Chapter 13 · Probability

Example 17: Three boxes of coins: the other coin is also gold, $P = \frac23$

ExamplesBayes' theorem · Section 13.5

The question

Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Exam answer

Let $E_1$ ⁠, $E_2$ ⁠, $E_3$ ⁠: boxes I, II, III are chosen, respectively, and $A$ ⁠: the coin drawn is of gold.

$E_1$ ⁠, $E_2$ and $E_3$ form a partition of the sample space (exactly one box is chosen).

The box is chosen at random, so $P(E_1) = P(E_2) = P(E_3) = \frac13$ ⁠.

$P(A|E_1) = \frac22 = 1$ ⁠, $P(A|E_2) = 0$ and $P(A|E_3) = \frac12$ ⁠.

The other coin is also of gold exactly when box I was chosen, so we need $P(E_1|A)$ ⁠.

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac13\cdot 1 + \frac13\cdot 0 + \frac13\cdot\frac12 \\ &= \frac13 + 0 + \frac16 = \frac12 \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac13 \div \frac12 \\ &= \frac13 \times 2 \end{aligned}

$\therefore$ The probability that the other coin is also of gold is $\dfrac23$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"The other coin is gold" is the same event as "the box is box I". Box I is the only box with two gold coins, so the question really asks: given that a gold coin came out, which box was it? That is Bayes' theorem.

Box I gives a gold coin every time, box III only half the time. So a gold coin is twice as likely to have come from box I as from box III, and the answer is $\frac23$ ⁠, not $\frac12$ ⁠.

Given, to find, and the plan

Given: Box I: gold, gold. Box II: silver, silver. Box III: gold, silver. A box is chosen at random and a coin taken out; it is gold.
To find: The probability that the other coin in that box is gold.
Plan: (1) Name the events and turn the question into $P(E_1|A)$ ⁠. (2) Find $P(E_i)$ and $P(A|E_i)$ ⁠. (3) Total probability. (4) Bayes' theorem. (5) Check by listing the six equally likely coins.

Basics you need

1Probability 1 and 0

A certain event has probability 1 (a gold coin from box I); an impossible one has probability 0 (a gold coin from box II). Basics: the classical definition

2Theorem of total probability (Section 13.5.2)

$P(A) = \sum P(E_i)P(A|E_i)$ over a partition. Lesson: total probability and Bayes' theorem

3Bayes' theorem (Section 13.5)

$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)}$ ⁠. Lesson: Bayes' theorem

4Counting equally likely outcomes

Each of the 6 coins is equally likely to be the one taken out. Basics: multiply the choices

Solution

Name the events and restate the question

Let
- $E_1$ ⁠, $E_2$ ⁠, $E_3$ ⁠: boxes I, II, III are chosen, respectively;
- $A$ ⁠: the coin drawn is of gold.
$E_1$ ⁠, $E_2$ ⁠, $E_3$ form a partition of the sample space: exactly one box is chosen.

If the coin drawn is gold, the other coin is gold only in box I (in box III the other coin would be silver). So the required probability is $P(E_1|A)$ ⁠.

The question asks about "the other coin", which is not one of our events. Translating it into an event about the box is the key move: the other coin is gold exactly when the box holds two gold coins, which is box I.

Bayes' theorem finds the probability of a cause given the effect. Lesson: Bayes' theorem
The probabilities
$P(E_1) = P(E_2) = P(E_3) = \frac13.$ $\begin{aligned} P(A|E_1) &= \frac22 = 1, \\ P(A|E_2) &= 0, \\ P(A|E_3) &= \frac12. \end{aligned}$
"At random" with three identical boxes gives each box probability $\frac13$ ⁠. Inside a box both coins are equally likely to be taken: box I has 2 gold of 2 coins, box II has 0 gold, box III has 1 gold of 2.

$P = \frac{\text{favourable}}{\text{total}}$ ⁠. Basics: the classical definition
The chance of a gold coin: total probability

By the theorem of total probability,
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac13\cdot 1 + \frac13\cdot 0 + \frac13\cdot\frac12 \\ &= \frac13 + 0 + \frac16 = \frac12. \end{aligned}$

The tree for Example 17. A gold coin comes through box I (rose path, $\frac13$ ⁠) or box III (green path, $\frac16$ ⁠); box II can never give gold. The box I path is twice the box III path.

The three paths to a gold coin are mutually exclusive, so they add. $\frac13 + \frac16 = \frac26 + \frac16 = \frac36 = \frac12$ ⁠. This also makes sense: there are 3 gold and 3 silver coins in all, and every coin is equally likely to be drawn.

Multiply along a path, add the paths. Basics: tree diagrams
Box I, given gold: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac13 \div \frac12 = \frac13 \times 2 = \frac23. \end{aligned}$
The top is the box I path, $\frac13 \cdot 1 = \frac13$ ⁠; the bottom is all gold draws, $\frac12$ ⁠. Dividing by $\frac12$ is multiplying by 2. The book writes the same fraction with the three products in the denominator and gets $\frac23$ ⁠.

Dividing by a fraction is multiplying by its flip. Basics: working with fractions
Check by listing the coins

Label the coins: box I has $g_1, g_2$ ⁠; box II has $s_1, s_2$ ⁠; box III has $g_3, s_3$ ⁠. Choosing a box at random and then a coin at random makes each of the 6 coins equally likely (each has probability $\frac13 \cdot \frac12 = \frac16$ ⁠).

A gold coin was drawn, so it is $g_1$ ⁠, $g_2$ or $g_3$ ⁠, equally likely. The other coin in the box is gold for $g_1$ (the other is $g_2$ ⁠) and for $g_2$ (the other is $g_1$ ⁠), but not for $g_3$ ⁠. So the probability is $\frac23$ ⁠.

Counting the actual coins shows why the tempting answer $\frac12$ ("it is box I or box III") is wrong: the two boxes are not equally likely once a gold coin is seen, because box I has two gold coins that could have been drawn and box III only one.

Equally likely outcomes: count. Basics: multiply the choices

Result

The probability that the other coin in the box is also of gold is $\frac23$ ⁠.

(Textbook answer: $\frac{2}{3}$ ⁠.)

Check it with numbers

Picture 600 tries. Each box is chosen about 200 times. Box I gives gold all 200 times; box II never; box III gives gold about 100 times. So about $200 + 100 = 300$ gold coins are drawn, and in 200 of those tries the other coin is gold: $\frac{200}{300} \approx 0.6667$ ⁠, the same as $\frac23 \approx 0.6667$ ⁠.

Common mistakes

Answering $\frac12$ ⁠. "Box II is ruled out, so it is box I or box III, each with chance $\frac12$ " ignores that box I produces a gold coin twice as often as box III. Bayes' theorem weights each box by $P(A|E_i)$ ⁠.
Answering $P(A|E_1) = 1$ ⁠. That is the chance of gold from box I, not the chance of box I given gold.
Dropping box II from the total. Its term is $\frac13 \cdot 0 = 0$ ⁠, so leaving it out does not change the number, but write it: the theorem of total probability runs over the whole partition.

What to take away

Restate a question about "the other coin" (or any hidden detail) as a question about the cause; then use Bayes' theorem.
Seeing an effect favours the causes that make it more likely, in proportion to $P(E_i)P(A|E_i)$ ⁠.
The same idea with coins that show heads: Exercise 13.3, Question 6.

Every fact used, and where it comes from

Fact	Used in	Source
Partition of the sample space	Step 1	Section 13.5.1 (Lesson: Bayes' theorem)
Probabilities 1 and 0; $P = \frac{\text{favourable}}{\text{total}}$	Step 2	Class XI, Probability (Basics: the classical definition)
Theorem of total probability	Step 3	Section 13.5.2 (Lesson: total probability)
Bayes' theorem	Step 4	Section 13.5 (Examples 15 to 21)
Counting equally likely outcomes	Step 5	Class XI, Probability (Basics: multiply the choices)

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Example 17: Three boxes of coins: the other coin is also gold, P=23P = \frac23P=32​

Basics you need

Solution

Name the events and restate the question

The probabilities

The chance of a gold coin: total probability

Box I, given gold: Bayes' theorem

Check by listing the coins

Every fact used, and where it comes from

Example 17: Three boxes of coins: the other coin is also gold, $P = \frac23$