Answer key · Unit 19

Patterns and chance.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 3

Lesson 1 of 3 · About 15 minutes

Chaos

Two almost identical double pendulums: motion fixed completely by its start that still cannot be predicted, a tiny gap that grows by multiplying, and why more precision buys very little time.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
The same start
Practice · Predict, then look
Question
The Difference in start angle is zero: both pendulums are released from exactly the same angle. What will happen during the run?

Correct answer
- Correct They will move as one the whole time
Setup, choices and feedback
The Scene at this step
Difference in start angle
0 °
Release angle
120 °fixed
Show the paths
Offfixed
Choices
- Correct They will move as one the whole time
- Wrong After a few seconds they will part, because the motion is chaotic
  Misconception: Thinks chaotic motion is random: that the same start could give a different motion.
  Feedback from the catalogue
  Chaos is not chance. The double pendulum is deterministic: exactly the same start always gives exactly the same motion, and two pendulums that start identical move as one for ever. What chaos does is multiply any difference, however small, so a long forecast needs a precision no one can reach.
- Wrong They will move in opposite directions
  Feedback
  The two pendulums are built the same and released the same, so there is no reason for one to turn the other way.
Once the prediction is locked in, the Scene changes
Show the paths
Onfixed
The Scene then shows: 0 ° · Double: the gap between the two pendulums at the end of the 20-second run/nodes/pendulums/outputs/endGap
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At the end of the run the gap is still $0^\circ$ ⁠. The motion is deterministic: the same start always gives the same motion. Chaos begins only when there is some difference to multiply.

Hint
The equations of motion are the same for both. What could tell them apart?

Worked solution
1. Both pendulums are computed from the same equations, with the same time step.
2. The starts are the same: the same angle, at rest.
3. So every step gives both the same state, and the gap stays $0$ ⁠ the whole run.
Step 2 of 4
Open this step
A tiny gap
Practice · Calculate
Question
The pink one starts $0.01^\circ$ ⁠ further round. Watch the timer on the board: after how many seconds does the gap between the pendulums reach $10^\circ$ ⁠?

Correct answer
4.051 s ± 0.4 s
Setup, choices and feedback
The Scene at this step
Difference in start angle
0.01 °fixed
Release angle
120 °fixed
Show the paths
On
Accepted within
± 0.4 s
Computed from the Scene at this step’s setting
Double: seconds after release until the gap between the two pendulums first reaches 10°, as the board's timer reads it; NaN if it stays under 10° for the whole 20-second run/nodes/pendulums/outputs/splitTime
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $4$ ⁠ seconds: a gap of $0.01^\circ$ ⁠ grows $1000$ ⁠ times over in four seconds. On the chart, on a log scale, it climbs in an almost straight line: exponential growth.

Hint
The timer shows the time since release and the gap. You can pause the motion and run it again.

Worked solution
1. Run from the release, and follow the gap on the board.
2. For the first seconds it is tiny, then it multiplies.
3. It passes $10^\circ$ ⁠ after about $4$ ⁠ seconds.
Step 3 of 4
Open this step
Ten times the precision
Practice · Predict, then look
Question
Now the starting gap is made ten times smaller, $0.001^\circ$ ⁠. When will the gap reach $10^\circ$ ⁠?

Correct answer
- Correct Only a little over a second later than before
Setup, choices and feedback
The Scene at this step
Difference in start angle
0.01 °
Release angle
120 °fixed
Show the paths
On
Choices
- Correct Only a little over a second later than before
- Wrong Ten times later, after about $40$ ⁠ seconds
  Misconception: Thinks a start gap ten times smaller keeps two chaotic systems together ten times as long.
  Feedback from the catalogue
  The gap grows by multiplying: about ten times over in each fixed stretch of time. A gap ten times smaller needs just one more such stretch to reach the same size, not ten times the time. As with a logarithm: the time grows with the $\log$ ⁠ of the precision, not in proportion to it.
- Wrong Never: so tiny a gap changes nothing
  Misconception: Thinks chaotic motion is random: that the same start could give a different motion.
  Feedback from the catalogue
  Chaos is not chance. The double pendulum is deterministic: exactly the same start always gives exactly the same motion, and two pendulums that start identical move as one for ever. What chaos does is multiply any difference, however small, so a long forecast needs a precision no one can reach.
- Wrong At exactly the same time
  Feedback
  A smaller gap has to grow more times over to reach $10^\circ$ ⁠, so it takes a little longer.
Once the prediction is locked in, the Scene changes
Difference in start angle
0.001 °
The Scene then shows: 5.357 s · Double: seconds after release until the gap between the two pendulums first reaches 10°, as the board's timer reads it; NaN if it stays under 10° for the whole 20-second run/nodes/pendulums/outputs/splitTime
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About $5.4$ ⁠ seconds, against about $4.1$ ⁠: only about a second and a quarter more. Ten times the precision does not buy ten times the time, only one more short, fixed stretch. That is why weather forecasts are limited to a few days.

Hint
The gap grows by multiplying. How long does it take to grow ten times over?

Worked solution
1. Released high, the gap grows about ten times over in a little more than each second.
2. A gap ten times smaller has to grow ten times over just once more to reach the same size.
3. So it reaches $10^\circ$ ⁠ about a second later than before, not ten times later.
If you expected ten times the time, you thought of growth by adding. With growth by multiplying, the time depends on the $\log$ ⁠ of the gap.
Step 4 of 4
Open this step
A gentle release
Practice · Predict, then look
Question
The pair is released low, from $30^\circ$ ⁠, with the same gap of $0.01^\circ$ ⁠. What will happen to the gap during the 20-second run?

Correct answer
- Correct It will grow only slowly, and still be tiny at the end of the run
Setup, choices and feedback
The Scene at this step
Difference in start angle
0.01 °
Release angle
120 °
Show the paths
On
Choices
- Correct It will grow only slowly, and still be tiny at the end of the run
- Wrong It will grow at the same rate: a double pendulum is always chaotic
  Feedback
  A double pendulum is chaotic only when released high. Below about $80^\circ$ ⁠ it swings regularly, and the gap grows only slowly.
- Wrong It will vanish, and the two pendulums will come together
  Feedback
  With no friction nothing shrinks the gap: it is kept, and here it grows a little.
Once the prediction is locked in, the Scene changes
Release angle
30 °
The Scene then shows: 0.03178 ° · Double: the gap between the two pendulums at the end of the 20-second run/nodes/pendulums/outputs/endGap
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
After 20 seconds the gap is only about $0.03^\circ$ ⁠: about three times over, instead of a thousand times in four seconds. One system can be regular in some conditions and chaotic in others.

Hint
Look at the chart: does it still climb in a straight line?

Worked solution
1. Released low, each pendulum swings to and fro in a pattern that repeats.
2. The gap grows only slowly, roughly with the time, not by multiplying.
3. At the end of the run it is still a few hundredths of a degree.

Lesson 2 of 3 · About 25 minutes

Fractals

Shapes whose every part is a small copy of the whole: how many parts each step makes, a perimeter that grows without end round a finite area, and a dimension that is not a whole number.

Open the lesson

The 8 steps and their answers

Step 1 of 8
Open this step
Copies of copies
Practice · Predict, then look
Question
At depth $3$ ⁠ the Sierpiński pyramid is made of $4^3 = 64$ ⁠ small pyramids. How many is it made of at depth $5$ ⁠?

Correct answer
1024 ± 0.5
A number prediction counts as right within ± 0.5
Setup, choices and feedback
The Scene at this step
Depth
3
Shape
Sierpiński pyramidfixed
Zoom in
Whole shape
Other settings
Readouts/nodes/fractal/props/readouts = false
Once the prediction is locked in, the Scene changes
Depth
5
The Scene then shows: 1024 · Pyramids (4ⁿ) or edges (3·4ⁿ) at the depth shown/nodes/fractal/outputs/pieces
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$4^5 = 1024$ ⁠: $16$ ⁠ times as many as at depth $3$ ⁠, because two steps have passed, each multiplying by four.

Hint
Each step swaps every pyramid for four.

Worked solution
1. Each step: every pyramid becomes $4$ ⁠.
2. From depth $3$ ⁠ to depth $5$ ⁠: two steps, $64 \cdot 4 \cdot 4$ ⁠.
3. $64 \cdot 16 = 1024 = 4^5$ ⁠.
Step 2 of 8
Open this step
A growing perimeter
Practice · Calculate
Question
At each step of the Koch snowflake every side becomes four sides, each a third as long. How many times longer is the perimeter at depth $5$ ⁠ than the starting triangle’s?

Correct answer
4.214 ± 1%
Setup, choices and feedback
The Scene at this step
Depth
5fixed
Shape
Koch snowflakefixed
Zoom in
Whole shape
Other settings
Readouts/nodes/fractal/props/readouts = false
Accepted within
± 1%
Computed from the Scene at this step’s setting
The snowflake’s perimeter, times the starting triangle’s, (4/3)ⁿ; NaN for the pyramid/nodes/fractal/outputs/perimeter
Wrong answers it recognises
- 2.667 ± 0.05
  Misconception: Adds the same amount at each step where each step multiplies: grows the Koch perimeter by a third of the start each step instead of × 4/3.
  Feedback from the catalogue
  At each step every side becomes four sides a third as long, so the whole perimeter is multiplied by $\tfrac{4}{3}$ ⁠, including what earlier steps added. After $n$ ⁠ steps that is $\left(\tfrac{4}{3}\right)^n$ ⁠, not $1 + \tfrac{n}{3}$ ⁠.
Feedback for a right answer
$\left(\tfrac{4}{3}\right)^5 \approx 4.21$ ⁠. After $50$ ⁠ steps it is already more than $1.7$ ⁠ million times, and the perimeter grows without end.

Hint
At each step the perimeter is multiplied by $\tfrac{4}{3}$ ⁠.

Worked solution
1. Each side of length $\ell$ ⁠ becomes $4$ ⁠ sides of length $\tfrac{\ell}{3}$ ⁠: $\tfrac{4}{3}\ell$ ⁠ in all.
2. That happens to every side, so the whole perimeter is multiplied by $\tfrac{4}{3}$ ⁠ each step.
3. After $5$ ⁠ steps: $\left(\tfrac{4}{3}\right)^5 = \dfrac{1024}{243} \approx 4.21$ ⁠.
If you got about $2.67$ ⁠, you added a third each step. But each step also multiplies what the earlier steps added.
Step 3 of 8
Open this step
Without end
Practice · Choose
Question
The snowflake’s construction goes on for ever. What happens to its perimeter and its area?

Correct answer
- Correct The perimeter grows without end, and the area comes close to a limit: $\tfrac{8}{5}$ ⁠ of the triangle
Setup, choices and feedback
The Scene at this step
Depth
5
Shape
Koch snowflake
Zoom in
Whole shape
Choices
Shown in a random order.
- Correct The perimeter grows without end, and the area comes close to a limit: $\tfrac{8}{5}$ ⁠ of the triangle
- Wrong Both grow without end
  Misconception: Thinks a curve of endless length must enclose an endless area.
  Feedback from the catalogue
  Length and area do not go together. Each step of the snowflake adds triangles $9$ ⁠ times smaller in area, and only $4$ ⁠ times as many of them, so the additions to the area shrink by $\tfrac{4}{9}$ ⁠ each step and add up to a limit: $\tfrac{8}{5}$ ⁠ of the triangle. The perimeter, by contrast, is multiplied by $\tfrac{4}{3}$ ⁠ each step and grows without end. The whole snowflake stays inside a circle round the triangle.
- Wrong Both come close to a limit
  Feedback
  The perimeter is multiplied by $\tfrac{4}{3}$ ⁠ each step, and repeated multiplying by a number above $1$ ⁠ grows without end.
- Wrong The area grows without end, and the perimeter comes close to a limit
  Feedback
  It is the other way round: the whole snowflake stays inside a circle round the triangle, so its area is bounded.
Feedback for a right answer
A line of endless length bounding a finite area: that is the mark of a fractal. The line is too crowded for dimension $1$ ⁠, and too thin for dimension $2$ ⁠.

Hint
Does the snowflake ever leave a circle drawn round the triangle?

Worked solution
1. The perimeter: $\left(\tfrac{4}{3}\right)^n$ ⁠, which grows without end.
2. The area: step $k$ ⁠ adds triangles each $\tfrac{1}{9^k}$ ⁠ of the triangle, and there are $3 \cdot 4^{k-1}$ ⁠ of them. The additions shrink by $\tfrac{4}{9}$ ⁠ each step.
3. A geometric series with ratio below $1$ ⁠ comes close to a limit: $1 + \dfrac{1/3}{1 - 4/9} = \dfrac{8}{5}$ ⁠.
Step 4 of 8
Open this step
A broken dimension
Practice · Calculate
Question
Each side of the Koch snowflake is made of $N = 4$ ⁠ copies of itself, each $\tfrac{1}{3}$ ⁠ its size. What is the snowflake’s fractal dimension?

Correct answer
1.262 ± 0.01
Setup, choices and feedback
The Scene at this step
Depth
3
Shape
Koch snowflakefixed
Zoom in
Whole shape
Other settings
Readouts/nodes/fractal/props/readouts = false
Other settings
Eq/callouts/shape-koch/when/eq = "none"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Similarity dimension, log(copies) ÷ log(scale factor)/nodes/fractal/outputs/dimension
Wrong answers it recognises
- 1.333 ± 0.01
  Misconception: Takes the fractal dimension as copies ÷ scale factor instead of log copies ÷ log scale factor.
  Feedback from the catalogue
  The dimension is the exponent $D$ ⁠ for which $s^D = N$ ⁠: how many times you multiply by $s$ ⁠ to get $N$ ⁠ copies. So $D = \dfrac{\log N}{\log s}$ ⁠, not $\dfrac{N}{s}$ ⁠. A square, say: $4$ ⁠ copies at half size, $2^2 = 4$ ⁠, dimension $2$ ⁠, though $\tfrac{4}{2}$ ⁠ happens to be $2$ ⁠ as well. For the snowflake $\tfrac{4}{3}$ ⁠ is not the dimension.
Feedback for a right answer
$D = \dfrac{\log 4}{\log 3} \approx 1.26$ ⁠: more than a line, less than a surface.

Hint
$D = \dfrac{\log N}{\log s}$ ⁠, where $s$ ⁠ is how many times smaller each copy is than the whole.

Worked solution
1. $N = 4$ ⁠ copies, each $s = 3$ ⁠ times smaller.
2. The dimension is the exponent $D$ ⁠ for which $3^D = 4$ ⁠.
3. $D = \dfrac{\log 4}{\log 3} \approx \dfrac{0.602}{0.477} \approx 1.26$ ⁠.
If you answered about $1.33$ ⁠, you divided $\tfrac{4}{3}$ ⁠. The dimension is a ratio of logarithms.
Step 5 of 8
Open this step
How many copies
Practice · Calculate
Question
Shape: Sierpiński pyramid, Depth $1$ ⁠. Into how many small pyramids, the same shape as the big one, does the pyramid break in one step?

Correct answer
4 ± 0
Setup, choices and feedback
The Scene at this step
Depth
1fixed
Shape
Sierpiński pyramidfixed
Zoom in
Whole shape
Accepted within
± 0
Computed from the Scene at this step’s setting
Smaller copies of the whole each step makes/nodes/fractal/outputs/copies
Wrong answers it recognises
- 8 ± 0
  Misconception: Expects a shape halved in size to break into 8 copies, as a cube does, whatever the shape.
  Feedback from the catalogue
  A half-size cube fits $8$ ⁠ times into the big one. But the Sierpiński pyramid does not keep all the volume: of each pyramid only $4$ ⁠ small ones are left, one at each corner, and the middle part is taken out.
Feedback for a right answer
$4$ ⁠ copies, one at each corner. The octahedron in the middle is taken out.

Hint
Count the pyramid’s corners: at each one a small pyramid is left.

Worked solution
1. The triangular pyramid, the tetrahedron, has $4$ ⁠ corners.
2. At each step a half-size pyramid is left at each corner, and the middle part is thrown away.
3. So $4$ ⁠ copies.
If you answered $8$ ⁠, you were thinking of a cube cut into eight. In the pyramid the middle part, an octahedron, is taken out, leaving only four.
Step 6 of 8
Open this step
How much smaller
Practice · Calculate
Question
At each step of the Sierpiński pyramid every pyramid is replaced by $4$ ⁠ smaller ones, each sitting in a corner. How many times shorter is each edge of a small pyramid than an edge of the big one?

Correct answer
2 ± 0
Setup, choices and feedback
The Scene at this step
Depth
1fixed
Shape
Sierpiński pyramidfixed
Zoom in
Whole shape
Accepted within
± 0
Computed from the Scene at this step’s setting
How many times smaller each copy is/nodes/fractal/outputs/scale
Wrong answers it recognises
- 4 ± 0
  Misconception: Mixes up how many copies a fractal step makes with how many times smaller each copy is.
  Feedback from the catalogue
  Each fractal step has two separate numbers: how many copies it makes, $N$ ⁠, and how many times smaller each is, $s$ ⁠. In the Sierpiński pyramid $N = 4$ ⁠ and $s = 2$ ⁠. The dimension, $\dfrac{\log N}{\log s}$ ⁠, needs both.
Feedback for a right answer
$2$ ⁠ times: half the edge. Four copies, each twice as small.

Hint
Two small pyramids sit side by side along each edge of the big one.

Worked solution
1. Along each edge of the big pyramid there are two small pyramids, corner to corner.
2. So each small edge is half the big one: $2$ ⁠ times shorter.
3. With $N = 4$ ⁠ and $s = 2$ ⁠: $D = \dfrac{\log 4}{\log 2} = 2$ ⁠.
If you answered $4$ ⁠, you mixed up the number of copies with how many times smaller each is. Those are the two numbers the dimension is made of.
Step 7 of 8
Open this step
The volume
Practice · Calculate
Question
At Depth $3$ ⁠ of the Sierpiński pyramid, what share of the starting pyramid’s volume is left? At each step every pyramid is replaced by $4$ ⁠ copies, each half the size.

Correct answer
0.125 ± 0.001
Setup, choices and feedback
The Scene at this step
Depth
3fixed
Shape
Sierpiński pyramidfixed
Zoom in
Whole shape
Accepted within
± 0.001
Computed from the Scene at this step’s setting
The pyramids’ volume, times the starting pyramid’s, (1/2)ⁿ; NaN for the snowflake/nodes/fractal/outputs/volume
Feedback for a right answer
$\dfrac{1}{8}$ ⁠, that is $0.125$ ⁠: the volume halves at each step.

Hint
A solid half the size in every direction has $\left(\dfrac{1}{2}\right)^3$ ⁠ of the volume.

Worked solution
1. Each copy is half the size, so its volume is $\dfrac{1}{8}$ ⁠ of the one before.
2. Four copies: $4 \times \dfrac{1}{8} = \dfrac{1}{2}$ ⁠ of the volume at each step.
3. After three steps: $\left(\dfrac{1}{2}\right)^3 = 0.125$ ⁠.
At each step the volume halves, so it heads for zero. But the area of all the faces stays the same at every step: $4$ ⁠ copies, each with a quarter of the area. It is a solid of dimension $2$ ⁠.
Step 8 of 8
Open this step
The snowflake’s area
Practice · Calculate
Question
Shape: Koch snowflake, Depth $1$ ⁠. On each of the triangle’s three sides a small triangle has grown, its side a third of the big triangle’s. How many times the starting triangle’s area is the snowflake’s?

Correct answer
1.333 ± 0.005
Setup, choices and feedback
The Scene at this step
Depth
1fixed
Shape
Koch snowflakefixed
Zoom in
Whole shape
Accepted within
± 0.005
Computed from the Scene at this step’s setting
The snowflake’s area, times the starting triangle’s, towards 8/5; NaN for the pyramid/nodes/fractal/outputs/area
Wrong answers it recognises
- 2 ± 0.005
  Misconception: Shrinks an area by the same factor as a length.
  Feedback from the catalogue
  When every length shrinks $3$ ⁠ times, the area shrinks $3^2 = 9$ ⁠ times, because it shrinks in width and in height. A triangle with a third of the side has a ninth of the area, not a third.
Feedback for a right answer
$\dfrac{4}{3} \approx 1.333$ ⁠. After endless steps: $\dfrac{8}{5}$ ⁠.

Hint
A triangle with a third of the side has $\left(\dfrac{1}{3}\right)^2$ ⁠ of the area.

Worked solution
1. Each new triangle: $\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$ ⁠ of the big triangle’s area.
2. Three of them: $3 \times \dfrac{1}{9} = \dfrac{1}{3}$ ⁠.
3. Together: $1 + \dfrac{1}{3} \approx 1.333$ ⁠.
At each step more triangles are added, but nine times smaller, so the additions shrink and the area heads for $\dfrac{8}{5} = 1.6$ ⁠, while the perimeter grows without end.

Lesson 3 of 3 · About 20 minutes

Game theory

The prisoner's dilemma: why betrayal pays in a single round even when cooperation is better for both, what happens when strategies compete over generations, and why a long game rewards tit for tat.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
The prisoner's dilemma
Practice · Choose
Question
A one-round game against a player you do not know. You cannot tell what they will do. What should you do, if you care only about your own points?

Correct answer
- Correct Betray: if they cooperate you get $5$ ⁠ instead of $3$ ⁠, and if they betray you get $1$ ⁠ instead of $0$ ⁠
Setup, choices and feedback
The Scene at this step
Defectors at the start
30 %
Tit for tat among the rest
50 %
Rounds per match
1
Generations
Starting mixfixed
Choices
Shown in a random order.
- Correct Betray: if they cooperate you get $5$ ⁠ instead of $3$ ⁠, and if they betray you get $1$ ⁠ instead of $0$ ⁠
- Wrong Cooperate, because then you both get $3$ ⁠
  Misconception: Thinks cooperating is each player's best move in a single round, because mutual cooperation is best for both together.
  Feedback from the catalogue
  Mutual cooperation, $3$ ⁠ and $3$ ⁠, is indeed better for both together than mutual betrayal. But each player chooses alone: if the other cooperates, betrayal gives $5$ ⁠ instead of $3$ ⁠; if they betray, betrayal gives $1$ ⁠ instead of $0$ ⁠. Either way betrayal is better, and that is the dilemma.
- Wrong There is no telling: it depends on what they do
  Feedback
  Check the two cases one at a time. In both, betrayal scores more, so you do not need to know what they will do.
Feedback for a right answer
Betrayal is a dominant strategy: it is better against every move of the other. But both think so, and both get $1$ ⁠ instead of $3$ ⁠: that is the prisoner’s dilemma.

Hint
Compare your points in each of the two cases: they cooperate, or they betray.

Worked solution
1. If the other cooperates: cooperating gives $3$ ⁠, betraying $5$ ⁠. Better to betray.
2. If the other betrays: cooperating gives $0$ ⁠, betraying $1$ ⁠. Better to betray.
3. Betrayal is better in both cases, so it is the choice, without knowing what the other will do.
Step 2 of 4
Open this step
Average against the table
Practice · Calculate
Question
At the table $30\%$ ⁠ betray, $35\%$ ⁠ always cooperate and $35\%$ ⁠ play tit for tat, and each match is one round. How many points a round does a defector get, on average against the whole table?

Correct answer
3.8 ± 0.05
Setup, choices and feedback
The Scene at this step
Defectors at the start
30 %fixed
Tit for tat among the rest
50 %fixed
Rounds per match
1fixed
Generations
Starting mixfixed
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Average points a round for a defector at the shown generation/nodes/table/outputs/scoreDefect
Wrong answers it recognises
- 5 ± 0.05
  Misconception: Gives a defector 5 points a round against everyone, forgetting the rounds against other defectors.
  Feedback from the catalogue
  A defector gets $5$ ⁠ only against a player who cooperates that round. Against another defector it gets $1$ ⁠. Its average is a weighted average: each payoff times the share of players it gets that payoff from.
Feedback for a right answer
$0.35 \cdot 5 + 0.35 \cdot 5 + 0.30 \cdot 1 = 3.8$ ⁠. In a single round tit for tat cooperates, so it is betrayed just like an unconditional cooperator.

Hint
What does tit for tat do in the first round?

Worked solution
1. Against an unconditional cooperator ( $35\%$ ⁠): $5$ ⁠.
2. Against tit for tat ( $35\%$ ⁠): it cooperates in the first round, so $5$ ⁠.
3. Against a defector ( $30\%$ ⁠): $1$ ⁠.
4. Weighted average: $0.35 \cdot 5 + 0.35 \cdot 5 + 0.30 \cdot 1 = 3.8$ ⁠.
If you answered $5$ ⁠, you forgot the rounds against other defectors.
Step 3 of 4
Open this step
Generation after generation
Practice · Predict, then look
Question
This mix plays generation after generation, and each strategy grows in proportion to the points it collects. Each match is still one round. What will be at the table after 30 generations?

Correct answer
- Correct Only defectors
Setup, choices and feedback
The Scene at this step
Defectors at the start
30 %fixed
Tit for tat among the rest
50 %fixed
Rounds per match
1fixed
Generations
Starting mix
Choices
- Correct Only defectors
- Wrong Mostly tit for tat
  Misconception: Thinks defectors take over any tournament, however many rounds each match lasts.
  Feedback
  In a single round tit for tat has no chance to respond: it cooperates and is betrayed, just like an unconditional cooperator.
- Wrong The same mix as at the start
  Feedback
  The defectors collect $3.8$ ⁠ a round and the others only $2.1$ ⁠, so their share grows each generation.
Once the prediction is locked in, the Scene changes
Generations
30 generations later
The Scene then shows: 1 · Share of players who always defect, 0 to 1, at the shown generation/nodes/table/outputs/defectors
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The defectors take over the whole table, and then they meet only each other and get $1$ ⁠ a round. Everyone would gain more from cooperating, but in a single round no one can afford it.

Hint
Who collects more points in this mix, and what happens to them in the next generation?

Worked solution
1. In one round a defector averages $3.8$ ⁠, and a cooperator or tit for tat only $0.70 \cdot 3 + 0.30 \cdot 0 = 2.1$ ⁠.
2. Each generation the defectors’ share grows, and the others’ shrinks.
3. With fewer cooperators the defectors earn less, but the others earn even less. In the end only defectors are left.
Step 4 of 4
Open this step
A repeated game
Practice · Predict, then look
Question
The same mix to start with, but now each match lasts ten rounds. Who will lead after 30 generations?

Correct answer
- Correct Tit for tat
Setup, choices and feedback
The Scene at this step
Defectors at the start
30 %fixed
Tit for tat among the rest
50 %fixed
Rounds per match
10
Generations
Starting mix
Choices
- Correct Tit for tat
- Wrong The defectors, as with one round
  Misconception: Thinks defectors take over any tournament, however many rounds each match lasts.
  Feedback from the catalogue
  In a one-round game betrayal always pays. In a long game, tit for tat loses to a defector only in the first round and then betrays back, but against cooperative players it collects $3$ ⁠ every round. The longer the game, the less the defector’s one good round is worth, and the defectors die out.
- Wrong The unconditional cooperators
  Feedback
  Unconditional cooperators are betrayed every round against a defector, so they lose most in the early generations. Tit for tat is betrayed only in the first round.
Once the prediction is locked in, the Scene changes
Generations
30 generations later
The Scene then shows: 0.8412 · Share of players who play tit for tat, 0 to 1, at the shown generation/nodes/table/outputs/titForTat
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Tit for tat reaches about $84\%$ ⁠ of the table, and the defectors almost vanish. Against tit for tat a defector gets $5$ ⁠ only in the first round and $1$ ⁠ in the next nine, while two cooperative players get $3$ ⁠ every round.

Hint
How many good rounds does a defector get against tit for tat in a ten-round match?

Worked solution
1. A defector against tit for tat over ten rounds: $\dfrac{5 + 9 \cdot 1}{10} = 1.4$ ⁠ a round.
2. Tit for tat against a cooperator or against itself: $3$ ⁠ a round.
3. At first the defectors still exploit the unconditional cooperators, but as those dwindle, tit for tat collects the most, and its share grows.

Across the unit · 11 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Chaos

The same start

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

A tiny gap

Question

Correct answer

The Scene at this step

Hint

Worked solution

Ten times the precision

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

A gentle release

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

Fractals

Copies of copies

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Hint

Worked solution

A growing perimeter

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Without end

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A broken dimension

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

How many copies

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

How much smaller

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The volume

Question

Correct answer

The Scene at this step

Hint

Worked solution