Group properties and structure

3 Rebecca and Claire repeatedly play a zero-sum game in which they each have a choice of three strategies, \(X , Y\) and \(Z\). The table shows the number of points Rebecca scores for each pair of strategies. \begin{table}[h] \end{table}

1. If both players choose strategy \(X\), how many points will Claire score?
2. Show that row \(X\) does not dominate row \(Y\) and that column \(Y\) does not dominate column \(Z\).
3. Find the play-safe strategies. State which strategy is best for Claire if she knows that Rebecca will play safe.
4. Explain why decreasing the value ' 5 ' when both players choose strategy \(X\) cannot alter the playsafe strategies.

5 The local rugby club has challenged the local cricket club to a chess match to raise money for charity. Each of the top three chess players from the rugby club has played 10 chess games against each of the top three chess players from the cricket club. There were no drawn games. The table shows, for each pairing, the number of games won by the player from the rugby club minus the number of games won by the player from the cricket club. This will be called the score; the scores make a zero-sum game.

1. How many of the 10 games between Sanjeev and Doug did Sanjeev win? How many of the 10 games between Sanjeev and Euan did Euan win? Each club must choose one person to play. They want to choose the person who will optimise the score.
2. Find the play-safe choice for each club, showing your working. Explain how you know that the game is not stable.
3. Which person should the cricket club choose if they know that the rugby club will play-safe and which person should the rugby club choose if they know that the cricket club will play-safe?
4. Explain why the rugby club should not choose Tom. Which player should the cricket club not choose, and why? The rugby club chooses its player by using random numbers to choose between Sanjeev and Ursula, where the probability of choosing Sanjeev is \(p\) and the probability of choosing Ursula is \(1 - p\).
5. Write down an expression for the expected score for the rugby club for each of the two remaining choices that can be made by the cricket club. Calculate the optimal value for \(p\). Doug is studying AS Mathematics. He removes the row representing Tom and then models the cricket club's problem as the following LP. $$\begin{array} { l l } \operatorname { maximise } & M = m - 4 \\ \text { subject to } & m \leqslant 4 x \quad + 6 z \\ & m \leqslant 2 x + 10 y + 4 z \\ & x + y + z \leqslant 1 \\ \text { and } & m \geqslant 0 , x \geqslant 0 , y \geqslant 0 , z \geqslant 0 \end{array}$$
6. Show how Doug used the values in the table to get the constraints \(m \leqslant 4 x + 6 z\) and \(m \leqslant 2 x + 10 y + 4 z\). Doug uses the Simplex algorithm to solve the LP problem. His solution has \(x = 0\) and \(y = \frac { 1 } { 6 }\).
7. Calculate the optimal value of \(M\).

5 Robbie received a new computer game for Christmas. He has already worked through several levels of the game but is now stuck at one of the levels in which he is playing against a character called Conan. Robbie has played this particular level several times. Each time Robbie encounters Conan he can choose to be helped by a dwarf, an elf or a fairy. Conan chooses between being helped by a goblin, a hag or an imp. The players make their choices simultaneously, without knowing what the other has chosen. Robbie starts the level with ten gold coins. The table shows the number of gold coins that Conan must give Robbie in each encounter for each combination of helpers (a negative entry means that Robbie gives gold coins to Conan). If Robbie's total reaches twenty gold coins then he completes the level, but if it reaches zero the game ends. This means that each attempt can be regarded as a zero-sum game.

1. Find the play-safe choice for each player, showing your working. Which helper should Robbie choose if he thinks that Conan will play-safe?
2. How many gold coins can Robbie expect to win, with each choice of helper, if he thinks that Conan will choose randomly between his three choices (so that each has probability \(\frac { 1 } { 3 }\) )? Robbie decides to choose his helper by using random numbers to choose between the elf and the fairy, where the probability of choosing the elf is \(p\) and the probability of choosing the fairy is \(1 - p\).
3. Write down an expression for the expected number of gold coins won at each encounter by Robbie for each of Conan's choices. Calculate the optimal value of \(p\). Robbie's girlfriend thinks that he should have included the possibility of choosing the dwarf. She denotes the probability with which Robbie should choose the dwarf, the elf and the fairy as \(x , y\) and \(z\) respectively. She then models the problem of choosing between the three helpers as the following LP. $$\begin{aligned} \text { Maximise } & M = m - 4 , \\ \text { subject to } & m \leqslant 3 x + 7 y + 5 z \\ & m \leqslant 5 y + 3 z \\ & m \leqslant 6 x + 5 z \\ & x + y + z \leqslant 1 , \\ \text { and } & m \geqslant 0 , x \geqslant 0 , y \geqslant 0 , z \geqslant 0 . \end{aligned}$$
4. Explain how the expression \(3 x + 7 y + 5 z\) was formed. Robbie's girlfriend uses the Simplex algorithm to solve the LP problem. Her solution has \(x = 0\) and \(y = \frac { 2 } { 7 }\).
5. Calculate the optimal value of \(M\).

5 Rose and Colin are playing a game in which they each have four cards. Each player chooses a card from those in their hand, and simultaneously they show each other the cards they have chosen. The table below shows how many points Rose wins for each combination of cards. In each case the number of points that Colin wins is the negative of the entry in the table. Both Rose and Colin are trying to win as many points as possible.

1. What is the greatest number of points that Colin can win when Rose chooses and which card does Colin need to choose to achieve this?
2. Explain why Rose should never choose and find the card that Colin should never choose. Hence reduce the game to a \(3 \times 3\) pay-off matrix.
3. Find the play-safe strategy for each player on the reduced game and show whether or not the game is stable. Rose makes a random choice between her cards, choosing with probability \(x\) with probability \(y\), and with probability \(z\). She formulates the following LP problem to be solved using the Simplex algorithm:
   maximise \(\quad M = m - 6\),
   subject to \(\quad m \leqslant 4 x + 10 y\),
   \(n \leqslant 9 x + 3 y + 11 z\),
   \(n \leqslant 2 x + 10 y + z\),
   \(x + y + z \leqslant 1\),
   and
   \(x \geqslant 0 , y \geqslant 0 , z \geqslant 0 , m \geqslant 0\).
   (You are not required to solve this problem.)
4. Explain how \(9 x + 3 y + 11 z\) was obtained. The Simplex algorithm is used to solve the LP problem. The solution has \(x = \frac { 7 } { 48 } , y = \frac { 27 } { 48 } , z = \frac { 14 } { 48 }\).
5. Calculate the optimal value of \(M\).

4 A group of rowers have challenged some cyclists to see which team is fitter. There will be several rounds to the challenge. In each round, the rowers and the cyclists each choose a team member and these two compete in a series of gym exercises. The time by which the winner finishes ahead of the loser is converted into points. These points are added to the score for the winner's team and taken off the score for the loser's team. The table shows the expected number of points added to the score for the rowers for each combination of competitors. Rowers \begin{table}[h] \end{table}

1. Regarding this as a zero-sum game, find the play-safe strategy for the rowers and the play-safe strategy for the cyclists. Show that the game is stable. Unfortunately, Wendy and Kath are needed by their coaches and cannot compete.
2. Show that the resulting reduced game is unstable.
3. Suppose that the cyclists are equally likely to choose Chris or Jamie. Calculate the expected number of points added to the score for the rowers when they choose Andy and when they choose Zac. Suppose that the cyclists use random numbers to choose between Chris and Jamie, so that Chris is chosen with probability \(p\) and Jamie with probability \(1 - p\).
4. Showing all your working, calculate the optimum value of \(p\) for the cyclists.
5. The rowers use random numbers in a similar way to choose between Andy and Zac, so that Andy is chosen with probability \(q\) and Zac with probability \(1 - q\). Calculate the optimum value of \(q\).

4 Ross and Collwen are playing a game in which each secretly chooses a magic spell. They then reveal their choices, and work out their scores using the tables below. Ross and Collwen are both trying to get as large a score as possible.

1. Explain how this can be rewritten as the following zero-sum game.

   		Collwen's choice
   		Fire	Ice	Gale
   \cline { 2 - 5 }	Fire	- 3	3	- 2
   \cline { 2 - 5 } Ross's choice	Ice	2	- 2	0
   \cline { 2 - 5 }	Gale	1	- 3	- 1
   \cline { 2 - 5 }

2. If Ross chooses Ice what is Collwen's best choice?
3. Find the play-safe strategy for Ross and the play-safe strategy for Collwen, showing your working. Explain how you know that the game is unstable.
4. Show that none of Collwen's strategies dominates any other.
5. Explain why Ross would never choose Gale, hence reduce the game to a \(2 \times 3\) zero-sum game, showing the pay-offs for Ross. Suppose that Ross uses random numbers to choose between Fire and Ice, choosing Fire with probability \(p\) and Ice with probability \(1 - p\).
6. Use a graphical method to find the optimal value of \(p\) for Ross.

6 At the final battle in a war game, the two opposing armies, led by General Rose, \(R\), and Colonel Cole, \(C\), are facing each other across a wide river. Each army consists of four divisions. The commander of each army needs to send some of his troops North and send the rest South. Each commander has to decide how many divisions (1,2 or 3) to send North. Intelligence information is available on the number of thousands of soldiers that each army can expect to have remaining with each combination of strategies. Thousands of soldiers remaining in \(R\) 's army \(C\) 's choice
\(R\) 's choice Thousands of soldiers remaining in \(C\) 's army
\(C\) 's choice
\(R\) 's choice

1. Construct a table to show the number of thousands of soldiers remaining in \(R\) 's army minus the number of thousands of soldiers remaining in \(C\) 's army (the excess for \(R\) 's army) for each combination of strategies. The commander whose army has the greatest positive excess of soldiers remaining at the end of the game will be declared the winner.
2. Explain the meaning of the value in the top left cell of your table from part (i) (where each commander chooses strategy 1). Hence explain why this table may be regarded as representing a zero-sum game.
3. Find the play-safe strategy for \(R\) and the play-safe strategy for \(C\). If \(C\) knows that \(R\) will choose his play-safe strategy, which strategy should \(C\) choose? One of the strategies is redundant for one of the commanders, because of dominance.
4. Draw a table for the reduced game, once the redundant strategy has been removed. Label the rows and columns to show how many divisions have been sent North. A mixed strategy is to be employed on the resulting reduced game. This leads to the following LP problem:
   Maximise \(\quad M = m - 25\)
   Subject to \(\quad m \leqslant 15 x + 25 y + 35 z\)
   \(m \leqslant 45 x + 20 y\)
   \(x + y + z \leqslant 1\)
   and
5. Interpret what \(x , y\) and \(z\) represent and show how \(m \leqslant 15 x + 25 y + 35 z\) was formed. A computer runs the Simplex algorithm to solve this problem. It gives \(x = 0.5385 , y = 0\) and \(z = 0.4615\).
6. Describe how this solution should be interpreted, in terms of how General Rose chooses where to send his troops. Calculate the optimal value for \(M\) and explain its meaning. Elizabeth does not have access to a computer. She says that at the solution to the LP problem \(15 x + 25 y + 35 z\) must equal \(45 x + 20 y\) and \(x + y + z\) must equal 1 . This simplifies to give \(y + 7 z = 6 x\) and \(x + y + z = 1\).
7. Explain why there can be no valid solution of \(y + 7 z = 6 x\) and \(x + y + z = 1\) with \(x = 0\). Elizabeth tries \(z = 0\) and gets the solution \(x = \frac { 1 } { 7 }\) and \(y = \frac { 6 } { 7 }\).
8. Explain why this is not a solution to the LP problem.

4 Rowan and Colin are playing a game of 'scissors-paper-rock'. In each round of this game, each player chooses one of scissors ( \(\$$ ), paper ( \)\square\( ) or rock ( \)\bullet$ ). The players reveal their choices simultaneously, using coded hand signals. Rowan and Colin will play a large number of rounds. At the end of the game the player with the greater total score is the winner. The rules of the game are that scissors wins over paper, paper wins over rock and rock wins over scissors. In this version of the game, if a player chooses scissors they will score \(+ 1,0\) or - 1 points, according to whether they win, draw or lose, but if they choose paper or rock they will score \(+ 2,0\) or - 2 points. This gives the following pay-off tables.
\includegraphics[max width=\textwidth, alt={}, center]{490ff276-6639-40a1-bffb-dc6967f3ab21-5_476_773_667_239}
\includegraphics[max width=\textwidth, alt={}, center]{490ff276-6639-40a1-bffb-dc6967f3ab21-5_478_780_667_1071}

1. Use an example to show that this is not a zero-sum game.
2. Write down the minimum number of points that Rowan can win using each strategy. Hence find the strategy that maximises the minimum number of points that Rowan can win. Rowan decides to use random numbers to choose between the three strategies, choosing scissors with probability \(p\), paper with probability \(q\) and rock with probability \(( 1 - p - q )\).
3. Find and simplify, in terms of \(p\) and \(q\), expressions for the expected number of points won by Rowan for each of Colin's possible choices. Rowan wants his expected winnings to be the same for all three of Colin's possible choices.
4. Calculate the probability with which Rowan should play each strategy.

1 The switching circuit in Fig. 1.1 shows switches, \(s\) for a car's sidelights, \(h\) for its dipped headlights and f for its high-intensity rear foglights. It also shows the three sets of lights. \begin{figure}[h] \end{figure} (Note: \(s\) and \(h\) are each "ganged" switches. A ganged switch consists of two connected switches sharing a single switch control, so that both are either on or off together.)

1. 1. Describe in words the conditions under which the foglights will come on. Fig. 1.2 shows a combinatorial circuit. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{ab28be76-9329-41c8-90fe-ff1bdb28f788-2_367_1235_1183_431} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
      \end{figure}
   2. Write the output in terms of a Boolean expression involving \(s , h\) and \(f\).
   3. Use a truth table to prove that \(\mathrm { s } \wedge \mathrm { h } \wedge \mathrm { f } = \sim ( \sim \mathrm { s } \vee \sim \mathrm { h } ) \wedge \mathrm { f }\).
2. A car's first gear can be engaged ( g ) if either both the road speed is low ( r ) and the clutch is depressed ( d ), or if both the road speed is low ( r ) and the engine speed is the correct multiple of the road speed (m).
   1. Draw a switching circuit to represent the conditions under which first gear can be engaged. Use two ganged switches to represent \(r\), and single switches to represent each of \(\mathrm { d } , \mathrm { m }\) and g .
   2. Draw a combinatorial circuit to represent the Boolean expression \(\mathrm { r } \wedge ( \mathrm { d } \vee \mathrm { m } ) \wedge \mathrm { g }\).
   3. Use Boolean algebra to prove that \(\mathrm { r } \wedge ( \mathrm { d } \vee \mathrm { m } ) \wedge \mathrm { g } = ( ( \mathrm { r } \wedge \mathrm { d } ) \vee ( \mathrm { r } \wedge \mathrm { m } ) ) \wedge \mathrm { g }\).
   4. Draw another switching circuit to represent the conditions under which first gear can be selected, but without using a ganged switch.

1

1. Use a truth table to prove \(\sim ( \sim \mathrm { T } \Rightarrow \sim \mathrm { S } ) \Leftrightarrow ( \sim \mathrm { T } \wedge \mathrm { S } )\).
2. Prove that \(( \mathrm { A } \Rightarrow \mathrm { B } ) \Leftrightarrow ( \sim \mathrm { A } \vee \mathrm { B } )\) and hence use Boolean algebra to prove that $$\sim ( \sim \mathrm { T } \Rightarrow \sim \mathrm {~S} ) \Leftrightarrow ( \sim \mathrm { T } \wedge \mathrm {~S} ) .$$
3. A teacher wrote on a report "It is not the case that if Joanna doesn't try then she won't succeed." He meant to say that if Joanna were to try then she would have a chance of success. By letting T be "Joanna will try" and S be "Joanna will succeed", find the real meaning of what the teacher wrote.

1

1. The Plain English Society presents an annual "Foot in Mouth" award for "a truly baffling comment". In 2004 it was presented to Boris Johnson MP for a comment on the \(12 ^ { \text {th } }\) December 2003 edition of "Have I Got News For You":
   "I could not fail to disagree with you less."
   1. Explain why this can be rewritten as:
      "I could succeed in agreeing with you more."
   2. Rewrite the comment more simply in your own words without changing its meaning.
2. Two switches are to be wired between a mains electricity supply and a light so that when the state of either switch is changed the state of the light changes (i.e. from off to on, or from on to off). Draw a switching circuit to achieve this. The switches are both 2-way switches, thus:
   \includegraphics[max width=\textwidth, alt={}, center]{88acde67-e22b-478a-9145-48abe931beff-2_127_220_895_1054}
3. Construct a truth table to show the following. $$[ ( a \wedge b ) \vee ( ( \sim a ) \wedge ( \sim b ) ) ] \Leftrightarrow [ ( ( \sim a ) \vee b ) \wedge ( a \vee ( \sim b ) ) ]$$

1

1. The following was said in a charity appeal on Radio 4 in October 2006.
   "It is hard to underestimate the effect that your contribution will make."
   Rewrite the comment more simply in your own words without changing its meaning.
2. A machine has three components, A, B and C, each of which is either active or inactive.
   • The machine is active if A and B are both active.
   • The machine is active if A is inactive and C is active.
   • The machine is active if B is inactive and C is active.
   • Otherwise the machine is inactive.
   The states (active or inactive) of the components and the machine are to be modelled by a combinatorial circuit in which "active" is represented by "true" and "inactive" is represented by "false". Draw such a circuit.
3. Construct a truth table to show the following. $$[ ( ( \mathrm { a } \wedge \mathrm {~b} ) \vee ( ( \sim \mathrm { a } ) \wedge \mathrm { c } ) ) \vee ( ( \sim \mathrm { b } ) \wedge \mathrm { c } ) ] \Leftrightarrow \sim [ ( ( \sim \mathrm { a } ) \wedge ( \sim \mathrm { c } ) ) \vee ( ( \sim \mathrm { b } ) \wedge ( \sim \mathrm { c } ) ) ]$$

1

1. Heard in Parliament: "Will the minister not now discontinue her proposal to ban the protest?"
   The minister replied "Yes I will."
   To what had the minister committed herself logically, and why might that not have been her intention?
2. In a cricket tournament an umpire might be required to decide whether or not a batsman is out 'lbw', ie 'leg before wicket'. The lbw law for the tournament refers to parts of the cricket pitch as shown in the diagram (assuming a right-handed batsman):
   \includegraphics[max width=\textwidth, alt={}, center]{52b8153f-e655-4852-a0f8-6f1c1e9c9170-2_254_1045_717_507} The umpire has to make a number of judgements:
   A Would the ball have hit the wicket?
   B Did the ball hit the batsman, or part of his equipment other than the bat, without hitting the bat?
   C Did the ball hit the batsman, or part of his equipment other than the bat, before hitting the bat?
   D Was the part of the batsman or his equipment which was hit by the ball, between the wickets when it was hit? E Was the part of the batsman or his equipment which was hit by the ball, outside of the wicket on the off side when it was hit? F Was the batsman attempting to play a stroke?
   The law can be interpreted as saying that the batsman is out lbw if \([ ( \mathrm { A } \wedge \mathrm { B } ) \vee ( \mathrm { A } \wedge \mathrm { C } ) ] \wedge [ \mathrm { D } \vee ( \mathrm { E } \wedge \sim \mathrm { F } ) ]\).
   The tournament's umpiring manual, in attempting to simplify the law, states that the batsman is out lbw if \(\mathrm { A } \wedge ( \mathrm { B } \vee \mathrm { C } ) \wedge ( \mathrm { D } \vee \mathrm { E } ) \wedge ( \mathrm { D } \vee \sim \mathrm { F } )\). For an lbw decision this requires 4 conditions each to be true.
   1. Use the rules of Boolean algebra to show that the manual's rule is logically equivalent to the law as stated above, naming the rules used at each step. A trainee umpire, using the manual, considers each condition in turn and judges that the following are true: A; B; E; D.
   2. What is her decision and why?
   3. What is odd about her judgement, and does this make the logic invalid?