 1 ) )( )
] [ ])1
ij i j
d i jE KO c d iE W E W c j
i j
(2.2)
Figure 3 Expected reward for different entering positions
5 10 15 20EntryPosition
4
2
2
4
6
8
ExpectedReward
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Figure 4 Expected waiting costs up to a certain position in queue, for different entering positions
2.2. Soup ModeWe observed from the episode that a large number of people gather outside the store
on the sidewalk waiting in line to be served soup, before they get inside the store
and into Soup Mode. Since the Soup Nazi cannot see these people he cannot be angry
at them either. Now we model this case with his power of kicking people out confinedto k-1 people alone (more precisely, k in shop implies k-1 in line).
This is again a birth death process as inFigure 5,
Figure 5 Transition rate diagram for ith stateAgain from the detailed balance equations (using the same variable symbols),
i =
15
i 15
i 3
2 4 6 8 10 12 14Exit position0.00
0.05
0.10
0.15
ExpectedCost upto
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0 0 ,
( )...( ( 1) )
( ) ,
i
i i
i k
i k
q q q i k i
q q i k
(2.3)
where
is defined as ( 1)k
. Rewriting,
( ) , 0ik i o k q q i
(2.4)
Using the normalizing we can solve (2.3)and (2.4)
1
0 0
0 0
( ) 1k
i
i kq q
(2.5)
This equation can be solved foro
q . The second half of the above equation represents a
geometric progression which is stable when the ratio is less than 1, and the first
half is a truncated expression of (1.2) and is always stable. Thus the stability
conditions lead to,
( 1)k (2.6)
(1 )k
(2.7)
We note that if we have 0 , as happens usually, then since at least 1 person
would be in the store, the queue is always stable. If then for stability there is a
threshold defining at least how many people can be allowed in the store.
We can numerically calculate the queue parameters, with the same input parameters.
However we are more interested in calculating the success probabilities. We note that
for the customers in the Soup Mode the success probabilities are the same as in the
previous section and for customers outside the shop success always happens. Thus,
( 2),
( 1)
1, 1
i
i
ip i k
i
p i k
(2.8)
The overall success probabilities are correspondingly defined as in(1.6) and are,
,( 1)
, 1( 1)
i
i
P i ki
P i kk
(2.9)
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Similarly we have the waiting time at the ith
position using the definition given
in(1.5),
~ exp(( 1) ),
~ exp(( 1) ),
i
i
T i i k
T k i k
(2.10)
This takes us back to the waiting time definition which we calculate as in(1.11). The
solution is the piecewise continuous function for [ ]i
E W ,
[ ] ,( 1)
[ ] ,( 1)
i
i
iE W i k
i
iE W i k
k
(2.11)
The associated rewards can also be calculated again using the same definition.
( ),( 1 )
(
[ ]
[ ])
,( 1 )
i
i
E R
E
di c i r i ki
di c k r kR i
k
(2.12)
Note that the expected reward is continuous at i=k due to the continuity in the Pi as
defined in(2.9). These are plotted inFigure 6.
Figure 6 Expected reward for different entry positions
2.3. I think you forgot my breadWe know that the Soup Nazi also had a cashier where customers were routed after
being served by him. This can be modeled in multiple ways as in (Hunt, 1965). However we
study this as a simple M/E2/1 queue. One change we make from the previous model
k=1
k=2
k=5
k=3
k=4
5 10 15 20EntryPosition
2
4
6
8
xpectedReward
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parameters is that now for both the Soup Nazi and the cashier the service rate are
exponentials with parameters 2, so that the overall service rate is Erlang with
parameter . We define the number of phases left in the system, m, when there are n
customers in the system as
2( 1) , {0,1}m n j j
(3.1)
Figure 7 Transition rate diagram for ith phase, i2
The transition rates are as follows (also represented inFigure 7:
12 : , 2
2
1: 2 , 12 : , 1
1: , 0
ii i i
i i ii i i
i i i
(3.2)
Note that an odd number of phases imply the customer in service is being served by
the cashier. The balance equations can be written as, with ri denoting the steady
state probability of i phases in the system,
1 2
1 1 2
1 2 2
2
2 2
(2 ) 2 , 2
o
i i i i
r r r
r r r
r r r r i
(3.3)
The solution to the above is messy but is computable in Mathematica. Using the
mapping process (3.1) we can get the steady state probabilities and the queue
parameters. So computing the phase parameters is sufficient. We can do the same
analysis for the rewards and waiting times as above, but now for each phase. Using
similar arguments as before, we have the time spent at the ith
phase alone, and the
success probability at the ith
phase as.
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1
[ ] , 31
22
iE T i
i
(3.4)
12 ( 1)
2
, 3122
i
i
p ii
(3.5)
Then using(1.10) we can model the expected reward at the ith
position.
3. Do I know you?When Jerry and his girlfriend Sheila are standing in line, with Sheila one position
ahead of Jerry, Sheila kisses Jerry but Jerry does not reciprocate. We take it that to
increase his overall success probability (and thus his reward) Jerry creates a
strategy to get Sheila kicked out and move to her spot3. This strategy could have
easily backfired as the Soup Nazi may have kicked both of them out, or instead onlyJerry out. Sheila, on the other hand, had no incentive in Jerry s strategy4. Thus in
general a customer who entered the queue at the ith
position and is currently at the
jth
position (with j i , and implicitly assuming that he is not kicked out between i and
j) may indulge in a game with the (j-1)th
customer in line. Of course, any customer has
an option to ignore anyone disturbing him/her, or refuse to play. We attempt to find
if both these types of groups can coexist or not and if so at what frequencies
(proportions). Before we proceed we provide some further definitions and some
simplifying characteristics
1. Player: A customer who takes part in this strategy and attempts to move aheadin line
a. Games between Players are very conspicuous and end only when one Playergets kicked out by the Soup Nazi (the one kicked out first is the one we
count).
2. Not Player: A customer who does not play and is contented to stay in line athis current position and wait for his turn
a. When a Not Player meets a Not Player, casual and discrete interactionfollows. None of them have an intention of throwing people out but their
talks may get noticed (although less conspicuously than the Player-Player)
The game ends when the Soup Nazi observes the one closer to him(the
opponent) and kicks him out or both maintain their stay.3. Assumptions on the game itself:
a. There is no cost to play the game itself
3As Elaine says later, So essentially you chose soup over a woman.4Although, this model may be grossly incorrect for most people.
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b. All Games are defined w.r.t. a player (without a capital P) at position j, whois one behind the opponent at (j-1). No games are initiated with a customer
behind our player.5
c. Every customer is either a Player or a Not Playerd. Games are instantaneous, so they do not interfere with the queueing input
output parameterse. A player can only be in one game at a timef. A player does not know beforehand who his opponent would beg. We still maintain that the expected kick out rate is lesser than the
expected service rate, . Later we would briefly discuss the implications
of relaxing this assumption
h. Probability of winning for a Player for a Player-Player game is zi. Probability of winning for a Player for a Player-Not Player game is y, y>zj. Probability of winning for a Not Player for a Not Player-Not Player game is
x
k. Probability of winning to a Not Player for a Not Player-Player game isundefined as he simply does not play.
4. We only model the Suffering for the Soup case, although the analysis is easilyextendable to other cases.
We define the corresponding payoff matrix (for a better understanding of these
definitions we refer the interested reader to (Prestwich) , -the web-page which was
immensely useful for this study- with Player denoting the jth
customer and Opponent
the (j-1)th
:
[ , ] [ , ']
[ ', ] [ ', ']
Player Opponent
Play NotPlayPlay E P P E P P
NotPlay E P P E P P
(3.6)
where P denotes Play and P denotes not Play. We attempt to model this as a hawk-dove
game (Alexander, 2009)where hawks behave aggressively and doves retreat immediately
if the opponent displays aggression. Thus hawk-hawk games end when one wins, hawk-
dove games win with the hawk, dove-hawk with the dove retreating and dove-dove with
sharing. We keep Players as the hawks and Not Players as the doves but make some
changes. For us, three events can happen at each position- move forward (Move),
maintain current position (Stay) or get kicked out (KO).
5. If a Player meets with a Player he risks being kicked out (KO) with themotivation of having a chance of moving forward (Move) quicker.
5There is a caveat here - if a customer is behind someone he is also ahead of someone.For large enough system size the assumption should be justified.
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6. If a Player meets with a Not Player then since the Not Player, realizing thatthe Player is cunning, would ignore him. Thus the reward is staying in this
case for the Player and the loss being a kick-out.
7. A Not Player being attracted to a Not Player would irresistibly andunknowingly start communicating with him, and if the Soup Nazi finds them he
would kick-out the opponent. Thus the reward is moving in this case for thePlayer and the loss being staying, although this may hurt our Not Player.
8. But our Not Player simply doesnt interact with a Player, then there is nospecial chance of being kicked out, other than the usual kick-out process and
he just stays there (Stay).
Thus we have the payoff matrix in(3.7), with the first entry denoting a win and the
second the loss.
/// /
Player Opponent
Play NotPlay
Play Stay KOMove KO
NotPlay Stay Stay Move Stay
(3.7)
Note that because of our assumption, if the customer is a Not Player then no matter
what the Opponent is, he stays. The expected payoff from a strategy (here Play and Not
play) is defined as:
[ ] ( ) ( ) ( )win loss
E Strategy p E Win p E Loss (3.8)
Here we should note that each of the four entries have different winning
expectations as well as probabilities. Specifically,
1
*
[ ] [ ]
[ ]
( ) ( )
(( 2 ) )(( 1 ) )
( )( )
(( 1 ) )(( 1 ) )[ ]
[ ] 0
j j
ij
c r d
j jE Move E R R
E KO E C c
E Sta
d i j
i j
y
(3.9)
where*ij
C is defined as the cost of exiting at j*if a customer entered at i. This is
equivalent to the insult cost plus the time spent waiting between i and j,
[ ] [ ]ij i jE C dE W W . (This is plotted inFigure 4using the parameter values we chose.) Note
again from(3.9) and also from Figure 8that for the Move reward is positive (thusmoving forward increases the expected reward). Also we observe from Figure 4that the
Kick-Out cost is greater as j is further away from i(thus if immediately kicked out
upon entry the wait is not so bad, as if kicked out at a position farther away).
Using (3.8)we have,
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1
1
[ , ] [ ] (1 ) [ ]
[ , '] (1 ) [ ]
[ ', ] 0
[ ', '] [ ]
j j ij
ij
j j
E P P zE R R z E C
E P P y E C
E P P
E P P xE R R
(3.10)
Figure 8 Expected Move reward for different initial positions
Further we let u be the frequency of people who Play and (1-u) that of those who do
not Play. Then the fitness of each strategy is defined as
( ) [ , ] (1 ) [ , ']
( ) [ ', ] (1 ) [ ', ']
W Play uE P P u E P P
W NotPlay uE P P u E P P
(3.11)
We can check if the Play and Not Play strategies are pure evolutionary stable
strategies (ESS) or not (Hamilton, 1967). For Not Play to be an ESS we should have,
( ) ( )W Play W NotPlay (3.12)
Expanding this using (3.10)and(3.11),
1 1( [ ] (1 ) [ ]) (1 )((1 ) [ ]) (1 )( [ ])j j ij ij j ju zE R R z E C u y E C u xE R R (3.13)
1 1( [ ] (1 ) [ ]) (1 )( [ ] (1 ) [ ])j j ij j j iju zE R R z E C u xE R R y E C (3.14)
Now we note that if 1[ ] (1 ) [ ] 0j j ijzE R R z E C then the above equation always holds and
Not Play is a pure ESS. This means that if the condition holds then regardless of itsproportion Not Play is immune to invasions by Players, and if Players arise in the
queue they would eventually be extinct. We can examine this condition further. For
Not Play to be a pure ESS,
1
[ ]
1 [ ]
ij
j j
E Cz
z E R R
(3.15)
5 10 15 20Movefrom
0.5
1.0
1.5
2.0
Move Reward
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Lets define the RHS of (3.15)as the adios-muchacho ratio, using the Soup-Nazis phrase
again (with the negative sign to make the whole term positive) at position j, ( , )r i j . For
brevity, we henceforth refer to it as simply adios ratio. We note that adios ratio
greater than unity implies the potential kick-out cost at position j exceeds the
potential move reward at j given an entry position i. Then (3.15)holds when
( , )
1 ( , )
r i jz
r i j
(3.16)
( , )1
zr i j
z
(3.17)
We can check when this equation holds true i.e. given an entry position i, at what
position j(if any) does Not Play become immune to Play(or Not Play becomes an ESS).
The mixed ESS case is also interesting i.e. 1[ ] (1 ) [ ] 0j j ijzE R R z E C , or,1
ij
zr
z
We
should note that the adios ratio is a function of the queue input parameters and
cost rates alone i.e. it does not depend on the game probabilities. Under our parameter
choices the ratio is always larger than 1 for any choice of i and j6. Then the above
condition yields that 0.5z which may not be very realistic as a Player-Player game
should have symmetric consequences.7
We do mention that if we take 0.5z (then normatively we should also take 0.5x to
have at least some sense of symmetry) then a solution to (3.14) with equality (equal
fitness of both Play and Not Play) may exist. For that, there must be a frequency, u,
such that
1
1
[ ] (1 ) [ ]
1 [ ] (1 ) [ ]
j j ij
j j ij
xE R R y E Cu
u zE R R z E C
(3.18)
which can be simplified using the adios ratio definition again.
(1 )
1 (1 )
ij
ij
x y ru
u z z r
(3.19)
For instance, if 0.5x z then some simple algebra leads to a conclusion that a
solution to u must exist. We can even specify bounds of u in this case satisfying the
condition that the adios ratio exceeds unity and is less than the odds of winning aPlayer-Player game. We have,
6The author does not expect this to be a general result. A simple excel goal seekexercise, while maintaining , r c shows that it is possible to construct many
example with the ratio less than 1.7It is possible though to extend to include cases where games are not symmetric.
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1 1
3 1.5
y yu
z y y
(3.20)
Note again that for 1ij
r (or rather1
ij
zr
z
) no solution exists to (3.19) since the RHS is
negative and the LHS is positive.
4. Pack it up. No more soup for you. Next!We conclude the following, for while remembering that our result maybe heavily
dependent on the queue parameters:
1. The Not-Play strategy is an ESS whenever the adios ratio exceeds the odds ofwinning in a Player-Player game.
2. A mixed ESS always exists when the adios ratio is less than the odds ofwinning the Player-Player game with player concentrations as defined above.
For the reader interested in the case of a brief discussion is provided in the
appendix.
References
Alexander, J. M. (2009). Evolutionary Game Theory. Retrieved from The Stanford Encyclopedia of Philosophy:
http://plato.stanford.edu/archives/fall2009/entries/game-evolutionary
Dixit, A. (2011). An Option Value Problem from Seinfeld. Economic Inquiry.
Hamilton, W. (1967). Extraordinary sex ratios. Science, 477-488.
Hunt, G. C. (1965). Sequential Arrays of Waiting Lines. Operations Research, 674-683.
Prestwich, K. N. (n.d.). Game Theory and EvolutionarilyStable Strategies . Retrieved from
http://college.holycross.edu/faculty/kprestwi/behavior/ESS/game_defs.html
APPENDIXAdios ratio Plots for entry positions i=2:15 versus exit positions with given
parameters. The x axis shows exiting positions and the y-axis shows the adios ratio.
Note that the ratio exceeds unity always and has a larger slope for exiting valuescloser to the entry position (visible more easily for larger i)
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,
,
,
2.2 2.4 2.6 2.8 3.0
1.4
1.6
1.8
2.0
2.2
2.0 2.5 3.0 3.5 4.0
1.5
2.0
2.5
3.0
3.5
2.5 3.0 3.5 4.0 4.5 5.0
2
3
4
5
3 4 5 6
2
3
4
5
6
7
3 4 5 6 7
2
4
6
8
3 4 5 6 7 8
2
4
6
8
10
12
3 5 6 7 8 9
2
4
6
8
10
12
14
4 6 8 10
5
10
15
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,
Figure 9 Adios ratio for different entering positions
We can compare this with when as inFigure 10. We keep the same parameters except
interchange the , values (so that now 0.5, 0.2 ). The ratio does go below unity
and is more convex than the corresponding graphs inFigure 9.
4 6 8 10
5
10
15
20
4 6 8 10 12
5
10
15
20
25
4 6 8 10 12
5
10
15
20
25
30
4 6 8 10 12 14
5
10
15
20
25
30
4 6 8 10 12 14
10
20
30
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,
,
,
,
2.0 2.2 2.4 2.6 2.8 3.0
1.0
1.5
2.0
2.5
3.0
2.5 3.0 3.5 4.0
2
4
6
8
2.5 3.0 3.5 4.0 4.5 5.0
2
4
6
8
0
2
4
3 4 5 6
5
10
15
20
3 4 5 6 7
5
10
15
20
25
30
3 4 5 6 7 8
10
20
30
40
3 5 6 7 8 9
10
0
0
0
0
0
4 6 8 10
20
40
60
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,
Figure 10 Adios ratio for different entering positions
4 6 8 10
20
40
60
80
4 6 8 10 12
20
40
60
80
100
4 6 8 10 12
20
40
60
80
100
120
140
4 6 8 10 12 14
50
100
150
4 6 8 10 12 14
50
100
150