Difference between revisions of "Aufgaben:Exercise 1.1: A Special Dice Game"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Mengentheoretische Grundlagen}}
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Some_Basic_Definitions}}
==A1.1 Würfelspiel Mäxchen==
 
[[File:P_ID3__Sto_A1_1.jpg|right|]]
 
Bei dem Würfelspiel Mäxchen wird jeweils mit zwei Würfeln geworfen. Die höhere Augenzahl der beiden Würfel wird mit 10 multipliziert und dann die niedrigere Augenzahl dazu addiert. Beispielsweise liefert eine „2“ und eine „4“ das Spielresultat 42 und eine „5“ und eine „6“ das Ergebnis 65. Das kleinstmögliche Resultat eines Wurfes ist somit 31.
 
  
Ein Pasch (zweimal die gleiche Augenzahl) wird im Allgemeinen höher bewertet als zwei ungleiche Würfel. So ist ein Einser-Pasch höher als 65, aber niedriger als jeder andere Pasch. Eine Sonderstellung nimmt bei diesem Spiel das Mäxchen (eine „1” und eine „2”) ein. Diese im Bild dargestellte Kombination steht noch über dem Sechser-Pasch.
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[[File:P_ID3__Sto_A1_1.jpg|right|frame|About the dice game "Max"]]
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In the dice game  "Max"  two dice are thrown.  The higher number of points of the two dice is multiplied by  $10$  and then the lower number of points is added to it.  For example,  a  "2"  and a  "4"  gives the game result  $42$  and a  "5"  and a  "6"  gives the result  $65$.  The smallest possible result of a roll is therefore  $31$.   The following rules also apply:
  
Der Spieler $X$ beginnt. Er gewinnt, wenn der Spieler $Y$ das vorgelegte Resultat nicht überbieten kann. Die weiteren vielfältigen Optionen dieses Spiels werden hier nicht berücksichtigt.
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*A  "double"  (the same number twice)  is generally valued higher than two unequal dice.
 +
*Thus,  a  "double one"  is higher than  $65$,  but lower than any other double.
 +
*A special position in this game is occupied by the  "Max"  (a  "1"  and a  "2").  This combination,  shown in the picture,  is higher than the   "double six".
  
'''Hinweis:''' Diese Aufgabe bezieht sich auf den gesamten Lehrstoff von Kapitel 1.1. Der Inhalt dieses Abschnitts ist in einem Lernvideo zusammengefasst:
 
  
===Fragebogen===
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Player  $X$  starts the game.  He wins if the player  $Y$  cannot beat the presented result.  The other multiple options of this game are not considered here.
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 +
 
 +
 
 +
 
 +
 
 +
Hints:
 +
*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Some_Basic_Definitions | Some basic definitions of probability theory]].
 +
 +
*The sample solution is summarized in a short (German language)   [[Würfelspiel_„Mäxchen”_(Lernvideo)|learning video]].
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Haben $X$ und $Y$ gleiche Gewinnchancen? Wenn Sie der Meinung sind, dass das Spiel unfair ist: Wie könnte man das Spiel fair gestalten?
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{Do&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; have equal chances of winning?&nbsp;  If you think the game is unfair:&nbsp; How could the game be made fair?
 
|type="[]"}
 
|type="[]"}
- Die Spieler $X$ und $Y$ haben gleiche Chancen.
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- Players&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; have equal chances.
+ Der Spieler $X$ ist im Vorteil.
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+ Player&nbsp; $X$&nbsp; has an advantage.
- Der Spieler $Y$ ist im Vorteil.
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- Player&nbsp; $Y$&nbsp; has an advantage.
  
{Wieviele unterschiedliche Resultate $I$ sind bei diesem Spiel möglich?
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{How many different outcomes&nbsp; $(I)$&nbsp; are possible in this game?
 
|type="{}"}
 
|type="{}"}
$I$ = { 21 }
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$I \ =\ $ { 21 }
  
{Wie groß ist die Wahrscheinlichkeit für einen Sechser-Pasch?
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{What is the probability of a&nbsp; "double six"?
 
|type="{}"}
 
|type="{}"}
$Pr[sechser-Pasch]$ = { 0.0278 3% }
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$\rm Pr\big[\text{"double six"}\big]\ =\ $ { 0.0278 3% }
  
{Wie groß ist die Wahrscheinlichkeit, dass man irgendeinen Pasch würfelt?
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{What is the probability of rolling&nbsp; "any double"?
 
|type="{}"}
 
|type="{}"}
$Pr[irgendein Pasch]$ = { 0.1667 3% }
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$\rm Pr\big[\text{"any double"}\big]\ =\ $ { 0.1667 3% }
  
{Mit welcher Wahrscheinlichkeit würfelt ein Spieler ein Mäxchen?
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{What is the probability that a player rolls a&nbsp; "Max"?
 
|type="{}"}
 
|type="{}"}
$Pr[Mäxchen]$ = { 0.0556 3% }
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$\rm Pr\big[\text{"Max"}\big]\ =\ $ { 0.0556 3% }
  
{Der Spieler $X$ hat eine „3” und eine „5” vorgelegt. Wie groß ist unter dieser Annahme die Gewinnchance von Spieler $Y$?
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{Player&nbsp; $X$&nbsp; has rolled a&nbsp; "3"&nbsp; and a&nbsp; "5".&nbsp; Under this assumption,&nbsp; what is player&nbsp; $Y$'s chance of winning??
 
|type="{}"}
 
|type="{}"}
$Pr[Y gewinnt]$ = { 0.5556 3% }Einige grundlegende Definitionen
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${\rm Pr}\big[Y\;  {\rm wins}\big]\ =\ $ { 0.5556 3% }
  
{Welches Spielergebnis $R_\min$ muss der Spieler $X$ mindestens erzielen, damit er eine größere Gewinnchance als 75% hat?
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{What is the minimum score&nbsp; $R_{\rm \min}$&nbsp; that player&nbsp; $X$&nbsp; must achieve to have a greater chance of winning than&nbsp; $75\%$?
 
|type="{}"}
 
|type="{}"}
$R_\min$ = { 65 }
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$R_{\rm \min}\ =\ $ { 65 }
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
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'''(1)'''&nbsp; <u>Player &nbsp; $X$&nbsp; has an advantage</u>,&nbsp; because player&nbsp; $Y$&nbsp; has to trump him.&nbsp;
'''2.'''
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*The game would be fair,&nbsp; for example,&nbsp; if it were scored as a&nbsp; "draw"&nbsp; when the rolls were exactly equal.&nbsp;
'''3.'''
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*However,&nbsp; over a longer period of time,&nbsp; even if&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; start alternating,&nbsp; there will be equal chances of winning.
'''4.'''
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'''5.'''
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'''6.'''
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'''7.'''
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'''(2)'''&nbsp;&nbsp; $\underline{I = 21}$&nbsp; different outcomes are possible.&nbsp; These are&nbsp; (starting with the lowest):
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:$$31, 32, 41, 42, 43, 51, 52, 53, 54, 61, 62, 63, 64, 65, 11, 22, 33, 44, 55, 66, 21.$$
 +
 
 +
 
 +
'''(3)'''&nbsp; However,&nbsp; the&nbsp; $21$&nbsp; possible results of this dice game are not equally probable,&nbsp; therefore the probabilities cannot be determined according to the classical definition&nbsp; $($the result would be&nbsp; $1/21)$.
 +
 
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*If one at least mentally makes a distinction between the dice,&nbsp; for example by a blue &nbsp;$(B)$&nbsp; and a red &nbsp;$(R)$&nbsp; one&nbsp; there are&nbsp; $6^{2} = 36$&nbsp; equally probable events,&nbsp; among others the event&nbsp; ${\rm Pr\big[\text{"double six"}\big]} = Pr\big[(B = 6) \cap (R = 6)\big]$.
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*For both dice,&nbsp; the probability of a &nbsp; "6"&nbsp; is equal to&nbsp; $1/6$.&nbsp; Since the numbers of the two dice are,&nbsp; of course,&nbsp; statistically independent,&nbsp; the following is true:
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 +
:$$Pr\big[\text{"double six"}\big] = {\rm Pr}(B = 6 \cap R = 6) = 1/36 \;\underline{= 0.0278}.$$
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 +
"Note": &nbsp; From the naming&nbsp; "result",&nbsp; one might incorrectly conclude that this is an outcome.&nbsp; However,&nbsp; according to the definitions in this chapter,&nbsp; the result is to be considered as an event&nbsp; (summary of outcomes).
 +
 
 +
 
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[[File:EN_Sto_A1_1_g.png|right|400px|Sum of two dice]]
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'''(4)'''&nbsp; This probability can be stated with&nbsp; $K = 6$&nbsp; and&nbsp; $M = 36$&nbsp; as follows:
 +
 
 +
:$${\rm Pr\big[\text{"any double"}\big] = Pr}(B = R) = K/M = 1/6 \; \underline{= 0.1667}.$$
 +
 
 +
 
 +
'''(5)'''&nbsp; Analogously,&nbsp; the probability for the&nbsp; "Max"&nbsp; can be calculated with&nbsp; $K = 2$&nbsp; and&nbsp; $M = 36$:
 +
 
 +
:$${\rm Pr\big[\text{"Max"}\big] = Pr}(B = 1 \cap R = 2) + {\rm Pr}(B = 2 \cap R = 1) = 1/18 \; \underline{= 0.0556}.$$
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 +
 
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'''(6)'''&nbsp; The probability that&nbsp; $Y$&nbsp; wins if&nbsp; $X$&nbsp; got&nbsp; "53"&nbsp; is&nbsp; $5/9\; \underline{\approx 0.555}$.
 +
 
 +
The sample solution of this subtask is summarized in a short (German language)&nbsp;  [[Würfelspiel_„Mäxchen”_(Lernvideo)|learning video]].
 +
 
 +
 
 +
'''(7)'''&nbsp; To solve the last sub-task, we again assume a two-dimensional representation and rank the matrix elements according to their values&nbsp; (see graph).&nbsp; From this we can see:
 +
*With the default value of&nbsp; $65$,&nbsp; the opponent has only a chance of winning of&nbsp; $8/36 = 0.222$.
 +
*This means that your own chance of winning is about&nbsp; $77.8\%$.
 +
*With&nbsp; $64$,&nbsp; on the other hand,&nbsp; player&nbsp; $X$'s&nbsp; chance of winning would be only about&nbsp; $72.2\%$.
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*The correct solution is therefore&nbsp; $R_\min\; \underline{ = 65}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Stochastische Signaltheorie|^1.2 Mengentheoretische Grundlagen^]]
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[[Category:Theory of Stochastic Signals: Exercises|^1.1 Some Basic Definitions^]]

Latest revision as of 17:47, 18 November 2021

About the dice game "Max"

In the dice game  "Max"  two dice are thrown.  The higher number of points of the two dice is multiplied by  $10$  and then the lower number of points is added to it.  For example,  a  "2"  and a  "4"  gives the game result  $42$  and a  "5"  and a  "6"  gives the result  $65$.  The smallest possible result of a roll is therefore  $31$.  The following rules also apply:

  • A  "double"  (the same number twice)  is generally valued higher than two unequal dice.
  • Thus,  a  "double one"  is higher than  $65$,  but lower than any other double.
  • A special position in this game is occupied by the  "Max"  (a  "1"  and a  "2").  This combination,  shown in the picture,  is higher than the  "double six".


Player  $X$  starts the game.  He wins if the player  $Y$  cannot beat the presented result.  The other multiple options of this game are not considered here.



Hints:

  • The sample solution is summarized in a short (German language)  learning video.

Questions

1

Do  $X$  and  $Y$  have equal chances of winning?  If you think the game is unfair:  How could the game be made fair?

Players  $X$  and  $Y$  have equal chances.
Player  $X$  has an advantage.
Player  $Y$  has an advantage.

2

How many different outcomes  $(I)$  are possible in this game?

$I \ =\ $

3

What is the probability of a  "double six"?

$\rm Pr\big[\text{"double six"}\big]\ =\ $

4

What is the probability of rolling  "any double"?

$\rm Pr\big[\text{"any double"}\big]\ =\ $

5

What is the probability that a player rolls a  "Max"?

$\rm Pr\big[\text{"Max"}\big]\ =\ $

6

Player  $X$  has rolled a  "3"  and a  "5".  Under this assumption,  what is player  $Y$'s chance of winning??

${\rm Pr}\big[Y\; {\rm wins}\big]\ =\ $

7

What is the minimum score  $R_{\rm \min}$  that player  $X$  must achieve to have a greater chance of winning than  $75\%$?

$R_{\rm \min}\ =\ $


Solution

(1)  Player   $X$  has an advantage,  because player  $Y$  has to trump him. 

  • The game would be fair,  for example,  if it were scored as a  "draw"  when the rolls were exactly equal. 
  • However,  over a longer period of time,  even if  $X$  and  $Y$  start alternating,  there will be equal chances of winning.


(2)   $\underline{I = 21}$  different outcomes are possible.  These are  (starting with the lowest):

$$31, 32, 41, 42, 43, 51, 52, 53, 54, 61, 62, 63, 64, 65, 11, 22, 33, 44, 55, 66, 21.$$


(3)  However,  the  $21$  possible results of this dice game are not equally probable,  therefore the probabilities cannot be determined according to the classical definition  $($the result would be  $1/21)$.

  • If one at least mentally makes a distinction between the dice,  for example by a blue  $(B)$  and a red  $(R)$  one  there are  $6^{2} = 36$  equally probable events,  among others the event  ${\rm Pr\big[\text{"double six"}\big]} = Pr\big[(B = 6) \cap (R = 6)\big]$.
  • For both dice,  the probability of a   "6"  is equal to  $1/6$.  Since the numbers of the two dice are,  of course,  statistically independent,  the following is true:
$$Pr\big[\text{"double six"}\big] = {\rm Pr}(B = 6 \cap R = 6) = 1/36 \;\underline{= 0.0278}.$$

"Note":   From the naming  "result",  one might incorrectly conclude that this is an outcome.  However,  according to the definitions in this chapter,  the result is to be considered as an event  (summary of outcomes).


Sum of two dice

(4)  This probability can be stated with  $K = 6$  and  $M = 36$  as follows:

$${\rm Pr\big[\text{"any double"}\big] = Pr}(B = R) = K/M = 1/6 \; \underline{= 0.1667}.$$


(5)  Analogously,  the probability for the  "Max"  can be calculated with  $K = 2$  and  $M = 36$:

$${\rm Pr\big[\text{"Max"}\big] = Pr}(B = 1 \cap R = 2) + {\rm Pr}(B = 2 \cap R = 1) = 1/18 \; \underline{= 0.0556}.$$


(6)  The probability that  $Y$  wins if  $X$  got  "53"  is  $5/9\; \underline{\approx 0.555}$.

The sample solution of this subtask is summarized in a short (German language)  learning video.


(7)  To solve the last sub-task, we again assume a two-dimensional representation and rank the matrix elements according to their values  (see graph).  From this we can see:

  • With the default value of  $65$,  the opponent has only a chance of winning of  $8/36 = 0.222$.
  • This means that your own chance of winning is about  $77.8\%$.
  • With  $64$,  on the other hand,  player  $X$'s  chance of winning would be only about  $72.2\%$.
  • The correct solution is therefore  $R_\min\; \underline{ = 65}$.