Chapter 13 · Probability

Example 23: A and B throw a die in turn until a 6: $\frac{6}{11}$ and $\frac{5}{11}$

ExamplesIndependent events and an infinite G.P. · Miscellaneous Examples

The question

A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first.

(The book prints "alternatively"; it means "alternately", that is, taking turns.)

Exam answer

Let $S$ ⁠: a throw gives a six, and $F$ ⁠: a throw does not give a six. Then $P(S) = \frac16$ and $P(F) = \frac56$ ⁠.

A throws 1st, 3rd, 5th, … and B throws 2nd, 4th, …; the throws are independent.

A wins at the 1st throw with probability $P(S) = \frac16$ ⁠. A wins at the 3rd throw if the first two throws fail, so by the multiplication rule for independent events

P(FFS) = \Big(\frac56\Big)^2\cdot\frac16

Similarly, $P(FFFFS) = \big(\frac56\big)^4\cdot\frac16$ ⁠, and so on.

\begin{aligned} &P(\text{A wins}) \\ &= \frac16 + \Big(\frac56\Big)^2\frac16 + \Big(\frac56\Big)^4\frac16 + \ldots \end{aligned}

This is an infinite G.P. with $a = \frac16$ and $r = \frac{25}{36}$ ⁠, where $|r| < 1$ ⁠. Using $S_\infty = \frac{a}{1 - r}$ ⁠,

\begin{aligned} P(\text{A wins}) &= \frac{\frac16}{1 - \frac{25}{36}} \\ &= \frac16 \times \frac{36}{11} \end{aligned}

B wins at the 2nd, 4th, … throws, so

\begin{aligned} &P(\text{B wins}) \\ &= \frac56\cdot\frac16 + \Big(\frac56\Big)^3\frac16 + \ldots \end{aligned}

This is an infinite G.P. with $a = \frac{5}{36}$ and $r = \frac{25}{36}$ ⁠, so

\begin{aligned} P(\text{B wins}) &= \frac{\frac{5}{36}}{1 - \frac{25}{36}} \\ &= \frac{5}{36} \times \frac{36}{11} \end{aligned}

$\therefore$ P(A wins) $= \dfrac{6}{11}$ and P(B wins) $= \dfrac{5}{11}$ ⁠.

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

The one idea this problem is built on

List the ways A can win, then add them. A wins on the 1st throw, or on the 3rd (after two failures), or on the 5th (after four failures), and so on. These are mutually exclusive, and each is a product of independent throws. Their probabilities

\frac16, \quad \Big(\frac56\Big)^2\frac16, \quad \Big(\frac56\Big)^4\frac16, \quad \ldots

form an infinite geometric progression, whose sum is $\frac{a}{1 - r}$ ⁠.

Given, to find, and the plan

Given: A and B take turns throwing a fair die, A first; the first to throw a six wins.
To find: P(A wins) and P(B wins).
Plan: (1) Probabilities of a six and of no six on one throw. (2) The probability that A wins on throw 1, 3, 5, …. (3) Add them as an infinite G.P. (4) B's probability. (5) Check another way.

Basics you need

1Independent throws (Section 13.4)

Separate throws of a die are independent, so the probability of a whole sequence of results is the product of their probabilities. Lesson: independent events

2Adding mutually exclusive events

"A wins on throw 1" and "A wins on throw 3" cannot both happen, so their probabilities add. Basics: rules of probability

3Infinite geometric progression

$a + ar + ar^2 + \ldots = \frac{a}{1 - r}$ when $|r| < 1$ (Class XI; the book's Remark after this example). Basics: infinite G.P.

4Powers of fractions

$\big(\frac56\big)^2 = \frac{25}{36}$ ⁠, $\big(\frac56\big)^4 = \frac{625}{1296}$ ⁠. Basics: working with fractions

Solution

One throw: success and failure

Let $S$ be a success (getting a six) and $F$ a failure (not getting a six) on a throw. Then
$P(S) = \frac16, \qquad P(F) = 1 - \frac16 = \frac56.$
Every throw, by A or by B, is a throw of the same fair die, so these two probabilities are the same at every throw. Only "six or not six" matters for the game.

$P = \frac{\text{favourable}}{\text{total}}$ ⁠: one face of six. Basics: the classical definition
When does A win?

A throws 1st, 3rd, 5th, …; B throws 2nd, 4th, …. A wins at the 1st throw if it is a six; at the 3rd throw if throws 1 and 2 fail and throw 3 is a six; and so on. Since the throws are independent,
$\begin{aligned} &P(\text{A wins at throw 1}) \\ &= P(S) = \frac16, \\ &P(\text{A wins at throw 3}) \\ &= P(F)P(F)P(S) = \Big(\frac56\Big)^2\frac16, \\ &P(\text{A wins at throw 5}) \\ &= \Big(\frac56\Big)^4\frac16, \end{aligned}$
and in general A wins at throw $2k + 1$ with probability $\big(\frac56\big)^{2k}\frac16$ ⁠.

The game, throw by throw. At every throw a six (green) ends the game; otherwise the die passes on. A's winning paths (blue) have probabilities $\frac16$ ⁠, $\big(\frac56\big)^2\frac16$ ⁠, $\big(\frac56\big)^4\frac16$ ⁠, …; B's (violet) have $\frac56\cdot\frac16$ ⁠, $\big(\frac56\big)^3\frac16$ ⁠, ….

To reach his 3rd-throw turn the game must still be going: A failed at throw 1 and B failed at throw 2. "A wins at throw 3" is the sequence F, F, S, and for independent throws the probability of a sequence is the product. Each extra round adds two more failures, one by A and one by B, which is the factor $\big(\frac56\big)^2$ ⁠.

For independent events, $P(E \cap F \cap G) = P(E)P(F)P(G)$ ⁠. Lesson: independent events
Add the ways A can win: an infinite G.P.
$\begin{aligned} &P(\text{A wins}) \\ &= \frac16 + \Big(\frac56\Big)^2\frac16 + \Big(\frac56\Big)^4\frac16 + \ldots \end{aligned}$
This is a G.P. with first term $a = \frac16$ and common ratio $r = \big(\frac56\big)^2 = \frac{25}{36}$ ⁠. Since $|r| < 1$ ⁠, its sum is $\frac{a}{1 - r}$ ⁠:
$\begin{aligned} P(\text{A wins}) &= \frac{\frac16}{1 - \frac{25}{36}} = \frac{\frac16}{\frac{11}{36}} \\ &= \frac16 \times \frac{36}{11} = \frac{6}{11}. \end{aligned}$
A can win at throw 1, or 3, or 5, …, and only one of these can happen (the game stops at the first six), so the probabilities add, without end. Each term is the one before times $\frac{25}{36}$ ⁠, so the sum is that of an infinite G.P. The ratio is less than 1, so the sum is finite. This is the book's $\frac{1/6}{1 - 25/36} = \frac{6}{11}$ ⁠.

$a + ar + ar^2 + \ldots = \frac{a}{1 - r}$ ⁠, $|r| < 1$ ⁠. Basics: infinite G.P.
B's chance of winning

The book uses the complement:
$\begin{aligned} P(\text{B wins}) &= 1 - P(\text{A wins}) \\ &= 1 - \frac{6}{11} = \frac{5}{11}. \end{aligned}$
The same value comes directly. B wins at throw 2, 4, 6, … with probabilities $\frac56\cdot\frac16$ ⁠, $\big(\frac56\big)^3\frac16$ ⁠, …, a G.P. with $a = \frac{5}{36}$ and $r = \frac{25}{36}$ ⁠:
$\begin{aligned} P(\text{B wins}) &= \frac{\frac{5}{36}}{1 - \frac{25}{36}} \\ &= \frac{5}{36} \times \frac{36}{11} = \frac{5}{11}. \end{aligned}$
"B wins = 1 − A wins" needs the game to end for certain. It does: the chance that the first $n$ throws all fail is $\big(\frac56\big)^n$ ⁠, which gets as small as we like, so a game that never ends has probability 0. The direct sum confirms it: $\frac{6}{11} + \frac{5}{11} = 1$ ⁠.

Complement rule $P(E') = 1 - P(E)$ ⁠. Basics: rules of probability
Check another way: B is A, one throw late

Let $p = P(\text{A wins})$ ⁠. If A fails at the first throw (probability $\frac56$ ⁠), B now throws first in a fresh game, so B then wins with probability $p$ ⁠. Hence $P(\text{B wins}) = \frac56 p$ ⁠, and
$\begin{aligned} p + \frac56 p &= 1, \\ \frac{11}{6}p &= 1, \\ p &= \frac{6}{11}. \end{aligned}$
After a failure the game looks exactly like the start, with the roles swapped, because the die has no memory (the throws are independent). This gives the answer without any series, and it explains the ratio $6 : 5$ ⁠: going first is worth a factor $\frac65$ ⁠.

Independent throws: the past does not change the next throw. Real world: first to roll a six

Result

P(A wins) $= \frac{6}{11}$ and P(B wins) $= \frac{5}{11}$ ⁠.

(Textbook answer: $\frac{6}{11}$ and $\frac{5}{11}$ ⁠.)

Check it with numbers

Decimals. $\frac{6}{11} \approx 0.5455$ and $\frac{5}{11} \approx 0.4545$ ⁠, which add to 1.

The first few terms. A's first three winning chances add to $\frac16 + \frac{25}{216} + \frac{625}{7776} = \frac{2821}{7776} \approx 0.3628$ ⁠; the later throws supply the rest of the $0.5455$ ⁠. The chance that the game is still going after 6 throws is $\big(\frac56\big)^6 \approx 0.3349$ ⁠, and after 30 throws only about $0.0042$ ⁠.

Common mistakes

Answering $\frac16$ for A. That is only the chance of winning at the first throw; A can also win later.
Using the ratio $\frac56$ ⁠. Between two of A's turns there are two failures (A's and B's), so the ratio is $\big(\frac56\big)^2 = \frac{25}{36}$ ⁠.
Answering $\frac12$ each. The game is not symmetric: A gets the first chance, so A is more likely to win.

What to take away

For "first to succeed" games, list the winning sequences; with independent repeats their probabilities form a G.P.
The trick in Step 5 (after a failure the game restarts) often gives the answer in one line.
The probabilities of all the ways the game can end add to 1: $\frac{6}{11} + \frac{5}{11} = 1$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$P(S) = \frac16$ ⁠, $P(F) = \frac56$	Step 1	Class XI, Probability (Basics: the classical definition)
Product rule for independent throws	Step 2	Section 13.4 (Lesson: independent events)
Adding mutually exclusive events	Steps 3, 4	Class XI (Basics: rules of probability)
Sum of an infinite G.P., $\frac{a}{1 - r}$	Steps 3, 4	Remark after Example 23; A.1.3 of the Class XI book (Basics: infinite G.P.)
Complement rule	Step 4	Class XI (Basics: rules of probability)
The game, worked as a story	Step 5	Examples 22 to 24; Real world: first to roll a six

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 23: A and B throw a die in turn until a 6: 611\frac{6}{11}116​ and 511\frac{5}{11}115​

Basics you need

Solution

One throw: success and failure

When does A win?

Add the ways A can win: an infinite G.P.

B's chance of winning

Check another way: B is A, one throw late

Every fact used, and where it comes from

Example 23: A and B throw a die in turn until a 6: $\frac{6}{11}$ and $\frac{5}{11}$