\documentclass[12pt]{amsart}
\usepackage{amssymb,amsmath}
\begin{document}
\title{Corrections to Naive Decision Making}
\author{T.W.K\"{o}rner}
\maketitle
This correction page (last modified August 30th 2022)
is based on corrections
by John Haigh, Robert MacKay, Nigel White
and Matthew Towers to whom many thanks

\emph{Page 4, line -3} Reverse inequality sign.

\emph{Page 5, line 2} Reverse inequality sign.

\emph{Page 6} In (iii) reverse both inequality signs.

\emph{Page 15} Second para of section, reverse inequality sign.

\emph{Page 53} Exercise~2.4.13, 5th line delete second `in'.
 
\emph{Page 58} Exercise~2.4.24  First sentence. `Players pay the
banker $1/40$ of a unit to take part in a game of Simplejack.
In this game a large pack of cards containing \ldots'

\emph{Page 65} Fifth line of footnote 23 `will be discussed'

\emph{Page 67} Example 2.5.16, first displayed inequality
should read

\[\Pr\left(\left|\frac{X_{1}+X_{2}+\dots+X_{n}}{n}
      \right|<\delta\right)\geq 1-\epsilon.\]




\emph{Page 69} Let $X_{jk}=1$ if $A_{k}$
chooses the $j$th grotto.


\[Z=\sum_{k=1}^{n}Y_{1}X_{1k}+\sum_{k=1}^{n}Y_{2}X_{2k}+\cdots
+\sum_{k=1}^{n}Y_{m}X_{mk}.\]

\emph{Page 74} Kelly recommends $t=1/12$.
Same correction line 1
of Exercise 2.6.7.

\emph{Page 75 First displayed formula}
\[\frac{11}{12}\times\left(\frac{11}{12}+\frac{11}{5}\frac{1}{12}\right)
=\frac{121}{120}\]

The figures in the next paragraph could be changed a little
but remain of the correct order.%Recalculate

\emph{Page 75} Exercise~2.6.7, line $1$. `This part requires' 


\emph{Page 77} Exercise 2.6.10. I fell asleep at the wheel.
Should read

I bet a fixed amount $t$.
Thus my fortune $X_{j+1}(t)$
after the $j+1$th throw is given by
\[X_{j+1}(t)=
\begin{cases}
X_{j}(t)+tu&\text{if the $j$th throw is heads,}\\
X_{j}(t)-t&\text{if the $j$th throw is tails.}
\end{cases}
\]

\emph{Page 79} Exercise~2.7.1
Space after (i)

The second (ii) should be (iii)

Correct suffix in second displayed equation to give
\[F(u_{1},u_{2})=
p_{1}\log p_{1}+p_{2}\log p_{2}+p_{1}\log u_{1}+p_{2} \log u_{2}.\]

\emph{Page 81} Exercise~2.7.2 part (iv) should begin

(iv) In the remaining Case C,  when $p_{1}u_{1}>1$ and
\[\frac{1}{u_{1}}+\frac{1}{u_{2}}+\dots+\frac{1}{u_{n}}>1,\]
show that


\emph{Page 90} Exercise 3.2.6 (ii).  
The final term of the sum should be $(m - 1)^{2}x^{m-2}$.

\emph{Page 104} First sentence of third line
`Thus $g'$ is increasing.'

\emph{Page 109} Exercise~3.5.10 displayed equation should read
\[c_{p}(|a|^{p}+|b|^{p})\geq |a|^{p}+|b|^{p}\]

\emph{Page 110} First line. I have the ranges of summation
in a twist. Should be
\[p_{i}=\sum_{j=1}^{m}\pi_{ij},\ q_{j}=\sum_{i=1}^{n}\pi_{ij}\]

\emph{Page 121} Lemma~4.3.4~(iv)'. I was asleep at the  wheel.
Should read 

(iv) If $a\equiv a'$ and $b\equiv b'$ then $a+b\equiv a'+b' \mod{n}$.

\emph{Page 124}
Proof of Lemma 4.3.10 (ii), third line should read.
\[y\equiv v_{1}y_{1}+v_{2}y_{2}
  \equiv v_{1}\times 0+v_{2}\times 1\equiv v_{2}\]

\emph{Page 155} Case (iv)

\begin{gather*}
A_{1}=\{X_{1}<X_{2},X_{3},X_{4},X_{5}\},\\
A_{2}=\{X_{2}<m\},\  A_{3}=\{X_{3}<m\},\ A_{4}=\{X_{4}<m\},
\ A_{5}=\{X_{5}<m\}
\end{gather*}

\emph{Page 159} Exercise~5.3.9~(ii) second sentence should read:-

Show that the probability
that we stop at the last card is at most $t$ and the probability
that none of the $k$ largest cards are turned over
before the $m$th card is at most $(1-t)^{m}$.

\emph{Page 161} Last line should read
\[An\log n\geq \log n!\geq Bn\log n\]

\emph{Page 170} Paragraph after Exercise 5.5.7.

What happens
if we try to work out the shortest paths between
every pair of towns?


\emph{Page 173} Exercise~5.5.18~(i) Replace the
meaningless `length a decreasing cycle' by
`a distance decreasing cycle'.

\emph{Page 179} Exercise~6.1.7. Last sentence
replaced by

If the process does
terminate, will  pairing it produces necessarily be stable?


\emph{Page 186} (Thanks to David Pauksztello) 5th line down
`everybody else's preferences $B>C>A$'


\emph{Page 192} Replace `$C$ beats $A$' by `$A$ beats $C$' 

\emph{Page 200} Third complete paragraph, second line.
Replace `choosing row $1$ heads with' by  `choosing row $1$ with'

\emph{Page 205} Second line of first complete paragraph

We shall say that Rowena adopts strategy ${\mathbf p}$
if she chooses row $i$ with probability $p_{i}$
and that  Calum adopts strategy ${\mathbf q}$
if he chooses column $j$ with probability $q_{j}$.
\emph{Page 217} Exercise 7.6.2. First sentence should read

Suppose that George has chosen $z>x_{0}$
and Fred knows the value of $z$.

\emph{Page 218} Exercise 7.6.2.

In (ii) replace $y<x_{0}$ by $z<x_{0}$

In (iii) replace $y=x_{0}$ by $z=x_{0}$

\emph{Page 228} Exercise 8.2.4, penultimate sentence.

`may be $(x_{1},y_{1})$ itself'

`for all $(x,y)\in{\tilde K}$'

\emph{Page 228} Exercise 8.2.2~(ii). The displayed alternatives should
read

\ \ \ (a) If $(x,y)\in K$ and $x=0$ then $y\leq 0$.

\ \ \ (b) If $(x,y)\in K$ and $y=0$ then $x\leq 0$.

\emph{Page 238} Penultimate line. Displayed formula
should read
\[\beta({\mathbf p}^{*},{\mathbf q})\leq
\beta({\mathbf p}^{*},{\mathbf q}^{*})\]

\emph{Page 243} First sentence should begin


Almost exactly the same arguments as before show that
there are exactly three Nash equilibrium points $(p,q)=(1,1)$
with expected value to Albert of $1$ and to Bertha of $0$,
$(p,q)=(0,0)$ with expected value to Albert of $0$ and to Bertha of $1$
and $(p,q)=(2/3,1/3)$

\emph{page 254} Matthew Towers informs me that my erratum
was in error and suggests

(5) If $C$ misses both, then, by (2), we know that $B$
will fire at $A$. If $B$ hits $A$, then  
the result is a duel
between $C$ and $B$ in which $C$ and $B$ fire alternately
and $C$ has first shot so the probability of $C$ winning
is
\[\frac{cb}{b+c-bc}.\]
If $B$ misses $A$ then $A$ will shoot at $B$ and hit him,
so $C$ must then shoot at $A$ and will win the game
if and only if he hits $A$. 

\begin{align*}                                                                  
Pr(\text{$C$ wins if he misses first shot})
&= Pr(\text{$C$ wins if            
$B$ hits $A$})b + Pr(\text{$C$ wins if $B$ misses $A$})(1-b) \\              
&=\frac{cb}{b+c-bc}+c(1-b)\\                              
&=c\frac{b^2c-b^2-2bc+c+2b}{b+c-bc}
\end{align*}


Thus $C$ has a better chance of winning if he misses the first shot
(and should therefore \emph{deliberately miss}) if
\begin{align*}                                                                  
    c\frac{b^2c-b^2-2bc+c+2b}{b+c-bc}> c\frac{1-b}{b+c-bc},                   
\end{align*}
which simplifies to
\[b(3-b) + (b-1)^2c > 1.\] 


\emph{Page 254} Replace first part of Exercise 9.2.3 as follows

Exercise 9.2.3 (i) Show that, if $3b-b^2>1$, then, whatever
the value of $c$, player
$C$ should always deliberately miss
with his first shot. Show that, if $b^{3}-3b^{2}+4b<1$
$C$ should always try to hit $B$
with his first shot. Explain to a non-mathematician why, if $B$
is strong, $C$ should deliberately miss his first shot
and, if $B$ (and therefore $C$) is very weak, why $C$ should
try to hit $A$ with his first shot.

\emph{Page 259} In the statement of Lemma~9.3.3 delete the
numbers $(i)$ and $(ii)$ and replace the starting words
`We have' in the last sentence by `Conclude that'.

\emph{Page 267} In the diagram there should be arrows
from THH to HHT and from HTT to TTH

\emph{Page 268} For those with good eyesight, last line
first $P_{TH}$ should be $p_{TH}$.

\emph{Page 274} Exercise 9.5.5 in table.
\[B(A,S)=\frac{1}{1-p}\]


{\emph Page 267}  The game {\bf HHH} was invented by 
Walter Penney (Journal of Recreational Mathematics, October 1969, p. 241)
and is referred to as Penney's game. Strong apologies.

{\emph Page 274} There is an error in the last line which 
affects the following paragraph


We now observe that
\begin{gather*}
B(I,N)-B(I,P)=\frac{2-p}{1-p}-3=\frac{2p-1}{1-p},\\
B(I,N)-B(I,O)=\frac{2(1+p)-3}{1-p^{2}}=\frac{2p-1}{1-p^{2}},\\
B(I,P)-B(I,O)=-\frac{3p^{2}}{1-p^{2}}+\frac{p}{1-p}=\frac{p(1-2p)}{1-p^{2}}.
\end{gather*}


Thus
\begin{gather*} 
B(I,N)>B(I,P)\ \text{for $p>1/2$},
\ B(I,P)>B(I,N)\ \text{for $1/2>p$},\\
B(I,N)>B(I,O)\ \text{for $p>1/2$},
\ B(I,O)>B(I,N)\ \text{for $1/2>p$},\\
B(I,O)>B(I,P))\ \text{for $p>1/2$},
\ B(I,P)> B(I,O)\ \text{for $1/2>p$}.
\end{gather*}
Looking at these results, we
advise Sonia to play the strategy `never press'
(or, what turns out to be exactly the same strategy,
`do the same as Tania') whenever $p\geq 1/2$,
to follow the strategy `do the opposite of Tania'
when $1/2\geq p\geq 0$.
(If $p=1/2$ there is a free choice between the two
recommended strategies.)

{\emph Page 274} Last line on page should read
\[B(I,P)-B(I,O)=-\frac{3p^{2}}{1-p^{2}}+\frac{p}{1-p}
  =\frac{p(1-2p)}{1-p^{2}}.\]

{\emph Page 283} Third displayed formula from the bottom of the page.
\[t=1\ \text{or}\ t=\frac{1-p}{p}.\]

{\emph Page 283} Exercise~10.1.4~(i) displayed formula
should read
\[q_{n}=A+B\left(\frac{1-p}{p}\right)^{n}\]

{\emph Page 285} Exercise 10.1.5 (ii)

Can you suggest three distinct strategies which are
as good as bold play?

{\emph Page 285}  Exercise 10.1.6 first line

stake of $1$ dollar

{\emph Page 286} Second displayed equation

Show, by induction, or otherwise that
\[p_{r}(f)=p_{r}(k2^{-r})\ \text{if $k2^{-r}\leq f<(k+1)2^{-r}$}\]
for integers $k$ with $0\leq k\leq 2^{r}$.


{\emph Page 290}  Exercise~10.2.7~(ii) line 3 delete second
`better' (or, if you prefer, first `better' but not both).

{\emph Page 290} Third line of second paragraph after Exercise~10.2.8.
replace `there' by `their'.

{\emph Page 290} Expression
\[10\,000\times{\mathbb E}(\text{expected number of months to bankruptcy}),\]
 should be replaced by
\[10\,000\times\text{expected number of months to bankruptcy},\]

{\emph Page 291} Last three sentences before Exercise~10.2.10
should read:-

The probability of bankruptcy in a particular
month is $(1-p)^{8}$ and so, by Lemma~9.3.3,
the expected number of times I enter the casino
is $1/(1-p)^{8}$. Thus the expected return
from an investment of $2\,560\,000$ is
\[10\,000\times(1-p)^{-8} .\]
Taking a typical value of $p=.49$ we get an
expected return of about
$2\,184\,935$ dollars.

{\emph Page 291} Exercise~10.2.10

(i) What is the expected return if $p=1/2$?
Why should we expect this?

(ii) What is the return if $p$ takes the extremely favourable
value $p=.495$?


{\emph Page 297} Exercise 10.3.11 There are two part~(ii)s.
Renumber appropriately.

The first (ii) should read

(ii) Write down the general solution of
\[(E-I)u_{n}=\binom{n}{r-1}.\]

{\emph Page 302} Three lines after $\bigstar$ undisplayed equation

`$(1-p)u_{n+2}-u_{n+1}+ue_{n}=1$' should be
`$(1-p)u_{n+2}-u_{n+1}+pe_{n}=0$'

{\emph Page 303} Last line should read

\[f_{n}=\frac{1-p}{1-2p}\left(\frac{1-p}{p}\right)^{N}
\left(1-\left(\frac{p}{1-p}\right)^{n}\right)\]

{\emph Page 304} First displayed formula should read
\[f_{N}=\frac{1-p}{1-2p}
\left(\left(\frac{1-p}{p}\right)^{N}-1\right).\]

{\emph Page 304} Exercise~10.4.7 First paragraph,
after first displayed equation add `with $r$ an integer' to read

`and $Y_{0}=r$ with $r$ an integer, explain'

{\emph Page 307} Exercise 10.4.10~(i) first displayed equation
\[e_{m}=\frac{\sum_{j=m}^{n} jp_{j}}{1-q\sum_{j=0}^{m-1}p_{j}}.\]

{\emph Page 307} Exercise 10.4.10~(i) last paragraph first
sentence second line. Reverse inequality to get `$e_{m+1}\leq e_{m}$'.



{\emph Page 307} Exercise 10.4.1, second paragraph, second sentence
`$V_{0}=p/q$'.

{\emph Page 308} The statement 

In the UK National Lottery
$50$\% of the cost of each ticket
is returned to the buyers as prizes.

should be modified by the insertion of the word `roughly'.

{\emph Page 313} Exercise 10.5.10 first sentence should read

(i) Show that $|x-\log(1+x)|\leq x^{2}$
for $1/2>|x|$.


{\emph Page 313} Exercise 10.5.10~(iv). Factor of $n$ missing.
Should read

Show that
\[\frac{2n}{N(n)^{2}}\left[
\log\left(1+\frac{a}{n}\right)+\log\left(1+\frac{2a}{n}\right)
+
\dots\log\left(1+\frac{N(n)a}{n}\right)\right]\rightarrow a.\]
Deduce that
\[\left[\left(1+\frac{a}{n}\right)\times\left(1+\frac{2a}{n}\right)
\times\left(1+\frac{3a}{n}\right)\times
\dots\times\left(1+\frac{N(n)a}{n}\right)\right]^{2n/N(n)^{2}}
\rightarrow e^{a}\]
as $n\rightarrow\infty$.

{\emph Page 318} L. F. Richardson not J. F. Richardson!

{\emph Page 322} Footnote clarification. 

and the elder Bush

{\emph Page 323} Exercise 11.1.1~(iv). At end of first
sentence insert `in a row'.

{\emph Page 323} Middle  

The probability of throwing 
ten heads in a row is now $(1/2)^{10}$, so, perhaps,
it would be good idea. 

{\emph Page 329} 4th line down. `and apply (iv)' [not (i)]



{\emph Page 330} Exercise~11.3.3~(vi). Third displayed equation
inequality should (of course) be reversed.
\[
\Pr\left(\left|\sum_{j=1}^{n}X_{j}\right|\leq y\right)
\geq 2\exp\big(-y^{2}/(2A)\big).\]

{\emph Page 331} Exercise 11.3.4, displayed formula should read 
\[\Pr\big(|Y_{n}-nq|\leq n\epsilon\big)
\leq 2\exp\big(-n\epsilon^{2}/2).\]

{\emph Page 332} Exercise~11.4.1, second line. Insert `the'
to get `both the following properties'.


{\emph Page 333} Exercise 11.4.2, second paragraph first
line. Replace `reject' by `accept'.


{\emph Page 334} Exercise 11.4.4~(iv). $R(a,p)$ instead of $R(P,a)$   


{\emph Page 356} Exercise~B6~(iii) last sentence. Exercise~B4
(not B.6).

{\emph Page 352} Exercise~C5~(ii) line one. Replace doubled `a'
by a singleton. 


{\emph Page 361} Exercise~C.4~(ii) second paragraph line 1.
Remove `of'.

\emph{Bibliography} The Martin Gardner book [23] is

\emph{The 2nd Scientific American Book of Mathematical Puzzles 
and Diversions}

\emph{index}
Should contain reference to Penney to go with
correction page 267.

{\emph Page 368} ref [41]: J. Maynard Smith.  (oops, thanks
to Charles Goldie) 

\end{document}












