Uniform Probability (by Outcomes)

Probability by outcomes is a probability obtained from a well-defined experiment in which all outcomes are equally likely. An example of this would be flipping a fair coin. It is known that there are two possible outcomes to this experiment: "heads" and "tails." It is also known that each outcome is equally likely, since the coin is fair.

A probability experiment is something that has an uncertain result. In the context of probability, this is often just referred to as an experiment.

An experiment could be rolling a fair 6-sided die, or flipping a fair coin. In either case, the result is random, and we cannot predict what it will be. In the field of science, we often think of "experiments" as things that we control in a lab. However, in probability theory, an experiment need not be something that we control. For example, a probability experiment could be the weather tomorrow.

Even though it's not possible to know what the result of an experiment will be, it can be very helpful to analyze and understand what could possibly happen. Something that could possibly happen is called an outcome.

An outcome is a possible result of an experiment.

A possible outcome of rolling a fair 6-sided die is \(4\). A possible outcome of flipping a fair coin is "heads." Both of these experiments have relatively few possible outcomes, and so it is easy to think of all the possible outcomes. However, there are some experiments with relatively massive amounts of potential outcomes. Think of a lottery as a probability experiment. There are millions of potential winners in a lottery, and so there are millions of potential outcomes of the experiment. It would be very difficult to consider every possible winner of the lottery individually.

The set of all outcomes in an experiment is called the sample space of that experiment.

The sample space of rolling a fair 6-sided die is \(\{1,2,3,4,5,6\}\). The sample space of flipping a fair coin is \(\{\text{heads},\text{tails}\}\). The sample space of a lottery would be too large to list out here.

In probability theory, we often group outcomes together in order to make analyzing the sample space more meaningful. As it was mentioned earlier, it would be impossible to list the sample space of a lottery with millions of participants. However, we could have a discussion about certain parts of that sample space. For example, the potential winners who are from a certain city, or the potential winners who are women over the age of 65. These "parts" of the sample space are called events.

An event is a subset of the sample space.

We could define \(E=\text{an even number is rolled on a fair 6-sided die}\). In this case, \(E=\{2,4,6\}\), which is a subset of the sample space, \(\{1,2,3,4,5,6\}\).

In the sample space of a fair coin flip, each outcome, "heads" or "tails," is just as likely as the other. Likewise, in the sample space of fair 6-sided die rolls, each roll is just as likely as the other. These sample spaces are called uniform.

A sample space is uniform if all outcomes are equally likely.

If the sample space of a given experiment is known to be uniform, then the probability of an event can be found with the sizes of the event and the sample space:

Suppose there is an experiment with sample space \(S\), and \(X\) is an event in that sample space. If the experiment is performed many times, then the probability of \(X\) is the expected proportion of times that any outcome in \(X\) will happen.

The probability of \(X\) is denoted by \(P(X)\).

If \(S\) is uniform (all outcomes in \(S\) are equally likely), then the probability of \(X\) is the size of \(X\) divided by the size of \(S\):

\(P(X)=\dfrac{|X|}{|S|}\)

A deck of ten cards labeled with numbers 1 through 10 is shuffled, and a card is drawn. What is the probability an even-numbered card is drawn?

---

The sample space of this experiment is \(S = \{1,2,3,4,5,6,7,8,9,10\}\), and \(|S|=10\).

Let \(X\) be the event of drawing an even-numbered card. Then \(X = \{2,4,6,8,10\}\), and \(|X|=5\).

All of the outcomes in this experiment are equally likely, so we can use the above formula:

\(P(X) = \dfrac{|X|}{|S|}=\dfrac{5}{10} = \dfrac{1}{2}\).

The probability of drawing an even-numbered card is \(\boxed{\dfrac{1}{2}}\).

A die is rolled and a coin is tossed. Find the probability that the die shows an odd number and the coin shows a head.

---

First, determine the sample space, or all of the possible outcomes of the problem. The sample space S of the experiment is as follows: \[S = \{ (1,H),(2,H),(3,H),(4,H),(5,H),(6,H), ~~ (1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}.\]

Let \(E\) be the event "the die shows an odd number and the coin shows a head." Then event \(E\) consists of the following outcomes: \[E=\{(1,H),(3,H),(5,H)\}.\] Thus, the probability \(P(E)\) equals \[P(E) = \frac{|E|}{|S|} = \frac{3}{12} = \frac{1}{4}. \]

The examples given thus far have had relatively small sample spaces. For small sample spaces, it's a simple exercise to list out all possible outcomes, and then count the size of the sample space and the events in it. However, it is common for sample spaces to be too large to list exhaustively. The binomial coefficient can often be used to count the sizes of large sample spaces and events without having to list out outcomes.

Problems involving a standard playing card deck typically use binomial coefficients to find the size of sample spaces and events.

A standard playing card deck, also called a poker deck, contains 52 distinct cards.

These cards are divided into four suits:

Hearts and Diamonds are the two red suits. These are sometimes abbreviated as H and D.

Clubs and Spades are the two black suits. These are sometimes abbreviated as C and S.

There are 13 ranks in each suit: An Ace, nine cards numbered \(2\) through \(10\), and three face cards: the Jack, the Queen, and the King.

The face cards are abbreviated as J, Q, and K. The Ace is abbreviated as A.

---

Image Credit: Last-Dino

Image Credit: Trainler

When counting possible outcomes using the binomial coefficient or other means, it's important to apply the rule of sum and rule of product counting principles.

A player is dealt five cards from a shuffled poker deck. What is the probability of getting four aces among those cards?

---

The sample space consists of all five card hands that can be drawn from 52 cards, without regard to order. This sample space has \(\binom{52}{5} = 2,598,960\) outcomes in it.

Let \(F\) be the event consisting of all hands with four aces. How many outcomes are in \(F\)? If four of the cards in the five-card hand are aces, the fifth card can be any of the remaining 48 cards. Therefore (without regard to order) \(F\) has 48 outcomes in it. Therefore,

\[P(F) = \frac{48}{2,598,960}.\]

This is approximately 0.0000185. (Very small, as we would expect.)

Two cards are drawn from a shuffled standard playing card deck. What is the probability the cards are the same suit?

---

We will solve this question in the following two ways:

Solution 1. There are \(\binom{52}{2} = 1326\) ways to choose 2 cards from the deck. There are \(\binom{13}{2} = 78\) ways to choose 2 cards that are both hearts. There are the same number of ways to choose 2 cards that are both diamonds, clubs, or spades. So the probability is \(\frac{78 + 78 + 78 + 78}{1326} = \frac{4}{17}.\)

Solution 2. The first card can be anything. No matter what it is, there are 12 cards that are the same suit with 51 total cards remaining. So, the probability the two cards have the same suit is \(\frac{12}{51} = \frac{4}{17}.\)

The principles learned from playing card problems can be applied to other problems.