Basic probability

Probability (P) is defined as the number of things we'd like to have happen (like a flipped coin coming up heads) divided by the total number of things that can happen.

$$P = \frac{\text{no. of desired outcomes}}{\text{no. of possible outcomes}},$$

where an outcome is one possible result of what we'll call a probability experiment.

In this section, we'll talk a lot about coins and dice, because they're good simple models or probability experiments.

If we'd like to have a fair coin (one with equal likelihood of landing heads or tails) come up heads, we have a 1 in 2, or $\frac{1}{2}$ chance of that happening. We're interested in one thing happening, and a total of two things or outcomes can happen.

The probability of each fair coin outcome is $\frac{1}{2}.$

Similarly, if we roll a six-sided die ("die" is the singular of dice), the probability of rolling a one, or any one of the six numbers, as long as we specify it first, is one out of 6, or 1:6 or $\frac{1}{6}.$

The probability of rolling a given number on a six-sided die is $\frac{1}{6}.$

Discrete probability

These are examples of discrete probability experiments. Discrete in this context means that the result can be one thing or another, but nothing in between. A coin can fall with heads up (H), or tails up (T). If a coin somehow magically landed on its edge and stayed that way, we be amazed, but we wouldn't cout that as a result. Likewise, a 6-sided die can show only one number at a time on its top face: 1, 2, 3, 4, 5 or 6, but never something like 3.4. Those are discrete results.

The other kind of probability is continuous, in which a probability variable can take on any value at all within some range. An example might be if we measured the heights of all adult men in the U.S. We would find a more or less continuous range between about 3 feet and about 8 feet. People aren't required to have heights rounded to the nearest inch.


Here is a list of some of the most important terms we'll use in our discussions of probability.

We'll generally refer to a probability with the letter P, and we'll define various outcomes with single letters or numbers, so the probability of flipping a coin and finding heads or tails will be P(H) and P(T), respectively. Likewise, the probability of rolling a 4 on a single die will be P(4), and so on.

Term Definition
Experiment A probability experiment is any process that can lead to any of two or more outcomes with the process being random, or unpredictable until the result is finalized, such as a tossed coin coming to rest flat on a table.
Outcome The outcome of a probability experiment is its result. For tossing a 6-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. Only one outcome is the result.

The probability of an outcome or set of outcomes is the number of those outcome(s) divided by the number of all possible outcomes:

$$P = \frac{\text{no. of desired outcomes}}{\text{no. of possible outcomes}}$$

Likelihood is often used instead of probability.

Discrete A discrete probability experiment can have only certain outcomes with nothing in between. A 6-sided die can have outcomes of 1, 2, 3, 4, 5 or 6, and no number in between, such as 2.6.
Continuous A continuous probability experiment (not discussed in this section) can take on a continuum of results. For example, if we measured lengths of 5.25 cm and 5.26 cm, it would also be possible to measure a length of 5.255 cm. Continuous probability variables can take on any value within a given range.
Random For our purposes, random means that all similar outcomes will have the same probability. The probabilities of getting heads or tails are the same. The probabilities of rolling digits 1-6 on a six-sided die are all the same.
Fair When we do a probability experiment, we assume it's "fair." That is, we don't use a loaded coin or loaded dice, doctored so that they might have a slightly higher probability of landing in a certain way. We'll refer to "fair coins" or "fair dice."
Total probability The total probability of any outcome at all of a probability experiment is 1. That is, if we do the experiment, something has to happen. If we toss a coin, unless it rolls down into the sewer, it has to come up heads or tails.
Independence Two outcomes of probability experiments are independent if the outcome of the second experiment is unrelated to the outcome of the first. For example, on flipping a coin, it makes no difference if heads was rolled before. The probability of heads on the current roll is still ½.

Coin flipping in detail

A coin flipping experiment can have two outcomes, heads (H) or tails (T). The probabilities of each are 1 in 2 or 1:2 or 50%, but we'll generally use fractions:

$$P(H) = \frac{1}{2} \; \; \text{ and } \; \; P(T) = \frac{1}{2}$$

Probability of sequential outcomes

Now let's do some more interesting experiments with a coin. What is the probability of flipping a coin and coming up heads both times? We'll call that P(HH), but there's no standard notation for it; you can use your own if you'd like. Here is a diagram of the possible outcomes:

The chart shows that the result of the first toss can be H or T. After the next toss, we have all of our possible outcomes, HH (two heads in a row), TH and HT (which we can consider to be the same if order doesn't matter to us), and TT. The probabilities are:

$$P(HH) = \frac{1}{4}, \; \; \; P(TH) = \frac{1}{2}, \; \; \; P(TT) = \frac{1}{4}$$

In this case, we take P(TH) to be the sum of P(TH) = ½ and P(HT) = ½, though order of outcomes can be important sometimes (more later). So the probability of tossing two heads in a row is $\frac{1}{4} = 25\%.$ How about three heads? Here's the next diagram:

Three coin flips lead to six possible outcomes. In this case, we'll again decide not to care about the order of the results (the second column from the right has them in order), so there are basically four (last column): HHH, HHT, TTH and TTT. Their probabilities are

$$ \begin{matrix} P(HHH) = \frac{1}{8} && P(TTT) = \frac{1}{8} \\ P(HHT) = \frac{3}{8} && P(HHT) = \frac{3}{8} \end{matrix}$$

Notice that the sum of all the probabilities is $\frac{8}{8} = 1.$ Now the probabilities we've calculated so far are

$$P(H) = \frac{1}{2}, \; \; \; P(HH) = \frac{1}{4}, \; \; \; P(HHH) = \frac{1}{8}$$

for one, two and three flips, respectively. Notice that each successive probability is obtained by multiplying the previous one by $P(H) = \frac{1}{2},$ so that the probability of tossing n heads in a row is

$$P(nH) = \left( \frac{1}{2} \right)^n.$$

Probability of sequential events

The probability that one event, E, will happen n times in a row is

$$P(nE) = \left[P(E)\right]^n$$

Another way to write this (for just two sequential events, E) is

$$P(E \cap E) = P(E) \times P(E)$$

where $\cap$ means "and." In this context, the word "and" means multiply.

Example 1

Calculate the probability of rolling a 2 and a 3 (in that order) with two rolls of a six-sided die.

Solution: The probability of rolling a 2 is 1 in 6, or $P(2) = \frac{1}{6}.$ The probability of rolling a 3 is the same, an independent event. The probability of doing both in the correct order is the product of those probabilities:

$$P(2\cap 3) = \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$$

If we were to specify two rolls that resulted in a 2 and a 3, but in either order (two ways of getting the result, we'd need to multiply the result by 2, for a probability of $\frac{1}{2}.$

Example 2

Calculate the probability of simultaneously rolling five dice and getting all sixes.

Solution: The probability of rolling a six on a single die is $P(6) = \frac{1}{2}.$ Rolling five dice at once is the same as rolling one die five times and collecting the results; all events are still independent. The probability of rolling five sixes is:

$$P(6 \cap 6 \cap 6 \cap 6 \cap 6) = \left(\frac{1}{6}\right)^5 = \frac{1}{1776}$$

Example 3

Calculate the probability of two flips of a fair coin coming up the same. How about three flips?

Solution: This is a tricky one. Notice that we didn't specify whether the first flip should be heads or tails. It could be either, so we get the first flip for free. Then the probability of the second flip mathching the first would be $P = \frac{1}{2}.$ For three flips, we still get the first for free, then the probability of the next two being the same is $P = \frac{1}{4}.$

Adding probabilities

Here's a different kind of question about probability: If a fair coin is flipped three times, what is the probability that exactly two of the flips will yield heads?

In this case, there are three possible outcomes that include exactly two heads, they are

Each of these outcomes, as written, has a probability of

$$P(HHT) = P(HTH) = P(THH) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$$

But the probability we seek is not one of those, but a combination of all three. In other words, we have three chances to reach any of those three outcomes.

Note: I'm not using the $\cap$ sign here, but it's implied.

The total probability in this case is

$$P = P(HHT) + P(HTH) + P(THH) = \frac{3}{8}$$

Now we can also think of this as the probability of getting any of the three specified outcomes, such as $P(HHT) = \frac{1}{8},$ and multiplying it by the number of ways of rearranging those three outcomes: HHT, HTH, THH.

$$P = \left( \frac{1}{8} \right) \times 3 = \frac{3}{8}$$

Now let's look further into this notion. It should be clear that with more and more coin flips, it could be cumbersome to figure out all of the ways of, say, obtaining five heads from ten flips.

When there is more than one way to obtain an outcome, sum the probability of obtaining one of them, or just multiply that probability by the number of ways.

Probability of $n$ outcomes out of $m$ tries

Now let's stick with coin flipping and ask about the probability of obtaining n heads from m flips of the coin. We've already done this experiment with two flips, and with three in the last example, but let's start with three and look at all possible outcomes. They are listed below, in a more-or-less sensible order from HHH to TTT.

There are 8 possible combinations of three coin tosses, which turns out to be the number of possible outcomes (2) raised to the power of the number of tosses:

$$2^3 = 8$$

The probability of rolling three heads in a row is just the one way of doing it divided by 8:

$$P(HHH) = \frac{1}{8}$$

If the order of the flips doesn't matter (that is, HHT is considered to be the same as HTH and THH), then the other probabilities are

$$ \begin{matrix} P(HHH) = \frac{1}{8} && P(HHT) = \frac{3}{8} \\ \\ P(HTT) = \frac{3}{8} && P(TTT) = \frac{1}{8} \end{matrix}$$

We could also write those as:

$$ \begin{matrix} P(3H) = \frac{1}{8} && P(2H) = \frac{3}{8} \\ \\ P(1H) = \frac{3}{8} && P(0H) = \frac{1}{8} \end{matrix}$$

Notice that the sum of the probabilities is 1. Now let's repeat the process for four flips of one coin and look at the probabilities. The results can be written like this:

The probabilities (again assuming that order doesn't matter) are:

$$ \begin{matrix} P(4H) = \frac{1}{16} && P(3H) = \frac{1}{4} \\ \\ P(2H) = \frac{3}{8} && P(1H) = \frac{1}{4} \\ \\ P(0H) = \frac{1}{16} \end{matrix}$$

You can check for yourself that these probabilities also sum to 1. That means that if we flip a coin four times something (meaning one of the 16 possible outcomes) has to happen.

Finally, let's look at five throws and the outcome of that experiment.

There are 32 possible combinations, which turns out to be the number of possible outcomes (2) raised to the power of the number of experiments, 5.

$$2^5 = 32$$

The probabilities of obtaining 5, 4, 3, 2, 1 and zero heads are:

$$ \begin{matrix} P(5H) = \frac{1}{32} && P(4H) = \frac{5}{32} \\ \\ P(3H) = \frac{10}{32} && P(2H) = \frac{10}{32} \\ \\ P(H) = \frac{5}{32} && P(0H) = \frac{1}{32} \end{matrix}$$

Notice that the probability of rolling the precise sequence HHTHH is $\frac{1}{32},$ and it's the same for every unique sequence of five flips. So there's a difference in probability if we care about order. For our purpose here, we'll continue ignoring the order.

What's the pattern?

OK, there has to be a pattern here, so what is it? We're interested in knowing the probability of throwing n heads in m rolls of a single die. We've seen that the probability of rolling all heads or all tails is just $P(nH)^m,$ the probability of rolling one head raised to the power of the number of throws. What about mixtures of heads and tails?

Notice that in all of our lists of outcomes above, the only difference in the groups, say three heads and one tail, for example, was the order of those, and we specified that the order doesn't matter. In other words, HHHT, HHTH, HTHH and THHH are all the same. The probability of these occuring is larger than pure H or T cases because they are more numerous. So the question we need to answer is:

How many ways can $n$ elements consisting of two types be arranged?

The solution to that has been worked out in the binomial coefficient, which you can learn more about here. The binomial coefficient is

$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$

It is the number of ways of arranging $k$ things within a group of $n$ things. Let's take a look at how it figures into our examples. When we toss our coin three times, the basic probability is

$$P(HHH) = \left( \frac{1}{2} \right)^3 = \frac{1}{8},$$

but the probability of getting HHT or one of its other permutations, HTH or THH, is different because there are more ways of arriving at the combination. Using the binomial coefficient, we can calculate the number of ways:

$$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2} = 3.$$

We already knew there were three combinations, but we'll eventually be able to apply this method to much larger groupings and not have to write out all of the combinations. So the proability of throwing two heads and one tail is

$$P(HHT) = \left(\frac{1}{2} \right)^3 \, \binom{3}{2} = \frac{3}{8},$$

which matches our table above.

Let's work all the way through our four-toss scenario using our formula,

$$P = [P(H)]^n \binom{n}{k}$$

where in this case, n = 4 and will be 0, 1, 2, 3 and 4. First P(HHHH):

$$P(HHHH) = \left( \frac{1}{2} \right)^4 \binom{4}{0} = \frac{1}{16} \cdot \frac{4!}{4!} = \frac{1}{16}$$

$$P(HHHT) = \left( \frac{1}{2} \right)^4 \binom{4}{1} = \frac{1}{16} \cdot \frac{4!}{1!(3!)} = \frac{1}{4}$$

$$P(HHTT) = \left( \frac{1}{2} \right)^4 \binom{4}{2} = \frac{1}{16} \cdot \frac{4!}{2!(2!)} = \frac{3}{8}$$

$$P(HTTT) = \left( \frac{1}{2} \right)^4 \binom{4}{3} = \frac{1}{16} \cdot \frac{4!}{3!(1!)} = \frac{1}{4}$$

$$P(TTTT) = \left( \frac{1}{2} \right)^4 \binom{4}{4} = \frac{1}{16} \cdot \frac{4!}{4!(0!)} = \frac{1}{16}$$

Here's a graphical representation of those probabilities. The height of each bar represents the proportion of the total probability, but also the number of ways of arranging the five outcomes, given that some are H and some T.

Probability vs. outcome, 4 tosses

Notice that the probabilities sum to one: $P_{total} = 2\left( \frac{1}{16} \right) + 2 \left( \frac{1}{4} \right) + \frac{3}{8}.$

The five-toss case

Click here to see the math and graph of the probabilites of outcomes of five tosses of a coin

The possibilities and probabilities for the outcomes of tossing a coin 5 times are*:

$$P_{TTTTT} = [P(H)]^5 \binom{5}{0} = \frac{1}{32} \cdot \frac{5!}{5!0!} = \frac{1}{32}$$

$$P_{HTTTT} = [P(H)]^5 \binom{5}{1} = \frac{1}{32} \cdot \frac{5!}{4!} = \frac{5}{32}$$

$$P_{HHTTT} = [P(H)]^5 \binom{5}{2} = \frac{1}{32} \cdot \frac{5!}{2!3!} = \frac{10}{32}$$

$$P_{HHHTT} = [P(H)]^5 \binom{5}{3} = \frac{1}{32} \cdot \frac{5!}{3!2!} = \frac{10}{32}$$

$$P_{HHHHT} = [P(H)]^5 \binom{5}{4} = \frac{1}{32} \cdot \frac{5!}{1!4!} = \frac{5}{32}$$

$$P_{HHHHH} = [P(H)]^5 \binom{5}{5} = \frac{1}{32} \cdot \frac{5!}{0!5!} = \frac{1}{32}$$

*I've used subscripts on the P's to save horizontal space, that's all.

Here are those probabilities in a graph

Probability vs. outcome, 5 tosses

Notice that the sum of the probabilities (each directly proportional to the number of ways of arranging 5 outcomes, some of which are heads, some tails) is 1:

$$P_{total} = 2\left( \frac{1}{32} \right) + 2\left( \frac{5}{32} \right) + 2\left( \frac{10}{32} \right) = 1$$

Two-dice combinations

Now let's look at throwing two dice. First let's break down the probabilities of throwing every possible two-dice sum, from 2 to 12. The probability of rolling any digit on a single die is $P = \frac{1}{6}.$ The probability of rolling a specified number on both dice, in the order given, such as P(1, 2) or P(6, 6), is

$$P = \left( \frac{1}{6} \right)^2 = \frac{1}{36}$$

There is only one way to roll a two, both dice have to show 1, so the probability is $P(2) = \frac{1}{36}.$

There are two ways to roll a three. 1,2 or 2,1, so we'll multiply the basic probability, $\frac{1}{36}$ by the appropriate binomial coefficient:

$$P(3) = \frac{1}{36} \cdot \binom{2}{1} = \frac{1}{36} \cdot \frac{2!}{1! 1!} = \frac{2}{36} = \frac{1}{18}$$

In a similar way we can find the probabilities of all other sums of two dice:

There are three ways to roll 4: 1+3, 2+2 and 3+1, so we have

$$P(4) = \frac{1}{36} \binom{3}{1} = \frac{1}{36} \cdot \frac{3!}{1!2!} = \frac{3}{36} = \frac{1}{12}$$

There are four ways to roll a sum of 5: 1+4, 2+3, 3+2 and 4+1, so P(5) is

$$P(5) = \frac{1}{36} \binom{4}{1} = \frac{1}{36} \cdot \frac{4!}{1!3!} = \frac{4}{36} = \frac{1}{9}$$

There are five ways to roll a sum of 6: 1+5, 2+4, 3+3, 4+2 and 5+1, so P(6) is

$$P(6) = \frac{1}{36} \binom{5}{1} = \frac{1}{36} \cdot \frac{5!}{1!4!} = \frac{5}{36}$$

There are six ways to roll a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2 and 6+1, so P(7) is

$$P(7) = \frac{1}{36} \binom{6}{1} = \frac{1}{36} \cdot \frac{6!}{1!5!} = \frac{6}{36} = \frac{1}{6}$$

The probabilities of rolling sums of 8, 9, 10, 11 and 12 are found similarly:

$$P(8) = \frac{1}{36} \binom{5}{1} = \frac{1}{36} \cdot \frac{5!}{1!4!} = \frac{5}{36}$$

$$P(9) = \frac{1}{36} \binom{4}{1} = \frac{1}{36} \cdot \frac{4!}{1!3!} = \frac{4}{36} = \frac{1}{9}$$

$$P(10) = \frac{1}{36} \binom{3}{1} = \frac{1}{36} \cdot \frac{3!}{1!2!} = \frac{3}{36} = \frac{1}{12}$$

$$P(11) = \frac{1}{36} \binom{2}{1} = \frac{1}{36} \cdot \frac{2!}{1!1!} = \frac{2}{36} = \frac{1}{18}$$

$$P(12) = \frac{1}{36} \binom{1}{1} = \frac{1}{36} \cdot \frac{1!}{1!0!} = \frac{1}{36}$$

Here is a chart of those probabilities

Probabilities of 2-dice sums, 2-12

Click here for a different table of two-dice probabilities.

Example 4

Calculate the probability of rolling two consecutive sums of seven in a roll of two dice.

Solution: As we established above, the probability of rolling a sum of 7 on two dice is $P(7) = \frac{1}{6}.$ So the probability of doing it twice is $P(77) = \left( \frac{1}{6} \right)^2 = \frac{1}{36}.$

Example 4

Calculate the probability of rolling a sum of 9 on two dice AND rolling a four on one of those dice.

Solution: Let's first look at a diagram showing all of the possible outcomes of rolling two dice, in which we highlight the events of interest.

In the diagram, the rolls with at least one four are highlighted in green. The probability that a four will come up is

$$P(4_1) \cup P(4_2) - P(4_1) \cap P(4_2) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{11}{36}$$

where $\cup$ means the "union" of two sets, and implies addition of probabilities, and $\cap$ means "intersection."

In this case, that's the 4,4 outcome. Unless we subtract one copy of that outcome, we'd over-count it. The probability of two dice adding to 9, highlighted in magenta in the chart, is

$$P(9) = \frac{4}{36} = \frac{1}{9}$$

Now what is the probability that both events will happen? That's easy to see from the chart.. The blocks holding dice that both sum to 9 and have at least one 4 are highlighted in both green and magenta. Those represent two out of 36 possible outcomes, so the probability is $P = \frac{1}{18}.$

Mathematical representation

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