fair bold coffee 4

The Fair Bold Gambling Functionis Simply Singular

Richard Neidinger

Davidson College

September 2013

(Davidson College) Fair Bold Singular Function September 2013 1 / 29

Gambling on a game like Roulette

A Roulette bet on "red"has a = 18

38 ≈ .474 probability of winning.Each bet B is either lost or pays 1:1,

so net either −B or +B.Play rounds of such bets until you go broke or reach a goal!


Strategy

Game: Start with value 0 < x < 1 and play until value 0 or 1.Strategy: Bet?

If x = 12 then bet???

Bet it all! Decent chance (a = 1838 ) make goal!

If divide into smaller bets, losses will come more often!If x = 1

4 then bet???Bet it all!


Bet enough to reach goal: 14 .


Strategy




If divide into smaller bets, losses will come more often!


Bet it all!If x = 3

4 then bet???Bet enough to reach goal: 1

4 .


Strategy





4 then bet???

Bet it all!If x = 3

4 then bet???Bet enough to reach goal: 1

4 .


Strategy





4 then bet???Bet it all!


Bet enough to reach goal: 14 .


Bold Gambling

Game: Start with value 0 < x < 1 and play until value 0 or 1.Rounds: Each bet has probability a of winning and pays 1:1.

with focus on 0 < a < 12 < 1− a < 1.

Play: For any value x ,bet x if x ≤ 1

2 ,bet 1− x if x > 1

2 .

The Bold Gambling Function:ga(x) = probability of winning this game starting with x .

g1/3(12 ) = ?

g1/3(14 ) = ?

ga(x) is continuous and strictly increasing from (0,0) to (1,1).


Graph of the Bold Gambling Function

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Bold Gambling Function g(x) w/ a=0.33333

x

prob

abili

ty o

f rea

chin

g go

al

↑x=0.73047


Happy Gamblers

For Roulette a = 1838 ≈ .474 with x = 0.75,

prob of win ga(x) = 0.733.

So if people start with $750 and play this game,

73.3% happily finish with $1000, while26.7% go broke.

But at most two rounds.

Expected value: 0.733 ∗ $1000+ 0.267 ∗ $0 = $733.


Fractal Property (as in paper of de Rahm)

To win the game:

If x ≤ 12 , must win round and win game with 2x , so

ga(x) = a · ga(2x).If x > 1

2 , mustwin round OR lose round and win game with x − (1− x), so

ga(x) = a+ (1− a) · ga(2x − 1).

Graphically, this means:If (x , y) is a point on the graph, then

both ( x2 , a y) and (x2 +

12 , (1− a) y + a) are on the graph.

Shrink vertical by a on left half, (1− a) on the right half.


Bold Gambling Function with Secant Slopes

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Bold Gambling Function g(x) w/ a=0.33333

x

prob

abili

ty o

f rea

chin

g go

al


Derivative values?

g ′ does not exist at corners of boxes, at x =12 ,

14 ,

34 , all eighths, sixteenths,... called dyadic rationals.


Derivative values?

x , not a dyadic rational, lies in a nest of boxes of heights Hn(x) where

Hn+1(x) ={

aHn(x)(1− a)Hn(x)

with slope of box diagonal

Sn(x) =Hn(x)( 12

)n = 2nHn(x).

IF g ′(x) = c 6= 0, then limn→∞ Sn(x) = c , so

limn→∞

Sn+1(x)Sn(x)

=cc= 1

ButSn+1(x)Sn(x)

=

{2a

2(1− a)IMPOSSIBLE for a 6= 1/2!


Bold Gambling Function Derivative

Theoremg ′a(x) exists =⇒ g ′a(x) = limn→∞ Sn(x) = 0.

Problemlimn→∞ Sn(x) = 0 ; g ′a(x) exists.

Will show a variation on this function where Sn(x) DOES characterizethe derivative!


Symmetry Property

You and Opponent have x and 1− x value, respectively.You and Opponent have a and 1− a prob of winning.On each bet, amount is transferred from loser to winner.

From opponent viewpoint:

g(1−a)(1− x) = 1− ga(x).or

g(1−a)(t) = 1− ga(1− t).

So graph of g(1−a) flips both x and y values of ga,equivalent to rotation of 180◦ about ( 12 ,

12 ).


Your chance in red; Opponent chance in blue.

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bold gambling function

a=0.2inverserotatea=0.8


Fair Bold Gambling

Alternate probabilities a and 1− a on each round!Odd rounds you win with prob a,Even rounds you win with prob (1− a).Bold gambling required: For any value x ,

bet x if x ≤ 12 ,

bet 1− x if x > 12

Game: Start with value 0 < x < 1 and play until value 0 or 1.

The Fair Bold Gambling Function:fa(x) = probability of winning this game starting with x .

f1/3(12 ) = ?

f1/3(14 ) = ?

fa(x) is continuous and strictly increasing from (0,0) to (1,1).


Graph of the Fair Bold Gambling Function

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Fair Bold Gambling Function f(x) w/ a=0.33333

x

prob

abili

ty o

f rea

chin

g go

al


Symmetry: again, Opponent given by Rotation.

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Fair Bold Gambling Function f(x) w/ a=0.1

x

prob

abili

ty o

f rea

chin

g go

al

a=0.1inversea=0.9


Fractal Property

To win the fair-bold game:

If x ≤ 12 , must win round and win flipped-game with 2x , so

fa(x) = a · f(1−a)(2x).

If x > 12 , must

win round OR lose round and win flipped-game with x − (1− x), so

fa(x) = a+ (1− a) · f(1−a)(2x − 1).

Graphically, this means:

Shrink vertical by a , horizontal by 12 , and rotate into lower-left,

Shrink vertical by (1− a), horizontal by 12 , and rotate into

upper-right.


Iterations that converge to Fair Bold Gambling Function

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original

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box heights H1(x)

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.00..., .01..., .10..., .11...


Box heights by binary representation

0 < x < 14 has binary representation x = (.00b1b2 . . .),

14 < x < 1 has binary representation x = (.11b1b2 . . .),and box height

H2(x) = a(1− a)

slopeH2(x)( 12

)2 = 4a(1− a) < 1


Nested boxes converging to Fair Bold Gambling Function

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Fair bold gambling function w/ a=0.33333


"Flattest" and "Steepest" Places on Graph for

0 < a < 1/2.

Steepest place is x = (.101010 . . .) = 2/3, where secant slopes

Hn(x)(1/2)n

=(1− a)n(1/2)n

= (2(1− a))n → ∞.

Flattest place is x = (.010101 . . .) = 1/3, where secant slopes

Hn(x)(1/2)n

=an

(1/2)n= (2a)n → 0.

By the same argument as for bold gambling,

f ′a (x) exists =⇒ f ′a (x) = limn→∞

Hn(x)(1/2)n

= 0


Computing with the Fair-Bold-Gambling function

Usingfa(x) = a · f(1−a)(2x) if x ≤ 0.5fa(x) = a+ (1− a) · f(1−a)(2x − 1) if x > 0.5

Suppose x = (.b1b2b3b4 . . .), a binary representation.fa(.0 b2b3b4 . . .) = a · f(1−a)(.b2b3b4 . . .)fa(.1 b2b3b4 . . .) = a+ (1− a) · f(1−a)(.b2b3b4 . . .)f(1−a)(.0 b3b4 . . .) = (1− a) · fa(.b3b4 . . .)f(1−a)(.1 b3b4 . . .) = (1− a) + a · fa(.b3b4 . . .)So zeros create only multiplicative factors, alternating a and 1− a,ones create additive terms plus alternating multiplicative factors.


Derivative Zero at all Dyadic Rationals

Any 0 < a < 1, a 6= 12 and any

x = (.b1 . . . b2m0 . . .),h = (. 0 . . . 0 0 . . . 0 d1d2d3 . . .) with 2j zeros beyond 2m,

x + h = (.b1 . . . b2m0 . . . 0 d1d2d3 . . .)

fa(x) = A+ Bfa(0) = A.fa(x + h) = A+ Bfa(.0 . . . 0 d1d2d3 . . .) with 2j zeros.

= A+ Baj (1− a)j fa(.d1d2d3 . . .) ≤ A+ Baj (1− a)j

fa(x + h)− fa(x)h

≤ Baj (1− a)j( 12 )

2(m+j+1)= 4m+1B (4a(1− a))j → 0

So derivative from the right, D+fa(x) = 0 at any dyadic rational.Derivative from the left follows from (1’s instead of 0’s or from)rotation D−fa(x) = D+f(1−a)(1− x) = 0.(Davidson College) Fair Bold Singular Function September 2013 23 / 29

Secant Slopes Characterize the Derivative

Theorem (Derivative Bounding Theorem)

For 0 < a < 1, a 6= 1/2, and fair bold gambling function fa, there existα, β > 0 such that for all x ∈ (0, 1) with x not dyadic rational, and for allh > 0 such that x − h > 0 and x + h < 1,

αHn(x)(1/2)n

≤{fa(x + h)− fa(x)

h,fa(x)− fa(x − h)

h

}≤ β

Hn(x)(1/2)n

where (1/2)n+1 < h ≤ (1/2)n.

f ′a (x) = 0 iff limn→∞

Hn(x)(1/2)n

= 0

f ′a (x) = ∞ iff limn→∞

Hn(x)(1/2)n

= ∞

Other possibility: Hn(x )(1/2)n oscillates with multiple accumulation points

(maybe incl. ∞).(Davidson College) Fair Bold Singular Function September 2013 24 / 29

Digit agreement with 1010... (where derivative is infinite)

Example x = 715 :

x =. 0 1 1 1 0 1 1 1 0 1 1 1 . . .2/3 =. 1 0 1 0 1 0 1 0 1 0 1 0 . . .agree? N N Y N N N Y N N N Y N . . .kn = 0 0 1 1 1 1 2 2 2 2 3 3 . . .

qn = knn = 0

102

13

14

15

16

27

28

29

210

311

312 . . .

This qn(x)→ 14 , so this x agrees only about 25% with .1010...

So x mostly agrees with the opposite .0101...,and f ′a (.010101 . . .) = 0.in the extreme.This suggests that f ′a (x) = 0.


Secant Slopes by Portion of Digits that Agree with 1010...

kn = number of digits in x = (b1b2 . . . bn . . .) that agree with 1010 . . .

Hn(x)(1/2)n

= (2a)n−kn (2(1− a))kn

So derivative boils down to

qn(x) =knn= running portion of digits that agree with 1010 . . .

Hn(x)(1/2)n

=((2a)1−qn(x )(2(1− a))qn(x )

)n=

((2a)

(1− aa

)qn(x ))nwhere border between → 0 and → ∞ is given by

(2a)(1− aa

)Q= 1 iff Q(a) =

ln(2a)ln(a)− ln(1− a)


Regions of Digit Agreement with 1010...

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a

runn

ing

porti

on q

n(x)

f '(x) = 0

f '(x) = infinity

f '(x) = 0

f '(x) = infinity

Q(a)


Characterization of Derivative at x

For 0 < a < 12 ,

If all but finitely many qn(x) < some c < Q(a), then f ′a (x) = 0.

If infinitely many qn(x) ≥ Q(a), then f ′a (x) DNE.First case covers any random x (binary digits chosen with equal probabilityof 0 or 1), since odd digits and even digits will both tend to half 0′s andhalf 1′s, making qn(x)→ 0.5. Thus f ′a (x) = 0 with probability one(almost everywhere).

The only case not covered by one of above is when:eventually qn(x) < Q(a) but they get arbitrarily close to Q(a),where either f ′a (x) = 0 or f

′a (x) DNE is possible.


Derivative at All Rational Points

If x is rational and

has an odd number of repeating digits (incl. dyadic rationals),then qn(x)→ 0.5, so f ′a (x) = 0.

has a smallest even number 2m of repeating digits,so qn(x)→ k

2m where k is the number of digits that agree w/1010 . . . in a repeating string starting at an odd index, then

k2m < Q(a) =⇒ f ′a (x) = 0.k2m > Q(a) =⇒ f ′a (x) = ∞.k2m = Q(a) =⇒ f ′a (x) DNE and, in fact,Hn(x )(1/2)n eventually cycles between 2m positive real values!


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