Hard Further Maths Decision analysis Questions

Challenging, exam-style Further Maths Decision analysis questions with worked solutions. Stretch yourself on the hardest decision-analysis, decision-tree, fold-back, averaging-out problems.

decision-analysisdecision-treefold-backaveraging-outmulti-stageevpi
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A decision-maker must commit to exactly one action while the future state of the market is uncertain. A choice must be made between the actions Forge\text{Forge}, Marble\text{Marble}, Ingot\text{Ingot}, Willow\text{Willow}, Sable\text{Sable}. Exactly one of the states S1S1, S2S2, S3S3, S4S4 will occur, with probabilities P(S1)=110P(S1)=\frac{1}{10}, P(S2)=35P(S2)=\frac{3}{5}, P(S3)=15P(S3)=\frac{1}{5}, P(S4)=110P(S4)=\frac{1}{10}. The payoffs are as follows: action Forge\text{Forge} yields 2626, 100100, 185185, 99 under S1S1, S2S2, S3S3, S4S4 respectively; action Marble\text{Marble} yields 1313, −13-13, 219219, 131131 under S1S1, S2S2, S3S3, S4S4 respectively; action Ingot\text{Ingot} yields 157157, 132132, 144144, 139139 under S1S1, S2S2, S3S3, S4S4 respectively; action Willow\text{Willow} yields 8181, 186186, 136136, 180180 under S1S1, S2S2, S3S3, S4S4 respectively; action Sable\text{Sable} yields 163163, 126126, 3232, 2020 under S1S1, S2S2, S3S3, S4S4 respectively. Which action has the greatest expected monetary value?
Show worked solution

Worked solution

  1. Set out the actions, states and probabilities

    actions=5, states=4\text{actions}=5,\ \text{states}=4

    Each action is evaluated against every state of nature, weighted by the probability of that state.

  2. Find the expected monetary value of action Forge\text{Forge}

    EMV(Forge)=110×26+35×100+15×185+110×9=2012\text{EMV}(\text{Forge})=\frac{1}{10}\times 26 + \frac{3}{5}\times 100 + \frac{1}{5}\times 185 + \frac{1}{10}\times 9=\frac{201}{2}

    The EMV of an action is the probability-weighted sum of its payoffs across the states.

  3. Find the expected monetary value of action Marble\text{Marble}

    EMV(Marble)=110×13+35×(−13)+15×219+110×131=2525\text{EMV}(\text{Marble})=\frac{1}{10}\times 13 + \frac{3}{5}\times (-13) + \frac{1}{5}\times 219 + \frac{1}{10}\times 131=\frac{252}{5}

    The EMV of an action is the probability-weighted sum of its payoffs across the states.

  4. Find the expected monetary value of action Ingot\text{Ingot}

    EMV(Ingot)=110×157+35×132+15×144+110×139=6885\text{EMV}(\text{Ingot})=\frac{1}{10}\times 157 + \frac{3}{5}\times 132 + \frac{1}{5}\times 144 + \frac{1}{10}\times 139=\frac{688}{5}

    The EMV of an action is the probability-weighted sum of its payoffs across the states.

  5. Find the expected monetary value of action Willow\text{Willow}

    EMV(Willow)=110×81+35×186+15×136+110×180=164910\text{EMV}(\text{Willow})=\frac{1}{10}\times 81 + \frac{3}{5}\times 186 + \frac{1}{5}\times 136 + \frac{1}{10}\times 180=\frac{1649}{10}

    The EMV of an action is the probability-weighted sum of its payoffs across the states.

  6. Find the expected monetary value of action Sable\text{Sable}

    EMV(Sable)=110×163+35×126+15×32+110×20=100310\text{EMV}(\text{Sable})=\frac{1}{10}\times 163 + \frac{3}{5}\times 126 + \frac{1}{5}\times 32 + \frac{1}{10}\times 20=\frac{1003}{10}

    The EMV of an action is the probability-weighted sum of its payoffs across the states.

  7. Check the state probabilities total 11

    110+35+15+110=1\frac{1}{10} + \frac{3}{5} + \frac{1}{5} + \frac{1}{10}=1

    Exactly one state occurs, so the probabilities must add up to 11.

  8. List the expected monetary values of the actions

    EMV(Forge)=2012, EMV(Marble)=2525, EMV(Ingot)=6885, EMV(Willow)=164910, EMV(Sable)=100310\text{EMV}(\text{Forge})=\frac{201}{2},\ \text{EMV}(\text{Marble})=\frac{252}{5},\ \text{EMV}(\text{Ingot})=\frac{688}{5},\ \text{EMV}(\text{Willow})=\frac{1649}{10},\ \text{EMV}(\text{Sable})=\frac{1003}{10}

    Each action has one EMV; the decision-maker without information takes the largest.

  9. Identify the best action without further information

    best=Willow (EMV =164910)\text{best}=\text{Willow}\ (\text{EMV }=\frac{1649}{10})

    This is the action a decision-maker would take knowing only the probabilities.

  10. Recall the meaning of EVPI

    EVPI=EVwPI−best EMV\text{EVPI}=\text{EVwPI}-\text{best EMV}

    The value of perfect information is what a perfect forecast would add over the best decision made without it.

  11. Note that EVPI cannot be negative

    EVwPI ≥ best EMV\text{EVwPI}\ \ge\ \text{best EMV}

    Perfect information can never make the decision-maker worse off, so the EVPI is at least zero.

  12. State how many states of nature there are

    states=4\text{states}=4

    Perfect information would reveal which of these states is going to occur.

  13. Find the best action if state S1S1 were known to occur

    S1: Sable (163)S1:\ \text{Sable}\ (163)

    Given state S1S1, the decision-maker would take the action with the largest payoff in that column.

  14. Find the best action if state S2S2 were known to occur

    S2: Willow (186)S2:\ \text{Willow}\ (186)

    Given state S2S2, the decision-maker would take the action with the largest payoff in that column.

  15. Select the action of greatest expected monetary value

    recommend Willow (EMV =164910)\text{recommend }\text{Willow}\ (\text{EMV }=\frac{1649}{10})

    The action with the largest EMV is the recommended choice.

Answer
Willow\text{Willow}
Question 2
9 markschallenging
A manufacturer must commit to one option now, modelled by the decision tree set out here. The tree is set out in words as follows. The root is the decision node D1D1. At the decision node D1D1 the options are: choose Marble\text{Marble} to reach C1C1; choose Ember\text{Ember} to reach C2C2; choose Anchor\text{Anchor} to reach C3C3; choose Umber\text{Umber} to reach C4C4; choose Dynamo\text{Dynamo} to reach C5C5. At the chance node C1C1 the outcomes are: with probability 13\frac{1}{3} reach 213213; with probability 23\frac{2}{3} reach −6-6. At the chance node C2C2 the outcomes are: with probability 710\frac{7}{10} reach 202202; with probability 310\frac{3}{10} reach 134134. At the chance node C3C3 the outcomes are: with probability 25\frac{2}{5} reach 33; with probability 15\frac{1}{5} reach 172172; with probability 25\frac{2}{5} reach 127127. At the chance node C4C4 the outcomes are: with probability 13\frac{1}{3} reach 220220; with probability 13\frac{1}{3} reach 8181; with probability 13\frac{1}{3} reach 123123. At the chance node C5C5 the outcomes are: with probability 14\frac{1}{4} reach 2525; with probability 14\frac{1}{4} reach 3333; with probability 12\frac{1}{2} reach 189189. The decision-maker assigns the following utilities to the possible payoffs: U(−6)=0U(-6)=0, U(3)=13U(3)=13, U(25)=25U(25)=25, U(33)=36U(33)=36, U(81)=46U(81)=46, U(123)=55U(123)=55, U(127)=63U(127)=63, U(134)=70U(134)=70, U(172)=76U(172)=76, U(189)=81U(189)=81, U(202)=85U(202)=85, U(213)=88U(213)=88, U(220)=90U(220)=90. The decision-maker judges options not by money but by utility. The expected utility of an option is the sum, over the branches, of probability multiplied by the utility of the payoff reached. Using the utilities given, which option gives the greatest expected utility?
Show worked solution

Worked solution

  1. Set out the structure of the decision tree

    root D1, 1 decision node(s), 5 chance node(s)\text{root }D1,\ 1\text{ decision node(s)},\ 5\text{ chance node(s)}

    Square decision nodes are folded back by choosing the best option; circular chance nodes are averaged out; work from the leaves to the root.

  2. Take the expected utility at the chance node C1C1

    EU(C1)=13×88+23×0=883\text{EU}(C1)=\frac{1}{3}\times 88 + \frac{2}{3}\times 0=\frac{88}{3}

    The expected utility at C1C1 is the sum of probability times the utility of the payoff on each branch.

  3. Take the expected utility at the chance node C2C2

    EU(C2)=710×85+310×70=1612\text{EU}(C2)=\frac{7}{10}\times 85 + \frac{3}{10}\times 70=\frac{161}{2}

    The expected utility at C2C2 is the sum of probability times the utility of the payoff on each branch.

  4. Take the expected utility at the chance node C3C3

    EU(C3)=25×13+15×76+25×63=2285\text{EU}(C3)=\frac{2}{5}\times 13 + \frac{1}{5}\times 76 + \frac{2}{5}\times 63=\frac{228}{5}

    The expected utility at C3C3 is the sum of probability times the utility of the payoff on each branch.

  5. Take the expected utility at the chance node C4C4

    EU(C4)=13×90+13×46+13×55=1913\text{EU}(C4)=\frac{1}{3}\times 90 + \frac{1}{3}\times 46 + \frac{1}{3}\times 55=\frac{191}{3}

    The expected utility at C4C4 is the sum of probability times the utility of the payoff on each branch.

  6. Take the expected utility at the chance node C5C5

    EU(C5)=14×25+14×36+12×81=2234\text{EU}(C5)=\frac{1}{4}\times 25 + \frac{1}{4}\times 36 + \frac{1}{2}\times 81=\frac{223}{4}

    The expected utility at C5C5 is the sum of probability times the utility of the payoff on each branch.

  7. State the fold-back (averaging-out) method

    chance→EMV,decision→max⁡\text{chance}\to\text{EMV},\quad\text{decision}\to\max

    Chance nodes are replaced by their EMV and decision nodes by their best option value.

  8. Record the value obtained at the root

    value(D1)=9085\text{value}(D1)=\frac{908}{5}

    The root value is the expected monetary value of following the optimal decisions.

  9. Count the leaves of the tree

    leaves=13\text{leaves}=13

    Each leaf of the tree carries a single payoff.

  10. Separate the two kinds of node

    decision=1, chance=5\text{decision}=1,\ \text{chance}=5

    Square nodes are decisions the decision-maker controls; circular nodes are chance events governed by probability.

  11. Note that an EMV need not equal any actual payoff

    EMV∈Q\text{EMV}\in\mathbb{Q}

    An expected value is a weighted average, so it can be a fraction that no single outcome achieves.

  12. Confirm the direction of the calculation

    leaves → root\text{leaves}\ \to\ \text{root}

    The tree is evaluated from its leaves back towards the single root.

  13. Note that the tree has a single root

    root=D1\text{root}=D1

    Every route through the tree begins at the one root decision node.

  14. Check the probabilities at chance node C1C1 total 11

    13+23=1\frac{1}{3} + \frac{2}{3}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  15. Choose the option of greatest expected utility

    recommend Ember (EU =1612)\text{recommend }\text{Ember}\ (\text{EU }=\frac{161}{2})

    A risk-sensitive decision-maker selects the option with the largest expected utility.

Answer
Ember\text{Ember}
Question 3
9 markschallenging
A project team is choosing a plan under uncertainty, modelled by the decision tree set out here. The tree is set out in words as follows. The root is the decision node D1D1. At the decision node D1D1 the options are: choose Quarry\text{Quarry} to reach C1C1; choose Cascade\text{Cascade} to reach C2C2; choose Falcon\text{Falcon} to reach C3C3; choose Sterling\text{Sterling} to reach C4C4; choose Ember\text{Ember} to reach C5C5. At the chance node C1C1 the outcomes are: with probability 14\frac{1}{4} reach 190190; with probability 14\frac{1}{4} reach 1313; with probability 12\frac{1}{2} reach −30-30. At the chance node C2C2 the outcomes are: with probability 38\frac{3}{8} reach 2121; with probability 12\frac{1}{2} reach 163163; with probability 18\frac{1}{8} reach 5757. At the chance node C3C3 the outcomes are: with probability 13\frac{1}{3} reach 196196; with probability 16\frac{1}{6} reach 6363; with probability 12\frac{1}{2} reach −16-16. At the chance node C4C4 the outcomes are: with probability 25\frac{2}{5} reach 7979; with probability 25\frac{2}{5} reach 191191; with probability 15\frac{1}{5} reach 7070. At the chance node C5C5 the outcomes are: with probability 25\frac{2}{5} reach 6868; with probability 15\frac{1}{5} reach 176176; with probability 25\frac{2}{5} reach 6262. The decision-maker assigns the following utilities to the possible payoffs: U(−30)=0U(-30)=0, U(−16)=15U(-16)=15, U(13)=29U(13)=29, U(21)=42U(21)=42, U(57)=54U(57)=54, U(62)=65U(62)=65, U(63)=75U(63)=75, U(68)=84U(68)=84, U(70)=92U(70)=92, U(79)=99U(79)=99, U(163)=105U(163)=105, U(176)=110U(176)=110, U(190)=114U(190)=114, U(191)=117U(191)=117, U(196)=119U(196)=119. The decision-maker judges options not by money but by utility. The expected utility of an option is the sum, over the branches, of probability multiplied by the utility of the payoff reached. Using the utilities given, which option gives the greatest expected utility?
Show worked solution

Worked solution

  1. Set out the structure of the decision tree

    root D1, 1 decision node(s), 5 chance node(s)\text{root }D1,\ 1\text{ decision node(s)},\ 5\text{ chance node(s)}

    Square decision nodes are folded back by choosing the best option; circular chance nodes are averaged out; work from the leaves to the root.

  2. Take the expected utility at the chance node C1C1

    EU(C1)=14×114+14×29+12×0=1434\text{EU}(C1)=\frac{1}{4}\times 114 + \frac{1}{4}\times 29 + \frac{1}{2}\times 0=\frac{143}{4}

    The expected utility at C1C1 is the sum of probability times the utility of the payoff on each branch.

  3. Take the expected utility at the chance node C2C2

    EU(C2)=38×42+12×105+18×54=75\text{EU}(C2)=\frac{3}{8}\times 42 + \frac{1}{2}\times 105 + \frac{1}{8}\times 54=75

    The expected utility at C2C2 is the sum of probability times the utility of the payoff on each branch.

  4. Take the expected utility at the chance node C3C3

    EU(C3)=13×119+16×75+12×15=1793\text{EU}(C3)=\frac{1}{3}\times 119 + \frac{1}{6}\times 75 + \frac{1}{2}\times 15=\frac{179}{3}

    The expected utility at C3C3 is the sum of probability times the utility of the payoff on each branch.

  5. Take the expected utility at the chance node C4C4

    EU(C4)=25×99+25×117+15×92=5245\text{EU}(C4)=\frac{2}{5}\times 99 + \frac{2}{5}\times 117 + \frac{1}{5}\times 92=\frac{524}{5}

    The expected utility at C4C4 is the sum of probability times the utility of the payoff on each branch.

  6. Take the expected utility at the chance node C5C5

    EU(C5)=25×84+15×110+25×65=4085\text{EU}(C5)=\frac{2}{5}\times 84 + \frac{1}{5}\times 110 + \frac{2}{5}\times 65=\frac{408}{5}

    The expected utility at C5C5 is the sum of probability times the utility of the payoff on each branch.

  7. State the fold-back (averaging-out) method

    chance→EMV,decision→max⁡\text{chance}\to\text{EMV},\quad\text{decision}\to\max

    Chance nodes are replaced by their EMV and decision nodes by their best option value.

  8. Record the value obtained at the root

    value(D1)=122\text{value}(D1)=122

    The root value is the expected monetary value of following the optimal decisions.

  9. Count the leaves of the tree

    leaves=15\text{leaves}=15

    Each leaf of the tree carries a single payoff.

  10. Separate the two kinds of node

    decision=1, chance=5\text{decision}=1,\ \text{chance}=5

    Square nodes are decisions the decision-maker controls; circular nodes are chance events governed by probability.

  11. Note that an EMV need not equal any actual payoff

    EMV∈Q\text{EMV}\in\mathbb{Q}

    An expected value is a weighted average, so it can be a fraction that no single outcome achieves.

  12. Confirm the direction of the calculation

    leaves → root\text{leaves}\ \to\ \text{root}

    The tree is evaluated from its leaves back towards the single root.

  13. Note that the tree has a single root

    root=D1\text{root}=D1

    Every route through the tree begins at the one root decision node.

  14. Check the probabilities at chance node C1C1 total 11

    14+14+12=1\frac{1}{4} + \frac{1}{4} + \frac{1}{2}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  15. Choose the option of greatest expected utility

    recommend Sterling (EU =5245)\text{recommend }\text{Sterling}\ (\text{EU }=\frac{524}{5})

    A risk-sensitive decision-maker selects the option with the largest expected utility.

Answer
Sterling\text{Sterling}
Question 4
9 markschallenging
An investor must decide between several courses of action, modelled by the decision tree set out here. The tree is set out in words as follows. The root is the decision node D1D1. At the decision node D1D1 the options are: choose Forge\text{Forge} at a cost of 1515 to reach C1C1; choose Pioneer\text{Pioneer} at a cost of 2525 to reach C2C2; choose Alpha\text{Alpha} at a cost of 3030 to reach C3C3; choose Willow\text{Willow} at a cost of 2525 to reach C4C4; choose Beacon\text{Beacon} at a cost of 3030 to reach C5C5. At the chance node C1C1 the outcomes are: with probability 23\frac{2}{3} reach 155155; with probability 13\frac{1}{3} reach 107107. At the chance node C2C2 the outcomes are: with probability 15\frac{1}{5} reach −17-17; with probability 25\frac{2}{5} reach 9696; with probability 25\frac{2}{5} reach 196196. At the chance node C3C3 the outcomes are: with probability 12\frac{1}{2} reach 5959; with probability 14\frac{1}{4} reach 1616; with probability 14\frac{1}{4} reach 117117. At the chance node C4C4 the outcomes are: with probability 12\frac{1}{2} reach −26-26; with probability 12\frac{1}{2} reach 3030. At the chance node C5C5 the outcomes are: with probability 310\frac{3}{10} reach 185185; with probability 710\frac{7}{10} reach 201201. At a chance node the expected monetary value is the sum, over the branches, of probability multiplied by the value reached along that branch; at a decision node the value is the largest option value, where an option's value is the value it leads to minus any cost of taking it. The tree is folded back from the leaves to the root. Find the expected monetary value of choosing Willow\text{Willow}, allowing for any cost of that option.
Show worked solution

Worked solution

  1. Set out the structure of the decision tree

    root D1, 1 decision node(s), 5 chance node(s)\text{root }D1,\ 1\text{ decision node(s)},\ 5\text{ chance node(s)}

    Square decision nodes are folded back by choosing the best option; circular chance nodes are averaged out; work from the leaves to the root.

  2. Average out at the chance node C4C4

    EMV(C4)=12×(−26)+12×30=2\text{EMV}(C4)=\frac{1}{2}\times (-26) + \frac{1}{2}\times 30=2

    The EMV at C4C4 is the sum of probability times the value reached along each branch.

  3. Subtract the cost of choosing Willow\text{Willow}

    value=2−25=−23\text{value}=2-25=-23

    The cost of taking the option is subtracted from the value it leads to.

  4. State the fold-back (averaging-out) method

    chance→EMV,decision→max⁡\text{chance}\to\text{EMV},\quad\text{decision}\to\max

    Chance nodes are replaced by their EMV and decision nodes by their best option value.

  5. Record the value obtained at the root

    value(D1)=8315\text{value}(D1)=\frac{831}{5}

    The root value is the expected monetary value of following the optimal decisions.

  6. Count the leaves of the tree

    leaves=12\text{leaves}=12

    Each leaf of the tree carries a single payoff.

  7. Separate the two kinds of node

    decision=1, chance=5\text{decision}=1,\ \text{chance}=5

    Square nodes are decisions the decision-maker controls; circular nodes are chance events governed by probability.

  8. Note that an EMV need not equal any actual payoff

    EMV∈Q\text{EMV}\in\mathbb{Q}

    An expected value is a weighted average, so it can be a fraction that no single outcome achieves.

  9. Confirm the direction of the calculation

    leaves → root\text{leaves}\ \to\ \text{root}

    The tree is evaluated from its leaves back towards the single root.

  10. Note that the tree has a single root

    root=D1\text{root}=D1

    Every route through the tree begins at the one root decision node.

  11. Check the probabilities at chance node C1C1 total 11

    23+13=1\frac{2}{3} + \frac{1}{3}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  12. Check the probabilities at chance node C2C2 total 11

    15+25+25=1\frac{1}{5} + \frac{2}{5} + \frac{2}{5}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  13. Check the probabilities at chance node C3C3 total 11

    12+14+14=1\frac{1}{2} + \frac{1}{4} + \frac{1}{4}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  14. Check the probabilities at chance node C4C4 total 11

    12+12=1\frac{1}{2} + \frac{1}{2}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  15. State the expected monetary value of Willow\text{Willow}

    EMV(Willow)=−23\text{EMV}(\text{Willow})=-23

    This is the expected monetary value of that option net of its cost.

Answer
−23-23
Question 5
9 markschallenging
A company faces a decision under uncertainty, modelled by the decision tree set out here. The tree is set out in words as follows. The root is the decision node D1D1. At the decision node D1D1 the options are: choose Jasper\text{Jasper} at a cost of 3030 to reach C1C1; choose Willow\text{Willow} at a cost of 1515 to reach C2C2; choose Forge\text{Forge} at a cost of 1010 to reach C3C3; choose Alpha\text{Alpha} at a cost of 2525 to reach C4C4; choose Dynamo\text{Dynamo} at a cost of 3030 to reach C5C5. At the chance node C1C1 the outcomes are: with probability 12\frac{1}{2} reach 137137; with probability 310\frac{3}{10} reach −23-23; with probability 15\frac{1}{5} reach 8888. At the chance node C2C2 the outcomes are: with probability 35\frac{3}{5} reach 3030; with probability 25\frac{2}{5} reach 66. At the chance node C3C3 the outcomes are: with probability 15\frac{1}{5} reach −14-14; with probability 25\frac{2}{5} reach −26-26; with probability 25\frac{2}{5} reach 177177. At the chance node C4C4 the outcomes are: with probability 14\frac{1}{4} reach −10-10; with probability 34\frac{3}{4} reach 9494. At the chance node C5C5 the outcomes are: with probability 12\frac{1}{2} reach −29-29; with probability 12\frac{1}{2} reach 219219. At a chance node the expected monetary value is the sum, over the branches, of probability multiplied by the value reached along that branch; at a decision node the value is the largest option value, where an option's value is the value it leads to minus any cost of taking it. The tree is folded back from the leaves to the root. Find the expected monetary value of choosing Alpha\text{Alpha}, allowing for any cost of that option.
Show worked solution

Worked solution

  1. Set out the structure of the decision tree

    root D1, 1 decision node(s), 5 chance node(s)\text{root }D1,\ 1\text{ decision node(s)},\ 5\text{ chance node(s)}

    Square decision nodes are folded back by choosing the best option; circular chance nodes are averaged out; work from the leaves to the root.

  2. Average out at the chance node C4C4

    EMV(C4)=14×(−10)+34×94=68\text{EMV}(C4)=\frac{1}{4}\times (-10) + \frac{3}{4}\times 94=68

    The EMV at C4C4 is the sum of probability times the value reached along each branch.

  3. Subtract the cost of choosing Alpha\text{Alpha}

    value=68−25=43\text{value}=68-25=43

    The cost of taking the option is subtracted from the value it leads to.

  4. State the fold-back (averaging-out) method

    chance→EMV,decision→max⁡\text{chance}\to\text{EMV},\quad\text{decision}\to\max

    Chance nodes are replaced by their EMV and decision nodes by their best option value.

  5. Record the value obtained at the root

    value(D1)=65\text{value}(D1)=65

    The root value is the expected monetary value of following the optimal decisions.

  6. Count the leaves of the tree

    leaves=12\text{leaves}=12

    Each leaf of the tree carries a single payoff.

  7. Separate the two kinds of node

    decision=1, chance=5\text{decision}=1,\ \text{chance}=5

    Square nodes are decisions the decision-maker controls; circular nodes are chance events governed by probability.

  8. Note that an EMV need not equal any actual payoff

    EMV∈Q\text{EMV}\in\mathbb{Q}

    An expected value is a weighted average, so it can be a fraction that no single outcome achieves.

  9. Confirm the direction of the calculation

    leaves → root\text{leaves}\ \to\ \text{root}

    The tree is evaluated from its leaves back towards the single root.

  10. Note that the tree has a single root

    root=D1\text{root}=D1

    Every route through the tree begins at the one root decision node.

  11. Check the probabilities at chance node C1C1 total 11

    12+310+15=1\frac{1}{2} + \frac{3}{10} + \frac{1}{5}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  12. Check the probabilities at chance node C2C2 total 11

    35+25=1\frac{3}{5} + \frac{2}{5}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  13. Check the probabilities at chance node C3C3 total 11

    15+25+25=1\frac{1}{5} + \frac{2}{5} + \frac{2}{5}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  14. Check the probabilities at chance node C4C4 total 11

    14+34=1\frac{1}{4} + \frac{3}{4}=1

    The probabilities on the branches leaving a chance node must add up to 11.

  15. State the expected monetary value of Alpha\text{Alpha}

    EMV(Alpha)=43\text{EMV}(\text{Alpha})=43

    This is the expected monetary value of that option net of its cost.

Answer
4343

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