Dutch book arguments

The subjective interpretation of probability is based on the premise that degrees of belief must satisfy the axioms of probability calculus. The so-called “Dutch book” arguments provide a philosophical basis for this assumption.

The arguments start by operationalizing¹ the (otherwise quantitatively vague) concept of “degree of belief”. The degree of belief of a subject into a given uncertain event² \(E\) is supposed to be a number \(p(E)\) that can be quantified as follows:

• The subject must be willing to pay an amount \(p(E)\), with the condition that he will be rewarded an amount \(1\) if event \(E\) obtains.
• The subject must also be willing to accept an amount \(p(E)\), with the condition that he will give up an amount \(1\) if event \(E\) obtains.

(the concrete unit of payment in these imaginary bets is irrelevant to the arguments, but is assumed to be a continuous quantity). The idea is that, for the subject not to incur into a sure loss, the rule \(E\mapsto p(E)\) must define a single valued function that follows the laws of probability.

Let us start by showing that the function \(E\mapsto p(E)\) must be single valued (i.e. the degree of belief is operatively well-defined). Suppose that someone assigns two different degrees of belief \(p\) and \(q\) to the same event \(E\), and suppose that \(p>q\), for instance. This means that this person must be willing to place a bet of \(p\) on \(E\) and, at the same time, accept a bet of \(q\) on \(E\). But this doesn’t make sense, because the net result of these combined bets is a sure loss (of an amount \(p-q\)). We can conclude that the degree of belief assigned by any individual into a given event must be a single number.

With a similar approach we can prove the three basic axioms of probability:

Positivity. Degrees of belief must be positive numbers, because no rational agent would be willing to pay a positive amount \(q = -p\) (which is the only reasonable way to interpret accepting a negative amount \(p = -q\)), for later having to pay an additional amount \(1\) if event \(E\) obtains.

Normalization. Suppose that the event \(E\) is certain, and let \(p\) be your degree of belief into it. If \(p < 1\), one can force you to accept an amount \(p\), and to give \(1\) in return, with certainty since \(E\) is always true. If \(p > 1\), one can force you to pay an amount \(p\) to get \(1\) in return. In both cases, you face a sure loss of an amount \(\vert 1 - p\vert\), unless \(p = 1\).

Additivity. Suppose that events \(A\) and \(B\) are mutually exclusive, and let \(p\), \(q\) and \(r\) be your degrees of belief into \(A\), \(B\) and \(A\cup B\) ( denoting \(A\) or \(B\)). If \(r < p + q\), one can force you to accept the following bets:

• You pay \(p\) and receive \(1\) if \(A\) obtains.
• You pay \(q\) and receive \(1\) if \(B\) obtains.
• You receive \(r\) and pay \(1\) if \(A\cup B\) obtains.

The fact that \(A\) and \(B\) are mutually exclusive, implies that the cash flow from the returns of these three bets is always zero, and the net flow is entirely set by the initial payments. For you, this is \(r - p - q < 0\), which is a sure loss. If \(r > p + q\), your opponent could also force you into a sure loss by reversing the bet directions in the previous argument.

There are also Dutch book arguments showing that when some rational agent has to update its degrees of belief based on new information, it must do so according to the standard conditioning rule:

\[ p(A\vert B)p(B)=p(A\cap B), \tag{1}\] where \(p(A\vert B)\) denotes the degree of belief on \(A\) after finding out that \(B\) is true, while \(A\cap B\) denotes the event that both \(A\) and \(B\) are true. It is worth to notice that the argument given below relies on the fact that the agent’s assessment of \(p(A\vert B)\) is fixed and declared a priori, and does not change according to whether \(B\) actually obtains or not.

The argument goes as follows: suppose that Equation 1 does not hold and, for instance \(p(B)\cdot p(A\vert B) < p(A\cap B)\). We can then force the agent to participate in the following bets:

• The agent accepts \(p(B)\cdot P(A\vert B)\), to give \(P(A\vert B)\) in return if \(B\) obtains³.
• The agent pays \(p(A\cap B)\) to get \(1\) in return if both \(A\) and \(B\) obtain.
• If \(B\) obtains, the agent accepts \(p(A\vert B)\) to give in return \(1\) if \(A\) obtains.

It is easy to see that, irrespective of the outcomes of the \(A\) and \(B\) events, the agent always ends up loosing an amount \(p(A\cap B) - p(B) \cdot P(A\vert B)\). If \(p(B)\cdot p(A\vert B) > p(A\cap B)\), reversing the direction of these bets also leads to a sure loss for the agent. Clearly, the only way out is to set conditional probabilities according to Eq. Equation 1.

Footnotes

1. Which means “giving a practical way to measure”.

2. I use the word “events” in a conventional way. Strictly speaking, to reflect the wider generality of the subjectivist point of view, I should rather talk about “uncertain propositions” or “hypothetical facts”.

3. This is a slight generalization of the type of bet used in the operative definition of “degree of belief”, that assumes a unitary return amount. The reason why this is equivalent to the original definition is that, given bets \(B_1,\,B_2,\,\cdots,\,B_N\) with unitary return amounts, and positive numbers \(x_1, \,x_2,\,\dots,\,x_N\), we can always find positive integers \(n_1,\,n_2,\dots,\,n_N\), such that the total gain from repeating \(n_i\) times bet \(B_i\) for all \(i=1,2,\dots,N\) always has the same sign of the total gain from “generalized bets” that result by changing the return amount of bet \(B_i\) to \(x_i\).