Chapter 13 · Probability

Exercise 13.3 · Q12: Lost card is a diamond, given two diamonds are drawn, $P = \frac{11}{50}$

Exercise 13.3Bayes' theorem · Section 13.5

The question

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Exam answer

Let $E_1$ ⁠: the lost card is a diamond, $E_2$ ⁠: the lost card is not a diamond, and $A$ ⁠: the two cards drawn are both diamonds.

$E_1$ and $E_2$ form a partition of the sample space (the lost card either is a diamond or is not).

$P(E_1) = \frac{13}{52} = \frac14$ and $P(E_2) = \frac{39}{52} = \frac34$

Here ${}^{12}C_2 = 66$ ⁠, ${}^{13}C_2 = 78$ and ${}^{51}C_2 = 1275$ ⁠.

If $E_1$ occurs, 51 cards remain and 12 of them are diamonds, so

\begin{aligned} P(A|E_1) &= \frac{{}^{12}C_2}{{}^{51}C_2} = \frac{66}{1275} \\ &= \frac{22}{425} \end{aligned}

If $E_2$ occurs, 51 cards remain and 13 of them are diamonds, so

\begin{aligned} P(A|E_2) &= \frac{{}^{13}C_2}{{}^{51}C_2} = \frac{78}{1275} \\ &= \frac{26}{425} \end{aligned}

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) \\ &= \frac14\cdot\frac{22}{425} + \frac34\cdot\frac{26}{425} \\ &= \frac{22}{1700} + \frac{78}{1700} \\ &= \frac{100}{1700} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{22}{1700} \div \frac{100}{1700} \\ &= \frac{22}{1700} \times \frac{1700}{100} \end{aligned}

$\therefore$ The probability that the lost card is a diamond is $\dfrac{22}{100} = \dfrac{11}{50}$ (dividing by 2).

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

The one idea this problem is built on

The lost card changes the pack that the two cards are drawn from. If a diamond was lost, only 12 diamonds remain; if not, all 13 remain, so two diamonds are a little more likely. Seeing two diamonds is therefore weak evidence against a lost diamond. Split by the lost card, count with combinations in each case, and let Bayes' theorem turn the question round.

Given, to find, and the plan

Given: A pack of 52 cards (13 diamonds). One card is lost at random. Two cards are drawn at random from the other 51, and both are diamonds.
To find: The probability that the lost card is a diamond: $P(E_1|A)$ ⁠.
Plan: (1) Name the events; check the partition. (2) $P(E_1)$ ⁠, $P(E_2)$ ⁠. (3) $P(A|E_1)$ ⁠, $P(A|E_2)$ by combinations. (4) Total probability. (5) Bayes' theorem.

Basics you need

1A pack of cards

52 cards in 4 suits of 13: diamonds, hearts, clubs and spades. So 13 cards are diamonds and 39 are not. Basics: classical probability

2Combinations (Class XI)

The number of ways to choose 2 cards from $n$ is ${}^{n}C_2 = \frac{n(n - 1)}{2}$ ⁠; the order of the two cards does not matter. Basics: multiply the choices

3Conditional probability (Section 13.2)

$P(A|E_1)$ is the chance of two diamonds when we know the 51 remaining cards hold 12 diamonds. Lesson: conditional probability

4Total probability and Bayes' theorem (Section 13.5)

For a partition $E_1, E_2$ ⁠: $P(A) = \sum P(E_i)P(A|E_i)$ and $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)}$ ⁠. Lesson: Bayes' theorem

Solution

Name the events

Let
- $E_1$ ⁠: the lost card is a diamond;
- $E_2$ ⁠: the lost card is not a diamond;
- $A$ ⁠: the two cards drawn are both diamonds.
$E_1$ and $E_2$ form a partition of the sample space: an event and its complement never happen together, one of them always happens, and both are possible.

The lost card is the hidden cause; the two diamonds are the result we observe. Section 13.5.1 notes that an event and its complement form a partition; the condition "nonzero probabilities" holds here, since $P(E_1) = \frac14$ and $P(E_2) = \frac34$ (Step 2).

Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem
The chances for the lost card
$\begin{aligned} P(E_1) &= \frac{13}{52} = \frac14, \\ P(E_2) &= \frac{39}{52} = \frac34. \end{aligned}$
The lost card is equally likely to be any of the 52 cards, and 13 of them are diamonds.

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability
Two diamonds in each case

In both cases 51 cards remain, and the number of ways to draw 2 of them is ${}^{51}C_2 = \frac{51 \times 50}{2} = 1275$ ⁠.

If a diamond was lost ( $E_1$ ⁠), 12 diamonds remain, and there are ${}^{12}C_2 = \frac{12 \times 11}{2} = 66$ ways to draw two of them:
$P(A|E_1) = \frac{{}^{12}C_2}{{}^{51}C_2} = \frac{66}{1275} = \frac{22}{425}.$
If the lost card was not a diamond ( $E_2$ ⁠), all 13 diamonds remain, and there are ${}^{13}C_2 = \frac{13 \times 12}{2} = 78$ ways:
$P(A|E_2) = \frac{{}^{13}C_2}{{}^{51}C_2} = \frac{78}{1275} = \frac{26}{425}.$
Every pair of the 51 remaining cards is equally likely to be drawn, so the chance of two diamonds is the number of diamond pairs over the number of all pairs. Both fractions simplify by dividing by 3. The same values come from the multiplication rule, one card after the other: $\frac{12}{51} \times \frac{11}{50} = \frac{132}{2550} = \frac{22}{425}$ ⁠.

${}^{n}C_2 = \frac{n(n - 1)}{2}$ counts the pairs. Basics: multiply the choices
The chance of two diamonds

Since $E_1$ and $E_2$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac14\cdot\frac{22}{425} + \frac34\cdot\frac{26}{425} \\ &= \frac{22}{1700} + \frac{78}{1700} = \frac{100}{1700}. \end{aligned}$

The tree for Question 12. The rose path (a diamond lost, then two diamonds drawn) has product $\frac{22}{1700}$ ⁠; the green path (another card lost, then two diamonds) has $\frac{78}{1700}$ ⁠. Together, $P(A) = \frac{100}{1700}$ ⁠.

Two diamonds are drawn after exactly one kind of lost card, so the two path probabilities add. As a check, $\frac{100}{1700} = \frac{1}{17}$ ⁠, which is $\frac{13}{52} \times \frac{12}{51}$ ⁠: the lost card is unseen, so the two drawn cards behave like two cards from a full pack.

Multiply along a path, add the paths. Basics: tree diagrams
Lost card a diamond, given two diamonds: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{22}{1700} \div \frac{100}{1700} \\ &= \frac{22}{1700} \times \frac{1700}{100} \\ &= \frac{22}{100} = \frac{11}{50}. \end{aligned}$
The top is the "diamond lost" path to two diamonds; the bottom is both paths. The 1700s cancel, and dividing the top and the bottom of $\frac{22}{100}$ by 2 gives $\frac{11}{50}$ ⁠.

Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction
A quick sense check

For the other case, $P(E_2|A) = \frac{78}{100} = \frac{39}{50}$ ⁠, and $\frac{11}{50} + \frac{39}{50} = 1$ ⁠.

The answer has a neat meaning. Once two diamonds are seen, 50 cards are unseen: the lost card and the 49 left in the pack. Of these, $13 - 2 = 11$ are diamonds, and the lost card is just as likely to be any of the 50. So the chance is $\frac{11}{50}$ ⁠, a little less than the $\frac14$ we started with.

Given two diamonds, the lost card either is or is not a diamond, so the two updated probabilities must add to 1. The 50-card argument works because nothing distinguishes the lost card from the cards still in the pack.

The posterior probabilities of a partition add to 1. Lesson: Bayes' theorem

Result

The probability that the lost card is a diamond is $\frac{11}{50}$ ⁠. (Textbook answer: $\frac{11}{50}$ ⁠.)

Check it with numbers

List every way it can happen: 52 choices of lost card, then ${}^{51}C_2 = 1275$ pairs, so $52 \times 1275 = 66300$ equally likely outcomes. Two diamonds are drawn in $13 \times 66 = 858$ of them with a lost diamond and in $39 \times 78 = 3042$ with another lost card. So $858 + 3042 = 3900$ outcomes show two diamonds, and $\frac{858}{3900} = 0.22$ ⁠, the same as $\frac{11}{50} = 0.22$ ⁠. The verify script makes this list and counts it.

Common mistakes

Answering $\frac14$ ⁠. That is $P(E_1)$ ⁠, before the two diamonds are seen; the drawn diamonds make a lost diamond a little less likely.
Using 52 cards for the draw. The two cards come from the 51 that remain after one is lost.
Using 13 diamonds in both cases. If a diamond was lost, only 12 diamonds remain.
Using $\frac{13}{51} \times \frac{12}{51}$ ⁠. The cards are drawn without replacement, so the second factor is out of 50: $\frac{13}{51} \times \frac{12}{50}$ ⁠.

What to take away

A hidden first step (a lost card, a moved ball) changes the counts for the second step; split by the hidden step and count again in each case.
For draws without replacement, $\frac{{}^{k}C_2}{{}^{n}C_2}$ and $\frac{k}{n} \times \frac{k - 1}{n - 1}$ give the same value.
An unseen card is like any other unseen card: here the lost card has the same chance of being a diamond as each of the 49 cards still in the pack.

Every fact used, and where it comes from

Fact	Used in	Source
Partition of a sample space; an event and its complement	Step 1	Section 13.5.1 (Lesson: Bayes' theorem)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Steps 2, 3	Classical probability, Class XI (Basics: classical probability)
${}^{n}C_2 = \frac{n(n - 1)}{2}$	Step 3	Combinations, Class XI (Basics: multiply the choices)
Multiplication rule for draws without replacement	Steps 3, 4	Section 13.3 (Lesson: the multiplication rule)
Theorem of total probability	Step 4	Section 13.5.2 (Lesson: Bayes' theorem)
Bayes' theorem	Steps 5, 6	Section 13.5 (Examples 15 to 21)

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Exercise 13.3 · Q12: Lost card is a diamond, given two diamonds are drawn, P=1150P = \frac{11}{50}P=5011​

Basics you need

Solution

Name the events

The chances for the lost card

Two diamonds in each case

The chance of two diamonds

Lost card a diamond, given two diamonds: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q12: Lost card is a diamond, given two diamonds are drawn, $P = \frac{11}{50}$