structural:summarization
Equally likely outcomes

Some treatments of probability assume that the various outcomes of an experiment are always defined so as to be equally likely. For any sample space with N {\displaystyle N} equally likely outcomes, each outcome is assigned the probability 1 N {\displaystyle {\frac {1}{N}}}. However, there are experiments that are not easily described by a sample space of equally likely outcomes—for example, if one were to toss a thumb tack many times and observe whether it landed with its point upward or downward, there is no physical symmetry to suggest that the two outcomes should be equally likely.

Though most random phenomena do not have equally likely outcomes, it can be helpful to define a sample space in such a way that outcomes are at least approximately equally likely, since this condition significantly simplifies the computation of probabilities for events within the sample space. If each individual outcome occurs with the same probability, then the probability of any event becomes simply:

For example, if two fair six-sided dice are thrown to generate two uniformly distributed integers, D 1 {\displaystyle D_{1}} and D 2 {\displaystyle D_{2}}, each in the range from 1 to 6, inclusive, the 36 possible ordered pairs of outcomes (D 1, D 2) {\displaystyle (D_{1},D_{2})} constitute a sample space of equally likely events. In this case, the above formula applies, such as calculating the probability of a particular sum of the two rolls in an outcome. The probability of the event that the sum D 1 + D 2 {\displaystyle D_{1}+D_{2}} is five is 4 36 {\displaystyle {\frac {4}{36}}}, since four of the thirty-six equally likely pairs of outcomes sum to five.

If the sample space was all of the possible sums obtained from rolling two six-sided dice, the above formula can still be applied because the dice rolls are fair, but the number of outcomes in a given event will vary. A sum of two can occur with the outcome { (1, 1) } {\displaystyle \{(1,1)\}}, so the probability is 1 36 {\displaystyle {\frac {1}{36}}}. For a sum of seven, the outcomes in the event are { (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) } {\displaystyle \{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\}}, so the probability is 6 36 {\displaystyle {\frac {6}{36}}}.