Dutch Book explained incorrectly in a book?

Question

In chapter 4 of Darrell P. Rowbottom's Probability, the author explains the Dutch Book:

Imagine we're going to have a bet together, you and I, about whether something will happen. It could be about whether your favourite sportsperson or team will win their next match, or something more trivial like whether it will rain tomorrow anywhere in your country. We then go on to agree on a stake, $S$, which is the maximum amount of money that could change hands.

I will choose whether to bet for or against the event happening. (Note that my talk of ‘events happening’ is just for convenience, and can easily be translated into talk of ‘statements being true’. So we can understand the bets to concern propositions or statements. For example, betting on Manchester United winning their next match is equivalent to betting that ‘Manchester United will win their next match’ is true.) But before I make that choice, you're going to pick a number, a betting quotient $b$, on the understanding that the bet will be conducted as follows:

1. If I bet against the event occurring, you will pay me $bS$. If the event occurs, I will give you $S$.
2. If I bet for the event occurring, you will pay me $(1 − b)S$. If the event doesn't occur, I will give you $S$.

To choose $b$ is to determine ‘the odds’ of the bet, which is usually expressed as a ratio, namely $\frac{b}{1 − b}$. To select a value of one half for $b$, for example, would be to give ‘even odds’ on the event; you'll double your initial payment if you win the bet, otherwise you'll lose it. (You may notice that if you select a value of one for $b$, there is no value for the odds when they're defined in the way above. This is an intended feature of the setup, as we'll see.)

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I think there is a mistake in points 1 and 2. It can be corrected in two ways:

1. You pay me $(1-b)S$ or I pay you $bS$.
2. You pay me $bS$ or I pay you $(1-b)S$.

Or using odds:

1. You pay me $\frac{1-b}{b} S$ or I pay you $S$
2. You pay me $\frac{b}{1-b} S$ or I pay you $S$

Is there indeed a mistake and are my proposed solutions correct?

Answer 1

Addendum added to the end of my answer, to respond to the comment of Dmitrii I.

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The confusion is actually between (for example) the quoted odds of

• $5$ to $1$.
• $6$ for $1$.

To a Mathematician, the first syntax above is the most rational, and expresses the amount that you receive upon winning, not including the return of your original bet.

Casinos like to seduce the betting public, who are generally non-Mathematicians. $6$ for $1$ sounds more attractive than its mathematically equivalent expression of $5$ to $1$.

When reading a book on odds, you have to be careful to decipher the author's intent. Authors (unfortunately) may be careless in not explicitly specifying whether the quoted odds of $n::1$ represent $n$ to $1$ or $n$ for $1$.

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Addendum
You are assuming that what you read accurately expressed the intent. It is my contention that the text that you quoted intends a Mathematically valid exchange but is very poorly written.

It is my contention that the most likely intent is that you are the bookmaker. The bettor automatically gives you either $bS$ or $(1-b)S$, before the outcome of the event is determined.

Then, if the bettor wins, you pay the bettor $S$.

This means that if the bettor wins, he will receive $S$ back total after first giving you either $bS$ or $(1-b)S$. This is in fact standard practice with sportsbooks in Casinos. That is, the Sportsbooks never trust the bettor, but instead make the bettor put up the money that he is risking, in advance.

As an example, suppose that a sportbook requires you to lay $11$ to $10$ on a pick-em bet. This means that regardless of which side that you bet, you have to put up $11\$$ to win $10\$$.

Then, if you win, you receive $21\$$, which would represent a net profit to you of $10\$$, if you win. Horse racing is the exact same. You have to purchase the ticket in advance. If the horse that you bet on wins, then you can cash your ticket.

So, my answer reflected my surmise that the author that you quoted is not Mathematically inept, but (more likely) simply writes very poorly (or carelessly).

That is, I am surmising that the intent was that the bettor gives you either $bS$ or $(1-b)S$, in advance. Then, if he wins, he gets back $S$. Otherwise, he gets back nothing.

Assuming that my surmise is correct, the effect is the same as the bettor giving you the $bS$ or $(1-b)S$ only if he loses, and collecting from you $(1-b)S$ or $bS$ if he wins.

This is because (for example) the bettor paying $bS$ and then getting back $S$ is equivalent to the bettor winning $S - bS = (1-b)S.$

So, the theory indicated in my original answer is tenable and likely. That is, it is more likely that the author writes poorly, than that the author does not understand arithmetic.

Answer 2

The confusion is arising because of two different meanings given to the same symbol $b$ in the two parts

In the first part, it talks of the betting quotient $b$ (say $3:1$)

In the second part, it talks of the odds ratio being $\frac{b}{1-b}$ where $b$ is a fraction (probability ) $ = \frac34$ for the example we took !

Once you decide what $b$ is to be, it is easy to sort out the payouts