Demonstration 1 of 4
Weigh each outcome by its chance
A fair coin pays +2 on heads and costs 3 on tails. Is the game fair?
Each light bar is what one play can pay. Each dark bar is that payment scaled by its probability. The dark bars add to E, which is not on the list of one-play results.
\[E = (1/2)(2) + (1/2)(-3)\]
\[E = -1/2\]
E is the expected net value per play. Each outcome has a probability (1/2 for each face of a fair coin) and a net outcome in units: +2 for a gain, -3 for a loss. A contribution is probability times net outcome.
Predict first. Lower the tails loss from 3 to 2. What does E become?
Choose an example
Constructed example: the chapter's fair-coin game and its fairness adjustment.
Calculated values
- Heads
- 1/2 x (+2) = +1 unit
- Tails
- 1/2 x (-3) = -3/2 units
- Expected net value E
- -1/2 unit per play
- Probabilities add to
- 1/2 + 1/2 = 1
- Possible one-play outcomes
- +2 or -3
E = (1/2)(2) + (1/2)(-3) = 1 + (-3/2) = -1/2 unit per play, so the game is on the loss side for the player. One play still pays +2 or -3; -1/2 is a weighted average, not an outcome on the list.
Use the idea
Before calling a game, deal or rule fair, write its outcomes with signs and probabilities and add the contributions.
Where the conclusion applies
An ideal fair coin, so each face has probability 1/2, and net outcomes that already include every cost of one play. E describes the model's long-run average, not the next play; a short run can land anywhere.
Check your understanding: If heads paid +4 and tails cost 3, what would the expected net value be?
Chapter 23 source: section "Expected value is a weighted average". Demonstration C23-D01.
Demonstration 2 of 4
An average no play can produce
A gain of 4 with chance 1/4 and a loss of 2 with chance 3/4: where does E land?
The template's total divided by 4 is the probability-weighted average. Because the weights add to 1, a fee on every play lowers each bar and the dashed average line by exactly the fee.
\[E=\frac14(4)+\frac34(-2)=1-\frac32=-\frac12.\]
The four places stand for the weights: one place in four for the gain, three for the loss. A fee c is a charge on every play, so it comes off every outcome.
Predict first. Charge a fee of 1 unit on every play. How far does E move?
Choose an example
Constructed example: the chapter's worked example with gain 4 and loss 2, plus a stated fee.
Calculated values
- Fee on every play
- 0 units
- Net outcomes
- +4 with 1/4, -2 with 3/4
- Template total
- 4 + 3 x (-2) = -2
- Expected value E
- -2/4 = -1/2
- Bound check
- -2 < -1/2 < 4
E = (1/4)(4) + (3/4)(-2) = 1 + (-3/2) = -1/2. The template total is 4 + 3 x (-2) = -2, and -2/4 = -1/2. No fee is charged, as in the book. No single play produces -1/2; it sits between -2 and 4, as a weighted average must.
Use the idea
Use the bound check before trusting a computed expectation: it must sit between the smallest and largest possible net outcomes.
Where the conclusion applies
The stated probabilities 1/4 and 3/4 and a fee charged once on every play. The template shows the weights only; it does not promise that any four actual plays contain one gain. A fee on only one branch would need its own row.
Check your understanding: With a fee of 2 units on every play, what is E, and does it pass the bound check?
Chapter 23 source: section "Worked example: an average between possible outcomes". Demonstration C23-D02.
Demonstration 3 of 4
Solve for the fair loss
A game gains 9 with chance 1/4. What loss on the other branch makes it fair?
The gain branch always contributes 9/4. The loss branch contributes 3/4 of -x. Only when the two bars are the same height does the sum bar vanish.
\[(1/4)(9) + (3/4)(-x) = 0\]
\[9 - 3x = 0\]
x is the size of the loss, so the loss branch's net outcome is -x. Fair means the expected net value E equals 0. Multiplying the equation by 4 clears the fractions.
Predict first. Try a loss of 2. Is the expected value above zero or below it?
Choose an example
Constructed example: Part B of the chapter's worked examples, with other loss sizes.
Calculated values
- Gain branch
- (1/4)(9) = 9/4
- Loss branch
- (3/4)(-3) = -9/4
- Expected value E
- 0
- Fair loss from 9 - 3x = 0
- x = 3
- Verdict
- fair (E = 0)
With x = 3: (1/4)(9) + (3/4)(-3) = 9/4 - 9/4 = 0. The two branches cancel, so this loss makes the game fair. Check: 9 - 3(3) = 0.
Use the idea
When a rule is supposed to be fair, write the expected value with the unknown amount as a letter, set it to zero, and solve, instead of guessing.
Where the conclusion applies
The stated probabilities 1/4 and 3/4 and a loss size x of at least 0. Fair here means zero expected net value; it says nothing about any single play, and it does not price any real game or product.
Check your understanding: If the gain were 6 with chance 1/4, what loss would make the game fair?
Chapter 23 source: section "Worked examples: charges and fair losses". Demonstration C23-D03.
Demonstration 4 of 4
Same center, different spread
Two loss models both have E = -1. Are they the same experience?
Moving Model B's two outcomes apart by equal amounts keeps their average at -1, so both triangles stay put. What changes is how far a single play can land from the center.
\[E = 1(-1) = -1\]
\[E = (1/2)(0) + (1/2)(-2) = -1\]
Model A always loses 1 unit. Model B has two net outcomes, each with probability 1/2, the same distance above and below -1; at larger spreads the upper one is a gain. The range is the largest outcome minus the smallest.
Predict first. Set Model B's spread to 0. Can you still tell the two models apart?
Choose an example
Constructed example: the chapter's Models A and B, with Model B's spread varied.
Calculated values
- Model A
- E = 1(-1) = -1, range 0
- Model B outcomes
- -2 or 0, each 1/2
- Model B expected value
- (1/2)(0) + (1/2)(-2) = -1
- Model B range
- 0 - (-2) = 2
- Same expected value?
- yes
Model A: E = 1(-1) = -1. Model B: E = (1/2)(0) + (1/2)(-2) = 0 + (-1) = -1. The centers agree at -1, but Model B can land anywhere from -2 to 0, a range of 2, while Model A always gives -1.
Use the idea
When two choices have the same expected value, list their possible outcomes too. The average answers one question and leaves the spread question on the desk.
Where the conclusion applies
Constructed loss models with stated probabilities. Expected value compares centers only; choosing between the models would need preferences and facts this model does not contain.
Check your understanding: Model C loses 10 with probability 1/10 and 0 otherwise. What is its expected net value?
Chapter 23 source: section "Same expected value, different possible outcomes". Demonstration C23-D04.