the pokemon collector's problem: what are the chansey?...single collection equal probability...

The Pokemon Collector’s Problem: What are the Chansey?

The Pokemon Collector’s Problem: What are theChansey?

Kevin Wang

School of Mathematics and StatisticsThe University of Sydney

October 28, 2016

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

Motivation

I Interview question

I Statistical toolsI Probability (theoretical)I Simulation (methodological)I Data analysis (applied)

I To keep myself entertained and keep my PhD supervisors “amused”.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

Motivation

I Interview question

I Statistical toolsI Probability (theoretical)I Simulation (methodological)I Data analysis (applied)

I To keep myself entertained and keep my PhD supervisors “amused”.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

Motivation

I Interview question

I Statistical toolsI Probability (theoretical)I Simulation (methodological)I Data analysis (applied)

I To keep myself entertained and keep my PhD supervisors “amused”.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

The Pokemon Collector’s Problem

Problem

There are 142 individual Pokemons on the game “Pokemon GO”.Assuming we can collect 1 Pokemon per unit time, what is the expectednumber of Pokemons do we need to complete the collection?

I According to report 1, one player caught 4,269 Pokemons tocomplete this collection.

I I am on 1,430 in total, 99 of them are unique. And all of my PhDsupervisors are amused/annoyed/urging me to go back to research.

1http:

//www.polygon.com/2016/7/21/12247678/pokemon-go-complete-pokedex

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

The Pokemon Collector’s Problem

Problem

There are 142 individual Pokemons on the game “Pokemon GO”.Assuming we can collect 1 Pokemon per unit time, what is the expectednumber of Pokemons do we need to complete the collection?

I According to report 1, one player caught 4,269 Pokemons tocomplete this collection.

I I am on 1,430 in total, 99 of them are unique. And all of my PhDsupervisors are amused/annoyed/urging me to go back to research.

1http:

//www.polygon.com/2016/7/21/12247678/pokemon-go-complete-pokedex

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

The Problem is hard

I Pokemon problem has:I unequal probabilitiesI multiple collection with differing termination point (evolution)I arrival in batches of different sizes (eggs)

A theoretical closed form does not exist in literature so far.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

The Coupon Collector’s Problem (CCP)

I It is a weaker version of the Pokemon problem.

I Classical elements of the problem include:I Equal/unequal probabilitiesI Single/multiple collectionI Arrival in batches

I Some references:I The Coupon Collector’s Problem, M. Ferrante, M. Saltalamacchia,

(2014)I Introduction to Probability Models. S.Ross.I A First Course in Probability. S. Ross.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

Key of this talk

1. A diverse range of possible solutions, combining a whole spectrum ofStatistics.

2. Characteristics of the problems, including probability sensitivity,capture/evolution tradeoff.

Figure: A Pikture without context.

The Pokemon Collector’s Problem: What are the Chansey?

Introduction

Introduction

Single CollectionEqual Probability Assumption

Basic Probability SolutionMarkov Chain Solution (Omitted)

Unequal Probability AssumptionBasic Probability SolutionSensitivity of Overall CollectionPoisson Process Solution (Omitted)

Other Generalisations

Pokemon Problem Simulations

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Equal probability CCP

Problem

Suppose we buy one coupon per unit time. Assuming each coupon isequally likely to appear, how many coupons we have to purchase (onaverage) to complete a collection of n different coupons?

I Let X be the (random) number of coupons we need to collect tocomplete collection, which has n distinct types of coupons.

I Let pi denote the probability of collecting coupon of type i, whichequals to 1/n according to our equal probability assumption.

I Let Xi denote the additional number of coupons we need to collect,to pass from i− 1 to i distinct types of coupons in our collection.i = 1, 2, . . . , n.

I Hence, X = X1 +X2 + · · ·+Xn.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Equal probability CCP

Problem

Suppose we buy one coupon per unit time. Assuming each coupon isequally likely to appear, how many coupons we have to purchase (onaverage) to complete a collection of n different coupons?

I Let X be the (random) number of coupons we need to collect tocomplete collection, which has n distinct types of coupons.

I Let pi denote the probability of collecting coupon of type i, whichequals to 1/n according to our equal probability assumption.

I Let Xi denote the additional number of coupons we need to collect,to pass from i− 1 to i distinct types of coupons in our collection.i = 1, 2, . . . , n.

I Hence, X = X1 +X2 + · · ·+Xn.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Equal probability CCP

Problem

Suppose we buy one coupon per unit time. Assuming each coupon isequally likely to appear, how many coupons we have to purchase (onaverage) to complete a collection of n different coupons?

I Let X be the (random) number of coupons we need to collect tocomplete collection, which has n distinct types of coupons.

I Let pi denote the probability of collecting coupon of type i, whichequals to 1/n according to our equal probability assumption.

I Let Xi denote the additional number of coupons we need to collect,to pass from i− 1 to i distinct types of coupons in our collection.i = 1, 2, . . . , n.

I Hence, X = X1 +X2 + · · ·+Xn.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Equal probability CCP

Problem

Suppose we buy one coupon per unit time. Assuming each coupon isequally likely to appear, how many coupons we have to purchase (onaverage) to complete a collection of n different coupons?

I Let X be the (random) number of coupons we need to collect tocomplete collection, which has n distinct types of coupons.

I Let pi denote the probability of collecting coupon of type i, whichequals to 1/n according to our equal probability assumption.

I Let Xi denote the additional number of coupons we need to collect,to pass from i− 1 to i distinct types of coupons in our collection.i = 1, 2, . . . , n.

I Hence, X = X1 +X2 + · · ·+Xn.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Equal probability CCP

Problem

Suppose we buy one coupon per unit time. Assuming each coupon isequally likely to appear, how many coupons we have to purchase (onaverage) to complete a collection of n different coupons?

I Let X be the (random) number of coupons we need to collect tocomplete collection, which has n distinct types of coupons.

I Let pi denote the probability of collecting coupon of type i, whichequals to 1/n according to our equal probability assumption.

I Let Xi denote the additional number of coupons we need to collect,to pass from i− 1 to i distinct types of coupons in our collection.i = 1, 2, . . . , n.

I Hence, X = X1 +X2 + · · ·+Xn.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

Problem (Reduced)

When i− 1 distinct types of coupons have already been collected, what isthe probability that a new coupon will be of a distinct type be?

I X1 = 1.

I Every time, since i− 1 of those were already chosen, we are left withonly n− (i− 1) possible types to choose from.

I Probability to escape from this state is ξi =n−(i−1)

n

Each Xi is a geometric distributed random variable withprobability parameter ξi.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

Problem (Reduced)

When i− 1 distinct types of coupons have already been collected, what isthe probability that a new coupon will be of a distinct type be?

I X1 = 1.

I Every time, since i− 1 of those were already chosen, we are left withonly n− (i− 1) possible types to choose from.

I Probability to escape from this state is ξi =n−(i−1)

n

Each Xi is a geometric distributed random variable withprobability parameter ξi.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

Problem (Reduced)

When i− 1 distinct types of coupons have already been collected, what isthe probability that a new coupon will be of a distinct type be?

I X1 = 1.

I Every time, since i− 1 of those were already chosen, we are left withonly n− (i− 1) possible types to choose from.

I Probability to escape from this state is ξi =n−(i−1)

n

Each Xi is a geometric distributed random variable withprobability parameter ξi.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

Problem (Reduced)

When i− 1 distinct types of coupons have already been collected, what isthe probability that a new coupon will be of a distinct type be?

I X1 = 1.

I Every time, since i− 1 of those were already chosen, we are left withonly n− (i− 1) possible types to choose from.

I Probability to escape from this state is ξi =n−(i−1)

n

Each Xi is a geometric distributed random variable withprobability parameter ξi.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

I First year statistics course: the expectation of a Geo(ξ) randomvariable is 1/ξ. Hence:

E(X) = E(X1) + E(X2) + · · ·+ E(Xn)

=1

ξ1+

1

ξ2+ · · ·+ 1

ξn

=n

n+

n

n− 1+ · · ·+ n

1

= n ·Hn,

where Hn is the n-th Harmonic number.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Solution for equal probability case

I First year statistics course: the expectation of a Geo(ξ) randomvariable is 1/ξ. Hence:

E(X) = E(X1) + E(X2) + · · ·+ E(Xn)

=1

ξ1+

1

ξ2+ · · ·+ 1

ξn

=n

n+

n

n− 1+ · · ·+ n

1

= n ·Hn,

where Hn is the n-th Harmonic number.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Some characteristics

I The waiting time for the very last coupon is always the longest.

I 2H2 − 1H1 = 2, 100H100 − 99H99 = 6.177.

I 142H142 ≈ 786.19.

I E(X) = nHn = n log n+ γn+ 12 + o(1).

I E(X) grows at the rate of O(n log(n)), which is not too bad.

I V ar(X) < π2

6 n2, Chebyshev inequality gives:

P(|X − nHn| ≥ cn) ≤π2

6c2. (1)

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Some characteristics

I The waiting time for the very last coupon is always the longest.

I 2H2 − 1H1 = 2, 100H100 − 99H99 = 6.177.

I 142H142 ≈ 786.19.

I E(X) = nHn = n log n+ γn+ 12 + o(1).

I E(X) grows at the rate of O(n log(n)), which is not too bad.

I V ar(X) < π2

6 n2, Chebyshev inequality gives:

P(|X − nHn| ≥ cn) ≤π2

6c2. (1)

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Equal Probability Assumption

Markov Chain (Omitted)

I Due to independence inherited from the equal probabilityassumption, we can construct a Markov Chain.

I Details omitted.I State space is S = {0, 1, . . . , n}, and the transition probability

matrix is:

P =

0 1 0 . . . . . . . . . 00 1/n (n− 1)/n 0 . . . . . . 00 0 2/n (n− 2)/n 0 . . . 0...

.... . .

. . .. . .

. . ....

0 . . . . . . . . . 0 (n− 1)/n 1/n0 . . . . . . . . . 0 0 1

.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Introduction

Single CollectionEqual Probability Assumption

Basic Probability SolutionMarkov Chain Solution (Omitted)

Unequal Probability AssumptionBasic Probability SolutionSensitivity of Overall CollectionPoisson Process Solution (Omitted)

Other Generalisations

Pokemon Problem Simulations

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Single collection: unequal probability

I Now, let’s assume that probability of collecting each coupon isdifferent. i.e. pi are distinct and

∑ni=1 pi = 1.

I If we redefine Xi to be the random number of coupons we need tobuy to obtain the first coupon of type i, then each Xi ∼ Geo(pi).

I X = max{X1, . . . , Xn} is the number of coupons we should buy tillcompletion of the set.

I Any guess for the expected total waiting time E(X)?

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Single collection: unequal probability

I Now, let’s assume that probability of collecting each coupon isdifferent. i.e. pi are distinct and

∑ni=1 pi = 1.

I If we redefine Xi to be the random number of coupons we need tobuy to obtain the first coupon of type i, then each Xi ∼ Geo(pi).

I X = max{X1, . . . , Xn} is the number of coupons we should buy tillcompletion of the set.

I Any guess for the expected total waiting time E(X)?

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Solution via basic probability

E(X) = E [max{X1, . . . , Xn}] (2)

=

n∑m=1

(−1)m−1∑

1≤j1<···<jm≤n

1

pj1 + · · ·+ pjm

(3)

=

∫ ∞0

(1−

n∏i=1

(1− e−pix

))dx.

(4)

I The complication here is that Xi’s are no longer independent

I min(Xi, Xj) ∼ Geo(pi + pj), for i 6= j. This generalises to anyfinite number of Xi’s.

I The Maximum-Minimum Identity:E [max(Xi, Xj))] = E(Xi) + E(Xj)− E [min(Xi, Xj)], the lastsummand is of the form 1/(pi + pj), which also generalises.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Solution via basic probability

E(X) = E [max{X1, . . . , Xn}] (2)

=

n∑m=1

(−1)m−1∑

1≤j1<···<jm≤n

1

pj1 + · · ·+ pjm

(3)

=

∫ ∞0

(1−

n∏i=1

(1− e−pix

))dx.

(4)

I The complication here is that Xi’s are no longer independent

I min(Xi, Xj) ∼ Geo(pi + pj), for i 6= j. This generalises to anyfinite number of Xi’s.

I The Maximum-Minimum Identity:E [max(Xi, Xj))] = E(Xi) + E(Xj)− E [min(Xi, Xj)], the lastsummand is of the form 1/(pi + pj), which also generalises.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Solution via basic probability

E(X) = E [max{X1, . . . , Xn}] (2)

=

n∑m=1

(−1)m−1∑

1≤j1<···<jm≤n

1

pj1 + · · ·+ pjm

(3)

=

∫ ∞0

(1−

n∏i=1

(1− e−pix

))dx.

(4)

I The complication here is that Xi’s are no longer independent

I min(Xi, Xj) ∼ Geo(pi + pj), for i 6= j. This generalises to anyfinite number of Xi’s.

I The Maximum-Minimum Identity:E [max(Xi, Xj))] = E(Xi) + E(Xj)− E [min(Xi, Xj)], the lastsummand is of the form 1/(pi + pj), which also generalises.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Solution via basic probability

E(X) = E [max{X1, . . . , Xn}] (2)

=

n∑m=1

(−1)m−1∑

1≤j1<···<jm≤n

1

pj1 + · · ·+ pjm(3)

=

∫ ∞0

(1−

n∏i=1

(1− e−pix

))dx.

(4)

I The complication here is that Xi’s are no longer independent

I min(Xi, Xj) ∼ Geo(pi + pj), for i 6= j. This generalises to anyfinite number of Xi’s.

I The Maximum-Minimum Identity:E [max(Xi, Xj))] = E(Xi) + E(Xj)− E [min(Xi, Xj)], the lastsummand is of the form 1/(pi + pj), which also generalises.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Solution via basic probability

E(X) = E [max{X1, . . . , Xn}] (2)

=

n∑m=1

(−1)m−1∑

1≤j1<···<jm≤n

1

pj1 + · · ·+ pjm(3)

=

∫ ∞0

(1−

n∏i=1

(1− e−pix

))dx. (4)

I The complication here is that Xi’s are no longer independent

I min(Xi, Xj) ∼ Geo(pi + pj), for i 6= j. This generalises to anyfinite number of Xi’s.

I The Maximum-Minimum Identity:E [max(Xi, Xj))] = E(Xi) + E(Xj)− E [min(Xi, Xj)], the lastsummand is of the form 1/(pi + pj), which also generalises.

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Some characteristics

I Relations to Stochastic Processes and Extreme Value Theory.

I Each prize in McDonald’s Monopoly competition is exactly like this.

I The expression is not simple.

I Sensitive to introduction of low pi’s.

I E(X)(1/3,1/3,1/3) = 5.5

I E(X)(0.1,0.1,0.8) = 15.03

I E(X)(0.02,0.49,0.49) = 50

I Visualisation of the family (p1, p2, p3,E(X)).

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Some characteristics

I Relations to Stochastic Processes and Extreme Value Theory.

I Each prize in McDonald’s Monopoly competition is exactly like this.

I The expression is not simple.

I Sensitive to introduction of low pi’s.

I E(X)(1/3,1/3,1/3) = 5.5

I E(X)(0.1,0.1,0.8) = 15.03

I E(X)(0.02,0.49,0.49) = 50

I Visualisation of the family (p1, p2, p3,E(X)).

The Pokemon Collector’s Problem: What are the Chansey?

Single Collection

Unequal Probability Assumption

Poisson Process (Omitted)

I Avoids the horrible summations and integral tricks. The trade-offbeing you need to know something about Poisson Processes(STAT3911).

I Come to SUMS Lightening Talk in Week 13!

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

Introduction

Single CollectionEqual Probability Assumption

Basic Probability SolutionMarkov Chain Solution (Omitted)

Unequal Probability AssumptionBasic Probability SolutionSensitivity of Overall CollectionPoisson Process Solution (Omitted)

Other Generalisations

Pokemon Problem Simulations

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

Other Generalisations (Omitted)

I Consider m−collection problem now, which intuitively, should beless than m multiplied by their Single Collection counterpart.

I Equal Probability Solution:

E(X) = n

∫ ∞0

[1− (1− Sm(t)e−t)n

]dt, (5)

where Sm(t) =∑m−1k=0

tk

k! . For large m, this approaches mn.

I Unequal Probability Solution:I Bounds exist, see paper.I Markov Chain approach should work, but must require software

computations.

I Other generalisations: arrival in batches of constant sizes.

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

Other Generalisations (Omitted)

I Consider m−collection problem now, which intuitively, should beless than m multiplied by their Single Collection counterpart.

I Equal Probability Solution:

E(X) = n

∫ ∞0

[1− (1− Sm(t)e−t)n

]dt, (5)

where Sm(t) =∑m−1k=0

tk

k! . For large m, this approaches mn.

I Unequal Probability Solution:I Bounds exist, see paper.I Markov Chain approach should work, but must require software

computations.

I Other generalisations: arrival in batches of constant sizes.

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

Other Generalisations (Omitted)

I Consider m−collection problem now, which intuitively, should beless than m multiplied by their Single Collection counterpart.

I Equal Probability Solution:

E(X) = n

∫ ∞0

[1− (1− Sm(t)e−t)n

]dt, (5)

where Sm(t) =∑m−1k=0

tk

k! . For large m, this approaches mn.

I Unequal Probability Solution:I Bounds exist, see paper.I Markov Chain approach should work, but must require software

computations.

I Other generalisations: arrival in batches of constant sizes.

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

Other Generalisations (Omitted)

I Consider m−collection problem now, which intuitively, should beless than m multiplied by their Single Collection counterpart.

I Equal Probability Solution:

E(X) = n

∫ ∞0

[1− (1− Sm(t)e−t)n

]dt, (5)

where Sm(t) =∑m−1k=0

tk

k! . For large m, this approaches mn.

I Unequal Probability Solution:I Bounds exist, see paper.I Markov Chain approach should work, but must require software

computations.

I Other generalisations: arrival in batches of constant sizes.

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

The Pokemon Collector’s Problem

I Pokemon problem has:I unequal probabilitiesI multiple collection with differing termination point (evolution)I arrival in batches of different sizes (eggs)

I A theoretical closed form does not exist in literature so far.

I I am sure if you tweak features in this model here and there, you canpotentially get a paper out of this and a job.

The Pokemon Collector’s Problem: What are the Chansey?

Other Generalisations

The Pokemon Collector’s Problem

I Pokemon problem has:I unequal probabilitiesI multiple collection with differing termination point (evolution)I arrival in batches of different sizes (eggs)

I A theoretical closed form does not exist in literature so far.

I I am sure if you tweak features in this model here and there, you canpotentially get a paper out of this and a job.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Introduction

Single CollectionEqual Probability Assumption

Basic Probability SolutionMarkov Chain Solution (Omitted)

Unequal Probability AssumptionBasic Probability SolutionSensitivity of Overall CollectionPoisson Process Solution (Omitted)

Other Generalisations

Pokemon Problem Simulations

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Real-world considerations

I The game is even more complicated, with eggs, evolutions andhuman irrationality.

I Recent updates may have fundamentally changed rate parameters.The main challenge of our simulation is to get some sensible ratesand see how PP compares to classical CCP models.

I Capture completion is highly sensitive to existence of low rates.I Data trustworthiness, untidiness and missingness.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Real-world considerations

I The game is even more complicated, with eggs, evolutions andhuman irrationality.

I Recent updates may have fundamentally changed rate parameters.The main challenge of our simulation is to get some sensible ratesand see how PP compares to classical CCP models.

I Capture completion is highly sensitive to existence of low rates.I Data trustworthiness, untidiness and missingness.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Real-world considerations

I The game is even more complicated, with eggs, evolutions andhuman irrationality.

I Recent updates may have fundamentally changed rate parameters.The main challenge of our simulation is to get some sensible ratesand see how PP compares to classical CCP models.

I Capture completion is highly sensitive to existence of low rates.I Data trustworthiness, untidiness and missingness.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Step 1: Estimate rates

In God we trust. All others bring data —

William Edwards Deming

I While someone published a sensible table of capture rates. However,simulations suggested these were underestimated.

I By running a very close simulation, this rate was transformed to getmore sensible expected number of captures. This sets thebenchmark around 4,700 Pokemons.

I Again, let me assure you this is not easy.

I See demonstration.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Step 1: Estimate rates

In God we trust. All others bring data —

William Edwards Deming

I While someone published a sensible table of capture rates. However,simulations suggested these were underestimated.

I By running a very close simulation, this rate was transformed to getmore sensible expected number of captures. This sets thebenchmark around 4,700 Pokemons.

I Again, let me assure you this is not easy.

I See demonstration.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Step 2: Compare this to other strategies

I Now, assume we can ONLY access higher level Pokemons viaevolution of lower level Pokemons. This on average, require roughly10,310 Pokemons, taking into account specific number of collections.

I Compare this to the benchmark of 4,700 in a simulation wherecapture of higher level Pokemons is possible.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Step 2: Compare this to other strategies

I Now, assume we can ONLY access higher level Pokemons viaevolution of lower level Pokemons. This on average, require roughly10,310 Pokemons, taking into account specific number of collections.

I Compare this to the benchmark of 4,700 in a simulation wherecapture of higher level Pokemons is possible.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Trivial lessons

Rates for higher stage Pokemons are in a sweet spot —non-zero capture probability made sure you don’t end up

playing the game via evolution for years.

Theorem

Skipping research to catch a Charizard is therefore a high returninvestment in your future.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Trivial lessons

Rates for higher stage Pokemons are in a sweet spot —non-zero capture probability made sure you don’t end up

playing the game via evolution for years.

Theorem

Skipping research to catch a Charizard is therefore a high returninvestment in your future.

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

Concluding remarks

I Theory sometimes only go so far, which leave room for explorations.I Theory and application are not mortal enemies.I Mathematics and Statistics are not and should not be dreadful

subjects. They exist outside the realm of tutorial sheets andassignments.

I Worst case: Mathematics and Statistics are prosecutors of our“intuitions” and “common sense”.

Figure:

The Pokemon Collector’s Problem: What are the Chansey?

Pokemon Problem Simulations

References

I The Coupon Collector’s Problem, M. Ferrante, M. Saltalamacchia,(2014)

I Introduction to Probability Models. S.Ross.

I A First Course in Probability. S. Ross.

I STAT3911: Stochastic Processes Lecture Notes. R. Kawaii.