In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

$\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}$

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

$\displaystyle P=\frac{1}{6}$

Now, let's convert this into a percentage:

$\displaystyle \frac{1}{6}=0.1666$

$\displaystyle 0.1666\times100\%=16.66\%$

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

$\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%$

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

$\displaystyle E(x)=x_{1}\times p_{1}+x_{2}\times p_{2}+. . . x_{n}\times p_{n}$

In this equation the variables are identified as the following:

$\displaystyle x=\textup{event}$

$\displaystyle p=\textup{probability of event occurring}$

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

$\displaystyle E(x)=x_{10}\times p_{10}+x_{8}\times p_{8}+ x_{6}\times p_{6}+x_{7}\times p_{7}+x_{0}\times p_{0}+x_{9}\times p_{9}$

$\displaystyle E(x)=10\times 0.01+8\times 0.20+ 6\times 0.30+7\times 0.25+0\times 0.09+9\times 0.15$

$\displaystyle E(x)=0.1+1.6+1.8+1.75+0+1.35$

$\displaystyle E(x)=6.6$

It is important to note that we cannot solve this equation using the following formula:

$\displaystyle E(x)=\frac{1}{6}\times (10+8+6+7+0+9)$

$\displaystyle E(x)=\frac{40}{6}$

$\displaystyle E(x) \neq 6.\bar{6}$

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.