ib mathematics hl coursework type ii – the dice game

!Internal!Assessment!Type!II!

!Mathematics!Higher!Level!

!!!!!!!!!!!!!!!!!!!!Marc!Wierzbitzki!Hockerill!Anglo<European!College!!International!Baccalaureate!IB!Session!May!2012! ! ! ! !00<0815<083!

Marc%Wierzbitzki% % Mathematics%HL% Hockerill%Anglo7European%College%

00708157083% Internal%Assessment%Type%II% IB%Session%May%2012%

% 1%

!

!

HL!TYPE!II!–!THE!DICE!GAME!!

!

Two%players,%Ann%and%Bob,%each%roll%a%die%once%and%note%the%number%on%the%upper%face.%If%

Ann’s%number%is%greater%than%Bob’s,%she%wins.%However,%if%Bob’s%number%is%greater%than%

or%equal%to%Ann’s,%he%wins.%This%can%be%summarized%as%follows:%

%

(number%Ann)%>%(number%Bob)%!%Ann%wins%

(number%Ann)%≤%(number%Bob)%!%Bob%wins%

% %

Now,%using%a%probability%tree%diagram%as%a%helpful% tool,% the%probability% for%Bob%or%Ann%

winning%when%they%play%the%game%can%be%found.%%

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Figure!1:!Tree!diagram!displaying!the!probabilities!when!first!Ann!and!then!Bob!throw!a!fair!die.!The!winner!

is!shown!in!colour!at!the!end!of!each!path.!!

%

%

%

Explanation:%%

The%above%tree%diagram%shows%the%probabilities%and%the%winner%of%the%game%when%first%

Ann% (left% half)% and% then% Bob% (right% half)% throw% the% same% die% once% each.% The% possible%

results% (1,%2,%3,%4,%5,%6),%displayed% in%green,% is% the%number%on% the%die% that% is% facing% the%

player.%We%assume%that%they%are%playing%with%a%fair%die,%which%means%that%the%probability%

of%getting%a%1%or%2%or%3%or%4%or%5%or%6%is%equal.%These%probabilities%for%these%events%are%

calculated%below:%

%

% %

%

1

2

4

5

6

3

Ann’s throw

123456 1

234561

23456 1

234561

23456 1

23456

Bob’s throw

Bob winsBob winsBob winsBob winsBob winsBob wins

Bob winsBob winsBob winsBob winsBob wins

Ann wins

Bob winsBob winsBob winsBob wins

Ann winsAnn wins

Bob winsBob winsBob wins

Ann winsAnn winsAnn wins

Ann winsAnn winsAnn winsAnn winsBob winsBob wins Ann wins

Ann winsAnn winsAnn winsAnn winsBob wins



% 3%

General%formula:%

%

“! ! = ! !! ! %% % % %where%n(E)%is%the%number%of%occurrences%of%the%event%E%and%n(S)%is%the%

total%number%of%possible%outcomes.”%1%%

! 1 = ! !! ! = 1

6%

! 2 = ! !! ! = 1

6%

! 3 = ! !! ! = 1

6%

! 4 = ! !! ! = 1

6%

! 5 = ! !! ! = 1

6%

! 6 = ! !! ! = 1

6%%

To%check%this,%the%sum%of%all%the%probabilities%should%be%calculated.%If%it%is%equal%to%1,%then%

the%calculation%can%be%right.%It%is%important%to%understand%that%this%is%not%a%guarantee,%but%

if% the% sum% is% not% 1,% then% something% is% definitely%wrong.% In% the% case% above,% the% sum% is%

equal%to%1.%%

%

Above%are%now%the%probabilities%for%the%first%throw%(Ann),%but%after%her,%Bob%is%allowed%to%

throw% the% die% once.% This%means% that% each% of% the% paths% of% the% tree% diagram% has% to% be%

extended,% as% in% figure% 1.% As% he%will% be% using% the% same% fair% die% that% Ann% has% used,% the%

probabilities%for%each%number%will%remain%the%same.%%

Lastly,% the% total% probabilities% of% an% event% are% calculated% by%multiplying% the% individual%

probabilities%along%one%path.%The%probability%of%any%event%to%occur%after%the%two%throws%

(first%Ann,%then%Bob)%is%always%equal%to:%! !"#!!"!#$!!"#$%!!"#!!ℎ!"#$ = !!".%%

Now% that% the% probability% tree% diagram% has% been% constructed,% it% has% to% be% determined%

who%the%winner%along%each%individual%path%would%be.%To%do%this,%the%general%rules%of%the%

game%from%above%should%be%reconsidered:%

%

(number%Ann)%>%(number%Bob)%!%Ann%wins%

(number%Ann)%≤%(number%Bob)%!%Bob%wins%

%

%

%

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1%Bill%Roberts,%Sandy%MacKenzie%–%“Mathematics%HIGHER%LEVEL%for%the%IB%Diploma”,%Oxford%2007,%page%

561%



% 4%

Ann%only%wins%when%her%number%is%greater%than%Bob’s.%This%happens%in%the%following%15%

cases:%%

%

Ann‘s!number! Bob‘s!number! Probability!of!event!

2% 1%136%

3% 1%136%

3% 2%136%

4% 1%136%

4% 2%136%

4% 3%136%

5% 1%136%

5% 2%136%

5% 3%136%

5% 4%136%

6% 1%136%

6% 2%136%

6% 3%136%

6% 4%136%

6% 5%136%

Figure!2:! Table! listing! the! cases! in!which!Ann!would! get! a! higher!number! than!Bob! and! therefore!win! the!

game.!The!column!on!the!right!shows!the!probability!that!this!event!will!occur.!!

%

Adding%up%all%the%probabilities%for%the%cases%in%which%Ann%wins,%gives%a%total%probability%

of:%%

! !""!!"#$ = !1536%%

This% is% less% than%50%,%which%makes%sense%since%she% loses% if%both%players%get% the%same%

number,%but%apart%from%that%the%chances%of%winning%are%equal%for%both.%The%probability%

that%Bob%wins%in%this%case%can%now%be%calculated%by:%%

%



% 5%

%

! !"#!!"#$ = 1− ! !""!!"#$ = !1− 1536 =2136%

%

%

%

%

%

%

%

%

%

Now,% the% two% choose% to% slightly% alter% the% game.% Ann% is% now% allowed% to% throw% the% die%

twice% and% note% down% the% bigger% of% the% two% numbers.% All% other% rules% from% above% still%

apply.% The% two% probabilities%! !""!!"#$ %and%! !"#!!"#$ %are% calculated% for% this%

scenario.%%

%

%

To% approach% this% problem,% it% is% first% found% out% what% the% probabilities% are% of% Ann%

throwing%a%1%or%2%or%3%or%4%or%5%or%6%as%her%highest%number%in%two%throws%are,%using%the%

sample%space.%%

%

All%possible%outcomes:%%

%

1% 1% % 1% 2% % 1% 3% % 1% 4% % 1% 5% % 1% 6%

2% 1% % 2% 2% % 2% 3% % 2% 4% % 2% 5% % 2% 6%

3% 1% % 3% 2% % 3% 3% % 3% 4% % 3% 5% % 3% 6%

4% 1% % 4% 2% % 4% 3% % 4% 4% % 4% 5% % 4% 6%

5% 1% % 5% 2% % 5% 3% % 5% 4% % 5% 5% % 5% 6%

6% 1% % 6% 2% % 6% 3% % 6% 4% % 6% 5% % 6% 6%

Figure!3:!Sample!space!displaying!all!possible!outcomes!if!Ann!throws!the!die!twice.!

%

In% the%sample%space%above,% the%numbers% in%purple% represent% the%possible%outcomes% in%

Ann’s%first%throw%and%the%numbers%in%orange%the%outcomes%on%the%second%throw.%As%only%

the%higher%number%is%important%for%the%game,%the%smaller%ones%can%be%ignored,%giving:%

%

Conclusion:%%

If%Ann%and%Bob%play%a%game%with%a%fair%die,%where%each%of%them%is%allowed%to%throw%it%

once,%the%number%facing%upwards%is%noted%and%Ann%only%wins%if%her%number%is%greater%

than% Bob’s,% then% the% probability% of% her% winning% this% game% is%!(!""!!"#$) = ! !"!" .%Therefore,% the% probability% of% Bob%winning% the% game% (in% case% his% number% is% greater%

than%or%equal%to%Ann’s)%is%!(!"#!!"#$) = !"!".%%

%



% 6%

%

1% % % % 2% % % 3% % % 4% % % 5% % % 6%

2% % % % 2% % % 3% % % 4% % % 5% % % 6%

3% % % 3% % % % 3% % % 4% % % 5% % % 6%

4% % % 4% % % 4% % % % 4% % % 5% % % 6%

5% % % 5% % % 5% % % 5% % % % 5% % % 6%

6% % % 6% % % 6% % % 6% % % 6% % % % 6%

Figure!4:!Sample!space!representing!the!same!situation!as!in!figure!3,!but!this!time!only!showing!the!higher!

number,!since!it!is!the!important!one!for!the!game.!!!

%

From%this,%the%probabilities%that%the%highest%number%is%a%1%or%2%or%3%or%4%or%5%or%6%can%be%

calculated%using%the%same%method%as%above.%%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 1 = ! !! ! = 1

36%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 2 = ! !! ! = 3

36%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 3 = ! !! ! = 5

36%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 4 = ! !! ! = 7

36%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 5 = ! !! ! = 9

36%

! !"!!!!ℎ!"ℎ!"!!"#$%&!!" = 6 = ! !! ! = 11

36%%

Now,% the%tree%diagram%from%earlier%can%be%adapted%to% fit% this%situations.%The%arms%and%

winner’s% can% be% copied% directly,% but% the% probabilities% for% Ann’s% throws% have% to% be%

changed%and%replaced%with%the%new%probabilities%that%have%just%been%found.%%

%

%



% 7%

%

Figure!5:!Probability!tree!diagram!for!the!situation!where!Ann!is!allowed!to!throw!the!die!twice!can!note!the!

highest!number,!whereas!Bob!can!only!throw!the!die!once.!!

%

%

Ann‘s!highest!number!in!

two!throws!Bob‘s!number!

Probability!of!event!

= !!(!"#$%!!"#$%!)∗ !! !"#$%&!!"#$%& %

2% 1%336 ∗

16 =

3216%

3% 1%536 ∗

16 =

5216%

3% 2%536 ∗

16 =

5216%

1

2

4

5

6

3

Ann’s highest number in 2 throws

123456 1

234561

23456 1

234561

23456 1

23456

Bob’s throw

Bob winsBob winsBob winsBob winsBob winsBob wins

Bob winsBob winsBob winsBob winsBob wins

Ann wins

Bob winsBob winsBob winsBob wins

Ann winsAnn wins

Bob winsBob winsBob wins

Ann winsAnn winsAnn wins

Ann winsAnn winsAnn winsAnn winsBob winsBob wins Ann wins

Ann winsAnn winsAnn winsAnn winsBob wins



% 8%

4% 1%736 ∗

16 =

7216%

4% 2%736 ∗

16 =

7216%

4% 3%736 ∗

16 =

7216%

5% 1%936 ∗

16 =

9216%

5% 2%936 ∗

16 =

9216%

5% 3%936 ∗

16 =

9216%

5% 4%936 ∗

16 =

9216%

6% 1%1136 ∗

16 =

11216%

6% 2%1136 ∗

16 =

11216%

6% 3%1136 ∗

16 =

11216%

6% 4%1136 ∗

16 =

11216%

6% 5%1136 ∗

16 =

11216%


alternated!game!where!she!was!allowed!to!throw!the!die!twice.!

Calculating%the%sum%of%these%probabilities%gives:%

%

! !""!!"#$ = !125216%%

The%probability%that%Bob%wins%in%this%case%can%now%be%calculated%by:%%

%

! !"#!!"#$ = 1− ! !""!!"#$ = !1− 125216 =91216%

%

%

%

Conclusion:%%

If%Ann%and%Bob%play%a%game%with%a%fair%die,%where%Ann%is%allowed%to%throw%it%twice%and%

use%the%highest%number%as%her%result,%but%Bob%is%only%allowed%to%throw%it%once,% then%

the% probability% of% Ann% winning% this% game% is%!(!""!!"#$) = !"#!"# .% Therefore,% the%

probability%of%Bob%winning%the%game%(in%case%his%number%is%greater% than%or%equal%to%

Ann’s)%is%!(!"#!!"#$) = !"!"#.%%

%



% 9%

After%this,%the%two%decide%that%both,%Ann%and%Bob,%can%throw%the%die%twice%while%noting%

only%the%highest%number%they%have%scored.%The%probability%that%Bob%scores%a%1%or%2%or%3%

or%4%or%5%or%6%as%his%highest%in%two%throws%is%the%same%as%for%Ann,%which%is%why%we%can%

proceed%straight%to%the%table%to%alter%the%probabilities%for%Bob.%%

The%probability%of%Ann%winning%this%time%is%shown%in%the%table%below:%

%

Ann‘s!highest!number!in!

two!throws!

Bob‘s!highest!number!in!

two!throws!

Probability!of!event!

= !!(!"#$%!!"#$%&)∗ !! !"#$%&!!"#$%& %

2% 1%336 ∗

136 =

31296%

3% 1%536 ∗

136 =

51296%

3% 2%536 ∗

336 =

151296%

4% 1%736 ∗

136 =

71296%

4% 2%736 ∗

336 =

211296%

4% 3%736 ∗

536 =

351296%

5% 1%936 ∗

136 =

91296%

5% 2%936 ∗

336 =

271296%

5% 3%936 ∗

536 =

451296%

5% 4%936 ∗

736 =

631296%

6% 1%1136 ∗

136 =

111296%

6% 2%1136 ∗

336 =

331296%

6% 3%1136 ∗

536 =

551296%

6% 4%1136 ∗

736 =

771296%

6% 5%1136 ∗

936 =

991296%


game!when!both!are!allowed!to!throw!the!die!twice.!

!

%

%

%

%



% 10%

Again,%calculating%the%sum%of%these%probabilities,%gives:%

%

! !""!!"#$ = ! 5051296%%

The%probability%that%Bob%wins%in%this%case%can%now%be%calculated%by:%%

%

! !"#!!"#$ = 1− ! !""!!"#$ = !1− 5051296 =

7911296!%

%%

%

%

%

%

%

%

%

It%should%now%be%considered%to%find%a%general%form%for%the%probabilities%for%Ann%and%Bob%

to%win%the%game%when%they%roll%the%die%multiple%times%(the%same%amount%of%throws%for%

both% of% them)% and% each% notes% the% highest% number% they% have% scored.% All% other% rules%

remain%the%same.%%

%

The%formula,%which%states%the%probability% for%a%number%!%to%be%the%highest% in%!%throws%is:%

%

! ! = !! − (! − 1)!6! %

%

This%formula%can%be%derived%from%looking%at%the%general%formula%for%probability:%

%

! ! = ! !! ! %

Conclusion:%%

If%Ann%and%Bob%play%a%game%with%a%fair%die,%where%both%are%allowed%to%throw%it%twice%

and%use%the%highest%number%as% their%result,% then%the%probability%of%Ann%winning%the%

game%is%!(!""!!"#$) = !"!!"#$.%Therefore,% the%probability%of%Bob%winning%the%game%(in%

case%his%number%is%greater%than%or%equal%to%Ann’s)%is%!(!"#!!"#$) = !"#!"#$!.%%

%



% 11%

If%we%start%by%looking%at%the%denominator,%! ! %is%defined%as%the%total%number%of%possible%outcomes.%In%the%case%of%a%die,%it%is%always%equal%to%6!.%This%can%be%explained%by%referring%back% to% the% probability% tree% diagram,% where% in% case% that% a% die% that% is% thrown,% the%

denominator%will%always%be%multiplied%by%6%when%the%die%is%thrown%an%additional%time.%

%

%

Figure! 8:! Extract! from! a! probability! tree! diagram! to! explain! the! formula! for! the! probability! that! Z! is! the!

highest!number!in!n!throws.!!!

%

The%nominator,%! ! = !! − (! − 1)!,% can%be%explained%by% looking%closely%at%what%has%to% be% found.% The% formula% is% there% to% help% us% find% the% probability% that%!%is% the% highest%number%in%!%throws.%This%is%not%easy%to%be%calculated,%unless%we%consider%the%following:%%

! ! = ! = ! ! ≤ ! − !(! ≤ (! − 1))%%

where%!%is%the%symbol%for%the%highest%number%and%!%any%number%from%1%to%6%for%which%the%probability%has%to%be%found.%%Or,%by%considering%a%specific%example:2%

%

! ! = 4 = ! ! ≤ 4 − !(! ≤ 3)%%

For%the%above%example,%we%want%to%find%the%probability%that%the%highest%number%(!)%is%4.%This% however% is% not% easy% to% calculate,% which% is% why%we% look% to% express% this% in% other%

terms.% If%we% look% at% the% right% hand% side,% it% can%be% said% that% a% number% smaller% than%or%

equal%to%4%(1,%2,%3,%4)%minus%the%numbers%that%are%smaller%than%or%equal%to%3%(1,%2,%3)%is%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2%“Wahrscheinlichkeit%maximal%eine%4%zu%Würfeln”%7%

http://www.matheboard.de/archive/250012/1/thread.html,%accessed%November%7,%2011%11:43%GMT%

1

2

4

5

6

3

First throw

123456

Second throw

123456 ...

123456Third throw nth throw



% 12%

equal% to% 4.% This%means% that%we% have% expressed% ! = 4 %as% ! ≤ 4 − (! ≤ 3).% The% two%probabilities%on%the%right7hand%side%are%much%easier%to%find%by%looking%at%the%probability%

tree%diagram%and%we%can%come%up%with%the%formula:%

%

! ! = !! − (! − 1)!6! %

%

This% however% only% describes% the% probability% that%!%is% the% highest% number% in%!%throws,%which%is%the%first%step%to%find%the%probability%that%Ann%wins,%when%she%and%Bob%throw%the%

die%the%same%number%of%times.%We%can%now%look%back%at%the%cases%when%Ann%wins:%

%

%

1% 1% % 1% 2% % 1% 3% % 1% 4% % 1% 5% % 1% 6%

2% 1% % 2% 2% % 2% 3% % 2% 4% % 2% 5% % 2% 6%

3% 1% % 3% 2% % 3% 3% % 3% 4% % 3% 5% % 3% 6%

4% 1% % 4% 2% % 4% 3% % 4% 4% % 4% 5% % 4% 6%

5% 1% % 5% 2% % 5% 3% % 5% 4% % 5% 5% % 5% 6%

6% 1% % 6% 2% % 6% 3% % 6% 4% % 6% 5% % 6% 6%

Figure!9:!All!possible!outcomes!if!Ann!and!Bob!throw!the!die!the!same!number!of!times!(only!the!highest!

number!is!displayed).!Bob’s!highest!number!is!displayed!in!violet,!Ann’s!in!orange.!!

%

The% cases% in%which%Ann%wins% are%market% in% red% and% the% cases% in%which%Bob%wins% are%

marked%in%blue.%%

What% we% notice% is% that% in% general,% the% order% does% not% matter.% We% can% see% that%

! !"#!!"#$ %is%half%of%the%matrix%above,%plus%the%cases%in%which%the%two%have%the%same%number.%Therefore,%we%can%say:%

%

! Ann!!"#$ = 1− ! !"#$%&'!!"#!!ℎ!!!"!"2 %

%

To%explain%this%again,%the%probabilities%for%the%cells:%%

%%

2% 1%

%

or%%

1% 2%

%

in%figure%9%to%occur%are%exactly%the%same%(provided%that%they%each%throw%the%die%the%same%

amount%of%times).%This%basically%means%that%the%two%halves,%marked%in%yellow%in%figure%

10%have%exactly% the%same%probability.%The% total%probability%of% the%system% is%1,% and% the%

cases%in%which%they%score%the%same%number%are%marked%in%green.%%

%

%

%



% 13%

1% 1% % 1% 2% % 1% 3% % 1% 4% % 1% 5% % 1% 6%

2% 1% % 2% 2% % 2% 3% % 2% 4% % 2% 5% % 2% 6%

3% 1% % 3% 2% % 3% 3% % 3% 4% % 3% 5% % 3% 6%

4% 1% % 4% 2% % 4% 3% % 4% 4% % 4% 5% % 4% 6%

5% 1% % 5% 2% % 5% 3% % 5% 4% % 5% 5% % 5% 6%

6% 1% % 6% 2% % 6% 3% % 6% 4% % 6% 5% % 6% 6%

Figure!10:!Cases!in!which!the!two!players!score!the!same!number!are!marked!in!green,!all!other!cases!in!

yellow.!

!!

Therefore%we%can%conclude%that%the%probability%that%Ann%wins%is%1%(total%probability%of%

the%whole% system)%minus% the% cases% in%which% they% score% the% same% number% (since% Bob%

would%win%in%that%case).%This%would%leave%the%cases%marked%in%yellow,%but%Ann%only%wins%

in%half%the%cases%which%is%why%we%have%divide%by%2%and%get%to%the%formula:%%

%

! Ann!!"#$ = 1− ! !"#$%&'!!"#!!ℎ!!!"#$2 %

%

%

In%order%to%use%it%though,%! !"#$%&'!!"#!!ℎ!!!"#$ %has%to%be%found.%%%

%

! !"#$%&'!!"#!!ℎ!!!"#$ = !(1)! + !(2)! + !(3)! + !(4)! + !(5)! + !(6)!%%

%

This%can%be%expressed%using%sigma%notation:%

%

! !"#$%&'!!"#!!ℎ!!!"#$ = (!(!)!!

!!!)%

%

Here,%

%

! ! = !! − (! − 1)!6! %

%

%

Finally,%when%this%is%substituted%into%the%formula,%the%probability%of%Ann%winning%is:%%

%

! Ann!!"#$ = 12−

12 (!(!)!

!

!!!)%

%



% 14%

This%can%be%displayed%on%a%graph%in%order%to%be%able%to%quickly%calculate%the%probability%

that% Ann% wins.% To% do% so,% we% will% plot%! Ann!!"#$ %on% the% y7axis% and% the% number% of%throws%that%both%players%have%(!)%on%the%x7axis.%%%

%

%

Figure!11:!Graph!showing!the!probabilities!that!Ann!wins!when!she!and!Bob!throw!the!die!the!same!number!

of!times.!!

%

Looking%at%the%graph,%it%has%to%be%considered%that%only%positive%integers%can%be%used%as%x%

values.%This%is%because%the%players%can’t%throw%the%die%a%negative%amount%of%times,%nor%to%

a%fraction%of%a%time.%In%other%words,%they%can%throw%it%2,%3,%4%times,%but%not%71,%2.1%or%3.42%

times.%

%%

We%can%now%check%on%the%graph%whether%the%two%probabilities%that%we%found%out%before%

were% right%or%not% (they%each% throw% it%once,% twice).%This% is%done%by% setting% the%x%value%

equal%to%1%and%2,%which%will%then%create%two%vertical%lines.%We%can%then%find%the%point%of%

intersection%to%get%the%probability%that%Ann%wins.%%

%



% 15%

%

Figure! 12:! Graph! showing! the! same! as! in! figure! 11,! this! time! including! the! two! lines! x=1! and! x=2.! The! y\

coordinate!of!the!point!of!intersection!will!give!the!probability!that!Ann!wins!in!this!case.!!

%

For%both%intersections%we%get%the%right%y7coordinates%(and%therefore%probabilities)%that%

Ann%wins%(≈ 42%%and%≈ 39%).%%

%

%

%

%

%

%

%

Conclusion:%%

The%general% form%for% the%probability% that%Ann%wins,%when%she%and%Bob%play%a%game%

with%a%fair%die%which%is%rolled%the%same%number%of%times%by%both%of%them%and%only%the%

highest%number%is%noted%as%a%result%is:%

%

!(Ann!!"#$) = 12−

12!(!(!)!

!

!!!)%

%

where%

%

!(!) = !! − (! − 1)!6! %

Again:%

!(!"#!!"#$) = 1− !(!""!!"#$)%



% 16%

%

Since%this%is%only%true%when%both%roll%the%die%the%same%number%of%times,%we%now%want%to%

find%out%what%the%probabilities%of%winning%for%both%are,%when%they%roll%the%die%multiple%

times,%but%not%necessarily%the%same%number%of%times.%%

%

To%start%with,%Ann%wins%in%the%cases%that%were%listed%in%figure%7.%All%that%has%to%be%done%

now%is%that%we%have%to%find%the%probability%that%those%cases%occur:%

%

%

! !""!!"#$ = !!! 2 ∗ !! 1 + !!! 3 ∗ !!! 1 + !! 3 ∗ !! 2 + !! 4 ∗ !! 3 + !! 4∗ !! 2 + !! 4 ∗ !! 1 + !! 5 ∗ !! 4 + !! 5 ∗ !! 3 + !! 5 ∗ !! 2+ !! 5 ∗ !! 1 + !! 6 ∗ !! 5 + !! 6 ∗ !! 4 + !! 6 ∗ !! 3 + !! 6∗ !! 2 + !! 6 ∗ !! 1 %

%

%

= !!! 2 ∗ !! 1 !!!!!+ !! 3 !!! 1 + !! 2 + !!!!!!! 4 !!! 1 + !! 2 + !! 3 !!!!!+ !! 5 !!! 1 + !! 2 + !! 3 + !! 4 !!!!!+ !! 6 !!! 1 + !! 2 + !! 3 + !! 4 + !! 5 %

%

%

In% the% formulae% above,%!! ! %is% the% probability% that% Ann% scores% the% number%!%as% her%highest% in% multiple% throws% and% !! ! %expresses% the% same% for% Bob.% Now% these%

probabilities%have%to%be%found%by%applying%the%formula%that%was%found%before:%

%

! ! = !! − (! − 1)!6! %

%

In%this%case,%the%number%of%throws%(!)%is%not%the%same%for%Bob%and%Ann.%The%number%of%times%that%Ann%throws%the%die%is%defined%as!!%and%the%number%of%times%that%Bob%throws%it%as%!.%Therefore,%we%can%say%that:%%

! !""!!"#$ = !!! 2 ∗ !! 1 + !! 3 !!! 1 + !! 2+ !! 4 !!! 1 + !! 2 + !! 3 !!!!!+ !! 5 !!! 1 + !! 2 + !! 3 + !! 4 !!!!!+ !! 6 !!! 1 + !! 2 + !! 3 + !! 4 + !! 5 %

%



% 17%

= 2! − 1!6! ∗ 1! − 0!

6! + 3! − 2!6! ∗ 2! − 1!

6! + 1! − 0!6! + 4

! − 3!6!

∗ 3! − 2!6! + 2

! − 1!6! + 1

! − 0!6! + 5

! − 4!6!

∗ 4! − 3!6! + 3

! − 2!6! + 2

! − 1!6! + 1

! − 0!6! + 6

! − 5!6!

∗ 5! − 4!6! + 4

! − 3!6! + 3

! − 2!6! + 2

! − 1!6! + 1

! − 0!6! %

Simplifying:%

%

= 2! − 16!6! + 3

!2! − 2!2!6!6! + 4

!3! − 3!3!6!6! + 5

!4! − 4!4!6!6! + 6

!5! − 5!5!6!6! %

%

%

= 2! − 1+ 3!2! − 2!2! + 4!3! − 3!3! + 5!4! − 4!4! + 6!5! − 5!5!6!6! %

%

This%can%be%expressed%using%sigma%notation:%

%

! !""!!"#$ = 16!!! (!! ! − 1 ! − ! − 1 !!!)

!

!!!%

%

%

%

Note:%! %and%! %have% to% be% positive% integers,% because% of% the% reasons% that% have% been%explained%before.%%

%

%

If%we%consider%that%both%roll%their%die%10%times,%this%would%mean%that%! = 10%and%! = 10,%which%makes%the%equation%not%very%easy%to%calculate.%Therefore,%we%can%use%technology%

to%plot%a%graph%of%the%equation,%so%that%the%probability%can%be%found%much%easier.%%

%



% 18%

%

Figure!13:!Graph!showing!the!equation!that!Ann!wins,!when!she!and!Bob!throw!the!die!multiple!times.!!

%

The%x7axis%above%displays%the%number%of%times%that%Ann%has%throws%her%die%(!).%The%y7axis% shows%!%and% the% z7axis% the% probability% that% Ann% wins% (from% 071).% (Note% that% the%negative%axes%have%not%been% included,%since% there% is%no%such%situation%where%someone%

throws% the% die% a% negative% amount% of% times% and% the% probability% cannot% be% negative%

either).%%

%

However,%by%looking%at%the%graph%we%can%see%that%there%is%an%error.%Considering%the%case%

that%Ann%throws%the%die%1%time%and%Bob%0%times,%Ann%should%theoretically%win.%Both,%the%

formula%and%the%graph%give%a%probability%of%Ann%winning%at%! !""!!"#$ = !!.%This%is%not%

right% and% therefore%we% have% to% say% that% the% formula% and% the% graph% only%work% if% both%

players%throw%the%die%a%minimum%of%one%time.%It%can%also%be%said%that%the%graph%and%the%

technology%behind%it%have%been%useful,%since%they%helped%find%a%limitation%of%the%formula%

that%otherwise%would%not%have%been%uncovered.%%

%



% 19%

We%can%then%plot%the%graph%including%the%three%points%in%the%case%that%they%both%throw%

one%die,%two%dice%and%when%Ann%is%allowed%to%throw%twice,%while%Bob%only%throws%once.%

%

%

Figure!14:!Same!graph!as!in!figure!13,!this!time!showing!the!points!that!represent!the!specific!probabilities!

that!have!been!found!above.!!

%

As%it%can%be%seen,%all%three%points%are%contained%within%the%surface,%just%as%they%should%be.%

%%

If%we%then%want%to%find%the%probability%that%Ann%wins%for%another%situation,%for%example%

when%she%throws%the%die%7%times%and%Bob%only%5%times,%we%can%do%this%by%plotting%two%

other%planes%where%! = !%and%! = !.%Afterwards%the%point%where%the%two%planes%and%the%surface% intersect% can% be% found% and% the% z7coordinate% of% that% point% will% equal% the%

probability%that%Ann%wins%in%this%case.%



% 20%

%

Figure!15:!The!z\coordinate!of! the!point!of! intersection!of! the!planes!and!the!surface!shows!the!probability!

that!Ann!wins!in!the!case!where!she!throws!the!die!7!times!and!Bob!only!5!times.!!

%

From% the% point% of% intersection% of% the% planes% and% the% surface% above,% we% get% that%

! !""!!"#$ ≈ 32%.%This% can%of% course%also%be%done%using% the% formula,%but%using% the%graph%is%quicker%once%it%has%been%set%up.%%

Finally,% we% now% want% to% consider% that% they% want% to% throw% more% than% 10% times% and%

therefore%take%a%broader%look%at%the%graph.%%

%

%

%

%

%

%



% 21%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

%

Figure!16:!Graph!showing!the!equation,!over!a!broader!range!(100!throws!for!both)!

Figure!17:!Same!equation!as!in!figure!14,!from!a!different!angle.!!



% 22%

%

%

%

%

%

%

This%situation%should%now%be%applied%to%a%casino,%where%the%player%and%the%casino%each%

throw%a%die%once.%The%casino%will%have%the%same%probability%of%winning%that%Bob%had%at%

the% beginning% (! !"#$%&!!"#$ = !"!")% and% the% player% has% a% probability% of% winning% at%

! !"#$%&!!"#$ = !"!".%%

This%means%that%if%they%play%a%near%infinite%amount%of%games,%the%casino%would%win%21%in%

every%36%games% that%were%played.%We%now%consider% the% following%with% the%help%of% an%

example:%

%

For%playing%one%game,%the%player%has%to%pay%$20%as%an%entry%fee,%which%we%will%get%back%if%

he%wins.%This%means%that%for%every%36%games,%%

%

36 ∗ $20 = $720%%

are%circulating.%If%the%player%wins,%they%will%not%only%get%their%$20%fee%back,%but%they%will%

also%make%some%profit.%Therefore,%we%divide%the%money%that%is%in%the%pot%by%the%number%

of%wins%(still%for%every%36%games):%

%

$72015 = $48%

%

Conclusion:%%

The%general% form%for% the%probability%that%Ann%wins,%where%she%and%Bob%play%a%game%

with%a%fair%die%which%is%rolled%any%number%of%times%by%both%of%them%(does%not%have%to%

be%the%same%amount)%and%only%the%highest%number%is%noted%as%a%result%is:%

%

!(!""!!"#$) = 16!!!!(!!(! − 1)! − (! − 1)!!!)

!

!!!%

%

where%!%is% the% number% of% times% that%Ann% throws% the% die% and!!%is% the% equivalent% for%Bob.%The%formula%is%only%valid%when%both%throw%the%die%a%minimum%of%one%time.%This%

can%be%displayed%on%a%graph%to%find%the%probability%even%easier,%just%as%shown%above.%%

%

Again:%

!(!"#!!"#$) = 1− !(!""!!"#$)%



% 23%

This%means% that% the%player%would%make%$28%profit% if% they%win.% In% this% case,% the%casino%

would%not%make%any%profit,%since%all%the%$720%that%has%been%paid%as%a%fee%(that%circulated)%is%given%out%to%players%again.%If%the%casino%also%wants%to%make%some%profit,%they%have%to%

pay%the%player%less%than%$28%of%profit,%which%means%that%they%can%keep%the%rest%that%was%

in%the%pot.%%

%

%

Figure!18:!Diagram!illustrating!the!cash!flow!in!the!casino!for!the!above!example.!!

%

It%should%be%considered%that%we%are%using%probabilities%and%that%this%does%not%mean%that%

it%will%always%happen%like%this.%It%might%as%well%be%that%one%time%the%casino%will%make%a%

loss%because%it%only%won%20%in%36%games,%and%another%time%the%casino%will%make%a%profit%

because% it%won%22% in%36%games.%However,%as%more%and%more%games%are%played,%reality%

will%equal%this%theoretical%model.%%

%

The%game%can%be%considered%worthwhile%for%both%parties,%since:%

%

7 The%casino%will%always%make%profit%as%long%as%enough%games%are%played%and%they%

carefully%calculate%how%much%they%can%give%the%player%as%a%reward.%In%the%example%

above,% this%would%be%$28% (if% they%pay%out%more,% they%would%make%a% loss).%How%

they%can%generally%calculate%the%pay%out%will%be%considered%later.%

7 The%player%has%got%a%reasonably%high%chance%of%winning%(~42%).%If%a%player%wins%

the%first%game%they%participated%in,%they%will%get%$48%from%the%casino.%This%means%

that%in%total%they%will%still%have%made%a%profit%of%$28.%As%this%is%higher%than%the%fee%

for%playing%one%game%($20),%they%will%probably%play%a%second%time,%since%even%if%

they%lose%this%time,%they%will%still%have%made%$8%profit.%%

%

%

%

%

Now,%a%general%formula%for%this%should%be%found%that%determines%how%much%money%the%

casino%can%pay%out%if%the%player%wins%so%that%they%just%break%even%(they%make%no%profit%or%

loss).% Therefore,%we% define% the% entry% fee% of% the% game% as% F% and% the% pay% out% (what% the%

player%gets%in%case%they%win,%including%F)%as%G.%%The%probabilities%remain%the%same.%%

PLAYER CASINO$20 fee (F) $20 every 21 in 36 games

POT

every 36 games:$720

$28 every 15 in 36 games

$20 every 15 in 36 games$48

ever

y 15

in

36

gam

es



% 24%

%

! ∗ 36 = 36!%%

This% determines% how%much%money% is% in% the% pot% (for% every% 36% games).% As% the% player%

should%theoretically%win%15%in%36%times,%we%divide%the%above%by%15%to%get%the%maximum%

pay%out,%G.%%

%

G = 36!15 = 36

15!%%

%

%

%

%

%

%

%

Now%that%this%formula%has%been%found,%we%can%also%consider%the%case%in%which%the%player%

and%the%casino%roll%the%die%multiple%times,%which%means%that%their%probability%of%winning%

changes.%To%do%this,%we%can%come%back%to%the%formula%

%

! !"#$%&!!"#$ = 16!!! (!! ! − 1 ! − ! − 1 !!!)

!

!!!%

%

where%!%is%the%number%of%times%that%the%player%throws%the%die%and%!%the%number%of%times%that% the%casino% throws% the%die.%We%say% that% the%player%wins% if%his%highest%number% in%!%

Conclusion:%%

For%the%above%game,%the%casino%can%maximally%pay%the%player:%%

%

G = 3615!%

%

where%G%is%the%maximum%pay%out%that%the%casino%can%give%the%player%when%they%paid%a%

fee%!%beforehand.% % In% this% case% the% game% can% always% be% considered% worthwhile% for%both%parties%because:%

%

7 The%probability%that%the%player%wins%the%first%game%is%relatively%high%at%42%.%If%

he%wins,%he%will%get%back%more%than%he%original%paid%as%a%fee%(!),%which%means%that%he%is%very%likely%to%play%again.%

7 As% long% as% the% casino% pays% the% player% less% than%G,% they% will% make% profit,%provided% that% enough% games% are% played% and% therefore% the% theoretical%

probability%can%be%applied%accurately%to%reality.%%



% 25%

throws%is%bigger%than%the%highest%number%of%the%casino%in%!%throws.%The%casino%wins%in%case%their%highest%number%in%!%throws%is%greater%than%or%equal%to%the%biggest%number%of%the%player%in%!%throws.%The%maximum%pay%out%G%is%then%determined%by:%%

G = 1! !"#$%&!!"#$ !%

%

This% is% since%! !"#$%&!!"#$ %gives% a% ratio% of% how%many% games% a% player%would%win% in%every% X% games.% We% want% to% multiply% the% X% by!! %to% determine% the% money% that% is%circulating%and%then%divide%it%by%the%number%of%games%that%a%player%theoretically%wins%in%

every%X%games.%This%would%determine%the%pay%out%G%when%the%casino%does%not%make%any%

profit.% In% theory,% for% the% game% to% be% considered% worthwhile,% two% things% have% to% be%

considered:%

%

1. G ≥ F%if% this% is% the% case,% then% the% player% is% very% likely% to% play% another% game,%provided%that%they%won%the%first%one.%They%will%have%more%money%than%before%and%

even%if%they%lose%the%second%game,%they%will%still%have%made%profit.%

2. G < !! !"#$%&!!"#$ !%only% in% this% situation% the% casino%will%make% profit,%which%will%

make%the%game%worthwhile%for%it.%%

%

%

%

%

END!OF!INTERNAL!ASSESSMENT!

Conclusion:%%

From%the%above%it%follows%that%%%

%

G = 1!(!"#$%&!!"#$)!%

%

Where% G% is% the%maximum% pay% out% and%!%the% initial% fee% the% player% has% to% pay%when%there%is%one%player%and%one%casino%and%%

%

!(!"#$%&!!"#$) = 16!!!!(!!(! − 1)! − (! − 1)!!!)

!

!!!%

%

For% the% game% to% be% considered%worthwhile% for% both% parties,% two% situations% have% to%

apply:%

1. G ≥ F%2. G < !

!(!"#$%&!!"#$)!%

ib mathematics hl coursework type ii – the dice game

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