Step 1: When we know that the defendant has been charged and brought to trial, both the probability of “Verdict” = “Guilty” and the probability of “guilty” $=$ “Yes” increase, as shown in Figure 16.20 (this also reflects the empirical knowledge that most defendants brought to trial for a serious crimes are eventually found guilty – the UK figure in 2011 was $81 \%$ ).

Step 2: We hear that the jury Verdict = “Innocent” (Figure 16.21). Note that the probability of not guilty jumps to $92.59 \%$.

Step 3: Now we discover that the defendant has a previous conviction for a similar crime, so enter “Yes” in that node, as in Figure 16.22. The probability of “not guilty” does not decreaserather, it increases slightly to $94.97 \%$. There is also a small decrease in the probability of the hard evidence. Essentially, the fact that the defendant was charged in the first place is explained away more by the previous conviction than the evidence. It can be shown that the fallacy “holds” for a wide range of sensitivity.

Suppose we know the following:
Fact 1 : item $\mathrm{X}$ was found at the crime scene that is almost certainly linked to the crime. Fact 2 : item $\mathrm{X}$ could belong to the defendant.
Fact 3: despite many public requests (including, e.g., one on the BBC Crimewatch UK program) for information for an innocent owner of item $\mathrm{X}$ to come forward and clear themselves, nobody has done so.

If we assumed that a fair die genuinely had exactly six equally likely elementary events $(1,2,3,4,5,6)$, then after many throws the proportion of times we throw a 4 will get increasingly close to $1 / 6$. Hence, we can reasonably state that the chance of getting a 4 in a throw of a fair die is 1 in 6 .

Consider an urn containing 50 equally sized and weighted balls of which 30 are known to be red, 15 are known to be white, and 5 are known to be black. Assume a ball is drawn randomly from the urn and then replaced. The event “next ball drawn from the urn” has three elementary events whose respective chances are
Red-(30 in 50$)$ or equivalently $60 \%$
White-(15 in 50 ) or equivalently $30 \%$
Black-(5 in 50$)$ or equivalently $10 \%$
With the assumptions of repeatability and independence, we can therefore state the frequentist definition of chance as follows:In the special case where all of the elementary events are equally likely (such as in the roll of a fair die) the frequentist definition of chance of an outcome is especially simple:

The rule for combinations (see Appendix A) explains why there are $13,983,816$ possible elementary events in a lottery that involves selecting 6 numbers from 49 (until 2016 the UK National Lottery had this format). Since it is reasonable to assume that each of these combinations of 6 numbers is equally likely to be chosen as the jackpot winner in any one draw, it follows that the chance of any single ticket winning the jackpot is
$$
\frac{1}{13,983,816} \text {. }
$$
Box 4.5 looks at a similar example with a pack of cards.

What is the chance that the jackpot winner in a 6 from 49 ball lottery consists entirely of even numbers?