Probability models: equally likely or not
A spinner has three colors, but its sections are different sizes. How do you say how likely each color is, and how can you tell whether a list of chances could possibly be right?
Where you are 14 in Basic probability
Why this exists
Counting the outcomes in an event and dividing by the total is a quick way to find a probability, but it rests on a claim that is easy to miss: every outcome must be exactly as likely as every other. A fair number cube earns that claim. Plenty of chance devices do not. The spinner on the right of the figure below lands on red, blue or yellow, yet red has the smallest section and yellow the largest, so nobody should expect each color a third of the time. A coin bent in a door hinge still lands heads or tails, but there is no longer any reason to think the two are equally likely.
What is needed is a way to write down the chances for any experiment, equal or not, and a test that any such list must pass before it can be trusted. That is a probability model.
Definition
A probability model for an experiment has two parts: the sample space, which lists every possible outcome, and a probability for each of those outcomes. The probabilities must obey two rules:
- every probability is between 0 and 1, with 0 and 1 themselves allowed;
- the probabilities of all the outcomes add to exactly 1.
A model is usually written as a table, with the outcomes in one row and their probabilities in the row below. For a fair coin:
| Outcome | heads | tails |
|---|---|---|
| Probability | \displaystyle \frac{1}{2} | \displaystyle \frac{1}{2} |
P(\text{heads}) is read “the probability of heads”. With outcomes o_1, o_2, \dots, o_n, the two rules say
The probability of an event in a model is the sum of the probabilities of the outcomes that make it up.
A model is uniform when every outcome has the same probability, and non-uniform otherwise. A chance device is fair when the uniform model fits it, and unfair (or biased) when it does not. A loaded number cube is one weighted so that some faces come up more often than others.
A model that obeys the two rules is valid. Whether a valid model fits a particular device is a separate question, and it is a question about the device, not about the numbers.
Intuition
Part 1: equally likely is an assumption
Nothing in a sample space says its outcomes are equally likely. The list 1, 2, 3, 4, 5, 6 is the sample space of a fair number cube and of a loaded one alike. Equal likelihood is a claim about the device, and the usual reason for making it is symmetry. The faces of a well-made cube are the same shape and size and the cube is balanced, so nothing favors one face, and the model gives none of them more. The same reasoning covers a spinner cut into equal sections, slips of paper folded alike and mixed in a hat, and a shuffled pile of cards with identical backs.
Once a model is assumed uniform, it builds itself. With n outcomes the n equal probabilities add to 1, so each is \displaystyle \frac{1}{n}, and an event made of k of the outcomes has probability \displaystyle k \times \frac{1}{n} = \frac{k}{n}. That is the counting rule again, now with its assumption in plain view. Even a fair device needs its outcomes chosen evenly: recorded as “six” or “not six”, a fair cube gives a model that is not uniform (Example 2).
Part 2: when the outcomes are not equally likely
The two spinners in the figure have the same sample space: red, blue and yellow. The left one has three equal sections, so symmetry gives \displaystyle \frac{1}{3} for each color. The right one has no such symmetry, and its model has to come from somewhere else. There are three common sources.
Geometry. A pointer flicked hard is equally likely to stop in any direction, so each section gets its angle’s share of the full turn. A section of \theta degrees has probability \displaystyle \frac{\theta}{360}. On the right-hand spinner, red gets \displaystyle \frac{90}{360} = \frac{1}{4}.
Counting something that is uniform underneath. In a raffle each ticket is equally likely to be drawn, but a person who bought more tickets is likelier to win. A uniform model on the small pieces gives a non-uniform model on the groups they form.
Data. A bent coin or a loaded cube has no symmetry to reason from and no smaller pieces to count. Its model can only be estimated from many trials, which is the work of experimental probability.
Wherever the numbers come from, the two rules still apply. They catch a model that cannot be right, and when every probability but one is known they supply the last.
A model can also be held up against data. Over N trials an outcome with probability p should happen about pN times, so its relative frequency should sit near p. Chance always leaves some gap. Over a few trials even a large gap says little, because each single trial moves the relative frequency a long way. Over many trials the relative frequency should have settled near p, so a large gap makes the model doubtful, and a small one means the data fit it, which supports the model without proving it.
Results
The uniform model. With n equally likely outcomes, each has probability \displaystyle \frac{1}{n}, and an event made of k of them has probability \displaystyle \frac{k}{n}.
Events in any model. The probability of an event is the sum of the probabilities of its outcomes. In particular P(\text{not } A) = 1 - P(A), the complement rule.
A missing probability. If every probability but one is known, the last one is 1 minus the sum of the others.
A spinner section. A section with an angle of \theta degrees at the center has probability \displaystyle \frac{\theta}{360}, provided the pointer is equally likely to stop in every direction.
What a model predicts. In N trials, an outcome with probability p is expected about pN times.
Why it is true
Run an experiment N times. Every trial gives exactly one outcome from the sample space, so each outcome’s count is between 0 and N and the counts add to N. Dividing by N, each outcome’s share of the trials lies between 0 and 1, and the shares add to 1. A probability is the number an outcome’s share settles near as the trials go on (the law of large numbers), so probabilities obey the same two rules. An event happens exactly when one of its outcomes does, and no trial gives two outcomes at once, so an event’s count, its share and its probability are each the sum over its outcomes.
For the uniform model, n equal probabilities p add to 1, so n \times p = 1 and \displaystyle p = \frac{1}{n}.
For a spinner, cut the full turn into 360 slices of 1^\circ. If the pointer is equally likely to stop in any direction, the slices are equally likely, each with probability \displaystyle \frac{1}{360}, and a section of \theta^\circ holds \theta of them, so its probability is \displaystyle \theta \times \frac{1}{360} = \frac{\theta}{360}. An angle that is not a whole number of degrees needs a finer cut, and the same share comes out.
Worked examples
Examples 1 to 3 belong to Part 1, uniform models. Examples 4 to 7 belong to Part 2, models that are not uniform.
A fair eight-sided die
An eight-sided die is a regular octahedron with faces numbered 1 to 8. Write a probability model for the number on top after a roll, say what assumption it rests on, check that it is valid, and find the probability of rolling at least 6.
SolutionThe sample space is 1, 2, 3, 4, 5, 6, 7, 8. The eight faces are identical triangles, each placed on the solid in the same way, so if the die is well made nothing favors one face. That symmetry is the assumption, and it makes the model uniform. Eight equal probabilities that add to 1 are each \displaystyle \frac{1}{8}, so every outcome from 1 to 8 has probability \displaystyle \frac{1}{8}.
Each probability lies between 0 and 1, and \displaystyle 8 \times \frac{1}{8} = 1, so the model is valid.
“At least 6” is the event made of 6, 7 and 8. Adding their probabilities:
A chipped corner would spoil the symmetry, and with it the uniform model.
Is a uniform model justified?
For each experiment, say whether a uniform model is justified, and why.
(a) A well-made number cube is rolled and the number on top is recorded.
(b) The same cube is rolled, and the result is recorded as “six” or “not six”.
(c) A coin that has been bent in a door hinge is tossed, and heads or tails is recorded.
(d) The right-hand spinner in the figure is spun, and the color is recorded.
Solution(a) Yes. The six faces are alike in shape, size and position, so by symmetry none is favored, and each gets \displaystyle \frac{1}{6}.
(b) No, even though the cube is fair. The outcome “not six” stands for five of the six equally likely faces and “six” for one, so the model is \displaystyle P(\text{six}) = \frac{1}{6} and \displaystyle P(\text{not six}) = \frac{5}{6}. Having two outcomes does not make the chances \displaystyle \frac{1}{2} each.
(c) No. Bending broke the symmetry between the two faces. The coin may now favor heads or tails, and nothing about its shape says by how much, so its model has to be estimated from tosses.
(d) No. There are three colors, but the sections have angles 90^\circ, 120^\circ and 150^\circ, so they are not alike. The uniform model, \displaystyle \frac{1}{3} for each color, would be valid, since its numbers obey both rules, but it would not fit this spinner. Example 4 builds the model that does.
Chance cards
A board game has a pile of 16 chance cards with identical backs, shuffled before play: 6 say “move forward”, 5 say “move back”, 3 say “pay a fine” and 2 say “collect a prize”. The top card is turned over.
(a) Write a model whose outcomes are the 16 cards, and then one whose outcomes are the four kinds of card.
(b) Find the probability that the card involves money, either paying or collecting.
Solution(a) The cards look the same from the back and have been shuffled, so no card is more likely than another to end up on top. The model with the 16 cards as outcomes is uniform, with probability \displaystyle \frac{1}{16} for each card.
The four kinds are not equally likely, because they hold different numbers of cards. “Move forward” is the event made of 6 of the cards, so its probability is \displaystyle 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}. Working the same way for each kind:
| Kind | Cards | Probability |
|---|---|---|
| forward | 6 | \displaystyle \frac{6}{16} = \frac{3}{8} |
| back | 5 | \displaystyle \frac{5}{16} |
| fine | 3 | \displaystyle \frac{3}{16} |
| prize | 2 | \displaystyle \frac{2}{16} = \frac{1}{8} |
Every entry is between 0 and 1, and in sixteenths they add to \displaystyle \frac{6 + 5 + 3 + 2}{16} = \frac{16}{16} = 1, so the model is valid. It is a non-uniform model built on a uniform one underneath.
(b) The event “money” is made of the outcomes fine and prize, so its probability is \displaystyle \frac{3}{16} + \frac{2}{16} = \frac{5}{16}.
A spinner with unequal sections
Find a probability model for the right-hand spinner in the figure, which has a red section of 90^\circ, a blue section of 120^\circ and a yellow section of 150^\circ. Then find the probability of not landing on yellow.
SolutionThe sections fill the circle, since 90 + 120 + 150 = 360. Flicked hard, the pointer is equally likely to stop in any direction, so each color gets its angle’s share of 360^\circ:
Each lies between 0 and 1. In twelfths they are \displaystyle \frac{3}{12}, \displaystyle \frac{4}{12} and \displaystyle \frac{5}{12}, which add to \displaystyle \frac{12}{12} = 1, so the model is valid. It is not uniform, and it differs from the \displaystyle \frac{1}{3} each that counting the colors would suggest.
“Not yellow” is the event made of red and blue, so \displaystyle P(\text{not yellow}) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}. The complement rule agrees: \displaystyle 1 - \frac{5}{12} = \frac{7}{12}.
Which proposed models are valid?
Three models are proposed. Say whether each is valid.
(a) For a bent coin: P(\text{heads}) = 0.62 and P(\text{tails}) = 0.38.
(b) For a loaded number cube: faces 1, 2 and 3 each \displaystyle \frac{1}{5}, and faces 4, 5 and 6 each \displaystyle \frac{1}{10}.
(c) For a spinner with red, blue and yellow sections: P(\text{red}) = 0.5, P(\text{blue}) = 0.6 and P(\text{yellow}) = -0.1.
Solution(a) Valid. Both numbers are between 0 and 1, and 0.62 + 0.38 = 1. Whether it fits this particular coin is a question for tossing it, not for arithmetic.
(b) Not valid. Each number is between 0 and 1, but the total is
A roll must show some face, so the probabilities have to account for every roll, and this model leaves \displaystyle \frac{1}{10} unaccounted for.
(c) Not valid, even though 0.5 + 0.6 + (-0.1) = 1. A probability cannot be negative, since no color can come up fewer than zero times.
The two rules are separate tests: (b) passes the range test and fails the sum test, and (c) passes the sum test and fails the range test.
A loaded cube
A number cube is loaded so that 6 is three times as likely as each other face, and the faces 1 to 5 are equally likely. Find the model, and the probability of rolling an even number.
SolutionLet p be the probability of each face from 1 to 5. Then P(6) = 3p. The six probabilities must add to 1, and five of them are p:
So 8p = 1 and \displaystyle p = \frac{1}{8}. Faces 1 to 5 each have probability \displaystyle \frac{1}{8}, and \displaystyle P(6) = 3 \times \frac{1}{8} = \frac{3}{8}. All six lie between 0 and 1, and \displaystyle 5 \times \frac{1}{8} + \frac{3}{8} = \frac{8}{8} = 1, so the model is valid.
The even faces are 2, 4 and 6. Adding their probabilities:
Counting faces would give \displaystyle \frac{3}{6} = \frac{1}{2}, which is right for a fair cube and wrong for this one, because its outcomes are not equally likely.
Does the data fit?
Nesta spins the right-hand spinner in the figure 360 times and gets red 87 times, blue 124 times and yellow 149 times. Oisin had proposed the uniform model, \displaystyle \frac{1}{3} for each color. Compare Oisin’s model and the angle model from Example 4 with the data.
SolutionThe counts add to 87 + 124 + 149 = 360, as they must. Each color’s relative frequency is its count over 360, such as \displaystyle \frac{87}{360} \approx 0.24 for red. The table sets these beside each model’s probabilities, all rounded to two places:
| Color | Data | Uniform | Angles |
|---|---|---|---|
| red | 0.24 | 0.33 | 0.25 |
| blue | 0.34 | 0.33 | 0.33 |
| yellow | 0.41 | 0.33 | 0.42 |
Against the angle model every relative frequency is within about 0.01. Against the uniform model, red is 0.09 low and yellow 0.08 high. Whether a gap matters depends on how many trials lie behind it. One spin out of 360 moves a relative frequency by less than 0.003, so red’s gap of 0.09 is 33 spins’ worth. After this many spins the relative frequencies should already have settled near the true probabilities, so gaps that large make the uniform model doubtful, while the angle model fits the data well.
Fitting is not proof: the model red 0.24, blue 0.34, yellow 0.42 is also valid and fits about as well.
Boundaries
Valid is not the same as right. The two rules only test whether a list of numbers could be a probability model at all. The uniform model for the unequal spinner passes both and still does not describe that spinner.
Equal likelihood needs a reason. Symmetry in the device, or a fair way of choosing such as shuffling, justifies a uniform model. The number of outcomes alone never does.
Favorable over total works only in a uniform model. In any other model, add the probabilities of the outcomes in the event (Example 6).
Data can only fit or fail to fit. A few trials say very little, since chance alone moves a relative frequency a long way when the count is small. Many trials can make a model doubtful, but no number of trials proves a model exactly. Saying precisely how large a gap must be before it counts as evidence needs the binomial distribution and significance tests, which come much later.
The outcomes here can be listed. Every model in this lesson has a sample space that can be written out. The exact direction a pointer stops in has endlessly many possible values, each with probability 0; there probability is attached to whole sections, the idea behind geometric probability.
Where this goes next
A probability model is the starting point for nearly everything that follows in probability. Attach a number to each outcome, such as the points a chance card is worth, and the model becomes a random variable with a distribution; weighting each value by its probability gives the expected value. Comparing a model with trials grows into the law of large numbers. To design a simulation is to choose a device whose model matches a real situation. And the spinner’s angle shares extend to areas, where a target region’s share of the whole is its probability.
Exercises
Part 1: uniform models
- A fair spinner has 5 equal sections labeled A, B, C, D and E. Write its probability model, and find the probability of landing on a vowel.
- A well-made twelve-sided die has faces numbered 1 to 12. What is the probability of each outcome? Find the probability of rolling a multiple of 4.
- A bag holds 18 fruit sweets: 9 strawberry, 6 lemon and 3 lime. One is taken out without looking. Write a model whose outcomes are the flavors. Is it uniform?
- A fair number cube has its faces labeled 1, 2, 2, 3, 3, 3. Write a model for the number on top, and find the probability of an odd number.
- Romilly writes the names of the 7 members of her group on identical slips, folds them all the same way, mixes them in a hat and draws one. What is the probability that any one member is drawn? If 4 of the members play an instrument, what is the probability that the member drawn plays one?
- Say whether a uniform model is justified, and why. (a) A coin with a lump of putty stuck to one face is tossed. (b) A card is drawn from a pile of 10 shuffled cards with identical backs, and the card is recorded. (c) A fair number cube is rolled, and “less than 3” or “3 or more” is recorded. (d) A spinner with four equal sections is flicked hard.
Part 2: models that are not uniform
- A spinner has a green section of 60^\circ, an orange section of 100^\circ and a purple section of 200^\circ. Check that the sections fill the circle, and write the model.
- A spinner has P(\text{red}) = 0.25 and P(\text{blue}) = 0.4, and the rest of it is yellow. Find P(\text{yellow}) and the angle of the yellow section.
- Each list is proposed as a model for a spinner with three sections. Say whether each is valid. (a) 0.3, 0.3, 0.3 (b) \displaystyle \frac{1}{2}, \displaystyle \frac{1}{3}, \displaystyle \frac{1}{6} (c) 0.7, 0.5, -0.2 (d) 0, 0.25, 0.75
- In a raffle 40 tickets were sold: Maelor bought 12, Nesta 8, Oisin 15 and Romilly 5. One ticket is drawn. Write a model for who wins, and find the probability that Maelor or Nesta wins.
- A loaded number cube has P(1) = P(2) = P(3) = P(4) = P(5) = 0.12. Find P(6) and the probability of rolling more than 4.
- A bent coin has P(\text{heads}) = 0.58. Find P(\text{tails}), and about how many tails to expect in 300 tosses.
- On a loaded cube the three odd faces are equally likely, the three even faces are equally likely, and each even face is twice as likely as each odd face. Find the model, and the probability of rolling 2 or less.
- Quentin is making 20 chance cards and wants the kind of card drawn to follow the model forward 0.4, back 0.25, fine 0.2, and prize for the rest. Find P(\text{prize}), and how many cards of each kind he should make.
- A spinner has four sections with probabilities \displaystyle \frac{1}{3}, \displaystyle \frac{1}{4}, \displaystyle \frac{1}{6} and one unknown. Find the unknown probability and the angle of every section.
Part 2: does the data fit?
- Prudence’s cube is said to be loaded, with \displaystyle P(6) = \frac{1}{4} and each other face \displaystyle \frac{3}{20}. Check that this model is valid. She rolls the cube 400 times and gets faces 1 to 6 these numbers of times: 58, 63, 61, 57, 62, 99. Do the data fit the model? Do they fit a fair cube?
- Maelor spins the right-hand spinner in the figure 240 times: red 82, blue 79, yellow 79. Compare the relative frequencies with the angle model from Example 4. Do the data fit it? Which model do they look more like?
- Oisin tosses a bent coin 50 times and gets 29 heads. Compare the relative frequency of heads with the fair model. Does that show the coin is unfair?
- Quentin rolls a cube 12 times and never gets a 6, and says the cube must be loaded against 6. How many sixes does the fair model predict in 12 rolls, and do his data show the cube is loaded?
Reasoning
- Explain why the probabilities in any probability model must add to 1.
- Prudence says, “My spinner has three colors, so each has probability \displaystyle \frac{1}{3}.” Explain what she is assuming, and describe a spinner for which she would be wrong.
- A model passes both rules. Explain, with an example, why that does not show it is the right model for a particular device.
Answers
- A, B, C, D and E each have probability \displaystyle \frac{1}{5}, since the five sections are alike. The vowels are A and E: \displaystyle P(\text{vowel}) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}.
- Each outcome has probability \displaystyle \frac{1}{12}. The multiples of 4 are 4, 8 and 12, so the probability is \displaystyle \frac{3}{12} = \frac{1}{4}.
- \displaystyle P(\text{strawberry}) = \frac{9}{18} = \frac{1}{2}, \displaystyle P(\text{lemon}) = \frac{6}{18} = \frac{1}{3}, \displaystyle P(\text{lime}) = \frac{3}{18} = \frac{1}{6}. It is not uniform: the 18 single sweets are equally likely, but the flavors group them unequally.
- \displaystyle P(1) = \frac{1}{6}, \displaystyle P(2) = \frac{2}{6} = \frac{1}{3}, \displaystyle P(3) = \frac{3}{6} = \frac{1}{2}. The odd numbers are 1 and 3: \displaystyle P(\text{odd}) = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}.
- Each member has probability \displaystyle \frac{1}{7}, since the slips are alike and mixed. The 4 who play make an event of 4 equally likely outcomes, so the probability is \displaystyle \frac{4}{7}.
- (a) No: the putty makes one face heavier, breaking the symmetry. (b) Yes: identical backs and shuffling give each card \displaystyle \frac{1}{10}. (c) No: “less than 3” covers 2 faces and “3 or more” covers 4, so the model is \displaystyle \frac{1}{3} and \displaystyle \frac{2}{3}. (d) Yes: four equal sections and a hard flick give \displaystyle \frac{1}{4} each.
- 60 + 100 + 200 = 360. \displaystyle P(\text{green}) = \frac{60}{360} = \frac{1}{6}, \displaystyle P(\text{orange}) = \frac{100}{360} = \frac{5}{18}, \displaystyle P(\text{purple}) = \frac{200}{360} = \frac{5}{9}; in eighteenths, 3 + 5 + 10 = 18.
- P(\text{yellow}) = 1 - 0.25 - 0.4 = 0.35. The yellow angle is 0.35 \times 360 = 126^\circ.
- (a) Not valid: the total is 0.9. (b) Valid: each is between 0 and 1, and \displaystyle \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = 1. (c) Not valid: -0.2 is negative. (d) Valid: 0 + 0.25 + 0.75 = 1, and 0 is allowed; that section never comes up.
- Each ticket is equally likely, so each person’s share of the tickets is their probability: \displaystyle P(\text{Maelor}) = \frac{12}{40} = \frac{3}{10}, \displaystyle P(\text{Nesta}) = \frac{8}{40} = \frac{1}{5}, \displaystyle P(\text{Oisin}) = \frac{15}{40} = \frac{3}{8}, \displaystyle P(\text{Romilly}) = \frac{5}{40} = \frac{1}{8}. Maelor or Nesta: \displaystyle \frac{12 + 8}{40} = \frac{20}{40} = \frac{1}{2}.
- P(6) = 1 - 5 \times 0.12 = 0.4. More than 4 means 5 or 6: 0.12 + 0.4 = 0.52.
- P(\text{tails}) = 1 - 0.58 = 0.42. In 300 tosses expect about 0.42 \times 300 = 126 tails.
- Each odd face has probability p and each even face 2p, so 3p + 6p = 1 and \displaystyle p = \frac{1}{9}. Faces 1, 3, 5 each \displaystyle \frac{1}{9}; faces 2, 4, 6 each \displaystyle \frac{2}{9}. \displaystyle P(2 \text{ or less}) = \frac{1}{9} + \frac{2}{9} = \frac{1}{3}.
- P(\text{prize}) = 1 - 0.4 - 0.25 - 0.2 = 0.15. Cards: 0.4 \times 20 = 8 forward, 0.25 \times 20 = 5 back, 0.2 \times 20 = 4 fine and 0.15 \times 20 = 3 prize.
- The unknown is \displaystyle 1 - \frac{1}{3} - \frac{1}{4} - \frac{1}{6} = \frac{12 - 4 - 3 - 2}{12} = \frac{1}{4}. The angles are 120^\circ, 90^\circ, 60^\circ and 90^\circ (each probability times 360), which add to 360^\circ.
- Valid: every number is between 0 and 1, and \displaystyle 5 \times \frac{3}{20} + \frac{1}{4} = \frac{15}{20} + \frac{5}{20} = 1. Faces 1 to 5 have relative frequencies from \displaystyle \frac{57}{400} \approx 0.14 to \displaystyle \frac{63}{400} \approx 0.16, all within 0.01 of 0.15, and six has \displaystyle \frac{99}{400} \approx 0.25. Over 400 rolls that is a close fit. A fair cube gives six \displaystyle \frac{1}{6} \approx 0.17, a gap of 0.08 over many rolls, so the data do not fit a fair cube.
- No. Red’s relative frequency is \displaystyle \frac{82}{240} \approx 0.34 against 0.25 in the angle model, and yellow’s is \displaystyle \frac{79}{240} \approx 0.33 against \displaystyle \frac{5}{12} \approx 0.42: gaps of about 0.09 over 240 spins, too large for that many trials. All three are within 0.01 of \displaystyle \frac{1}{3}, so the data look like the uniform model, as if the sections were equal. The data make the angle model doubtful; they do not say why it fails.
- No. The relative frequency is \displaystyle \frac{29}{50} = 0.58 against 0.5; the fair model predicts \displaystyle \frac{1}{2} \times 50 = 25 heads, so the gap is 4 tosses. With only 50 tosses each toss moves the relative frequency by 0.02, and a gap of a few tosses is ordinary. The data fit a fair coin and also a coin that slightly favors heads; 50 tosses cannot tell them apart.
- The fair model predicts \displaystyle \frac{1}{6} \times 12 = 2 sixes, and he got 0. With 12 rolls each roll moves the relative frequency by \displaystyle \frac{1}{12} \approx 0.08, so the gap is just two rolls’ worth and says little. His data do not show the cube is loaded; far more rolls are needed.
- Every trial gives exactly one outcome, so in N trials the counts add to N and the shares add to 1. The probabilities are the numbers those shares settle near, so they add to 1 too. Put another way, the event “one of the outcomes happens” is certain.
- She assumes the colors are equally likely, which needs three equal 120^\circ sections and a fair pointer. The right-hand spinner in the figure has three colors with sections of 90^\circ, 120^\circ and 150^\circ, so its colors have probabilities \displaystyle \frac{1}{4}, \displaystyle \frac{1}{3} and \displaystyle \frac{5}{12}.
- The rules check only the numbers, not the device. For the right-hand spinner in the figure, \displaystyle \frac{1}{3} for each color passes both rules, yet the spinner gives \displaystyle \frac{1}{4}, \displaystyle \frac{1}{3} and \displaystyle \frac{5}{12}. Whether a model fits a device is settled by its symmetry or by data from it.