How to create a Dutch Book?

I found this question from Ian Hacking's book on induction and probability. I understand that a Dutch Book is a gambling term wherein everyone wins. Although, the last part of the question 'Describe a Dutch Book for Dave' is confusing. How is this prediction one where everyone wins?

Dave thinks that the probability of an early Spring if Wiarton Willie predicts an Early Spring is 4/5, but that the probability of not having an early Spring if Wiarton Willie predicts an Early Spring is 2/5. Describe a Dutch Book for Dave.

asked Mar 17, 2017 at 1:39 Andrew Raleigh Andrew Raleigh 161 8 8 bronze badges

$\begingroup$ If you offer Dave a bet on an early spring at $100$ to win $125$, and a late spring at $50$ to win $125$, then you'll make $150$ on the two bets and pay out $125$. That's right at the threshold, so you could sweeten the payout to anywhere between $125$ and $150$. $\endgroup$

Commented Mar 17, 2017 at 1:50

$\begingroup$ A Dutch Book, as I understand it, is a book plus a set of bets where every outcome is a win for the bookie—not one where everyone wins. $\endgroup$

Commented Mar 17, 2017 at 1:51

1 Answer 1

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Dave has managed to create a scenario in which he thinks if WW predicts an Early Spring, something will happen with a probability of $\frac65\gt1$.

This scenario is called a Dutch book - everybody knows that the maximum sum of probabilities can only be $1$, but the odds offered don't match with this, and hence there is a guaranteed profit for someone.

Let's assume WW predicts an Early Spring, Dave has two decisions, to go with WW or to reject WW's guess.

So Dave places £100 on both events at the odds he thinks he will get, but his mistake is in the way he calculated the odds. Elementary analysis reveals Dave's mistake:

A percentage $p\%$ to win equals $\frac1$ where $w$ are the odds offered.

So $0.8=\frac1\to1+w=\frac54\to 4w=1\to w=0.25$ and this gives odds of $4$ to $1$ on (stake £1, return £1.25).

And, $0.4=\frac1\to1+w=\frac52\to 2w=3\to w=1.5$ and this gives odds of $3$ to $2$ against (stake £1, return £2.50).

An obvious scenario would be $P(\text)=0.2$:

And, $0.2=\frac1\to1+w=5\to w=4$ and this gives odds of $4$ to $1$ against (stake £1, return £5.00).

So Dave's on a loser because if he had correctly evaluated the probabilities, he could have got much better odds.