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7.6: Basic Concepts of Probability

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Learning Objectives

After completing this section, you should be able to:

• Define probability including impossible and certain events.
• Calculate basic theoretical probabilities.
• Calculate basic empirical probabilities.
• Distinguish among theoretical, empirical, and subjective probability.
• Calculate the probability of the complement of an event.

It all comes down to this. The game of Monopoly that started hours ago is in the home stretch. Your sister has the dice, and if she rolls a 4, 5, or 7 she’ll land on one of your best spaces and the game will be over. How likely is it that the game will end on the next turn? Is it more likely than not? How can we measure that likelihood? This section addresses this question by introducing a way to measure uncertainty.

Introducing Probability

Uncertainty is, almost by definition, a nebulous concept. In order to put enough constraints on it that we can mathematically study it, we will focus on uncertainty strictly in the context of experiments. Recall that experiments are processes whose outcomes are unknown; the sample space for the experiment is the collection of all those possible outcomes. When we want to talk about the likelihood of particular outcomes, we sometimes group outcomes together; for example, in the Monopoly example at the beginning of this section, we were interested in the roll of 2 dice that might fall as a 4, 5, or 7. A grouping of outcomes that we’re interested in is called an event . In other words, an event is a subset of the sample space of an experiment; it often consists of the outcomes of interest to the experimenter.

Once we have defined the event that interests us, we can try to assess the likelihood of that event. We do that by assigning a number to each event ( E E ) called the probability of that event ( P ( E ) P ( E ) ). The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur.

Example 7.16

Determining certain and impossible events.

Consider an experiment that consists of rolling a single standard 6-sided die (with faces numbered 1-6). Decide if these probabilities are equal to zero, equal to one, or somewhere in between.

• P ( roll a 4 ) P ( roll a 4 )
• P ( roll a 7 ) P ( roll a 7 )
• P ( roll a positive number ) P ( roll a positive number )
• P ( roll a 1 3 ) P ( roll a 1 3 )
• P ( roll an even number ) P ( roll an even number )
• P ( roll a single-digit number ) P ( roll a single-digit number )

Let's start by identifying the sample space. For one roll of this die, the possible outcomes are {1, 2, 3, 4, 5,6}. We can use that to assess these probabilities:

• We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So, 0 < P ( roll a 4 ) < 1 0 < P ( roll a 4 ) < 1 .
• Notice that 7 is not in the sample space. So, P ( roll a 7 ) = 0 P ( roll a 7 ) = 0 .
• Every outcome in the sample space is a positive number, so this event is certain. Thus, P ( roll a positive number ) = 1 P ( roll a positive number ) = 1 .
• Since 1 3 1 3 is not in the sample space, P ( roll a 1 3 ) = 0 P ( roll a 1 3 ) = 0 .
• Some outcomes in the sample space are even numbers (2, 4, and 6), but the others aren’t. So, 0 < P ( roll an even number ) < 1 0 < P ( roll an even number ) < 1 .
• Every outcome in the sample space is a single-digit number, so P ( roll a single-digit number ) = 1 P ( roll a single-digit number ) = 1 .

Your Turn 7.16

Three ways to assign probabilities.

The probabilities of events that are certain or impossible are easy to assign; they’re just 1 or 0, respectively. What do we do about those in-between cases, for events that might or might not occur? There are three methods to assign probabilities that we can choose from. We’ll discuss them here, in order of reliability.

Method 1: Theoretical Probability

The theoretical method gives the most reliable results, but it cannot always be used. If the sample space of an experiment consists of equally likely outcomes, then the theoretical probability of an event is defined to be the ratio of the number of outcomes in the event to the number of outcomes in the sample space.

For an experiment whose sample space S S consists of equally likely outcomes, the theoretical probability of the event E E is the ratio

P ( E ) = n ( E ) n ( S ) , P ( E ) = n ( E ) n ( S ) ,

where n ( E ) n ( E ) and n ( S ) n ( S ) denote the number of outcomes in the event and in the sample space, respectively.

Example 7.17

Computing theoretical probabilities.

Recall that a standard deck of cards consists of 52 unique cards which are labeled with a rank (the whole numbers from 2 to 10, plus J, Q, K, and A) and a suit ( ♣ ♣ , ♢ ♢ , ♡ ♡ , or ♠ ♠ ). A standard deck is thoroughly shuffled, and you draw one card at random (so every card has an equal chance of being drawn). Find the theoretical probability of each of these events:

• The card is 10 ♠ 10 ♠ .
• The card is a ♡ ♡ .
• The card is a king (K).

There are 52 cards in the deck, so the sample space for each of these experiments has 52 elements. That will be the denominator for each of our probabilities.

• There is only one 10 ♠ 10 ♠ in the deck, so this event only has one outcome in it. Thus, P ( 10 ♠ ) = 1 52 P ( 10 ♠ ) = 1 52 .
• There are 13 ♡ s ♡ s in the deck, so P ( ♡ ) = 13 52 = 1 4 P ( ♡ ) = 13 52 = 1 4 .
• There are 4 cards of each rank in the deck, so P ( K ) = 4 52 = 1 13 P ( K ) = 4 52 = 1 13 .

Your Turn 7.17

It is critical that you make sure that every outcome in a sample space is equally likely before you compute theoretical probabilities!

Example 7.18

Using tables to find theoretical probabilities.

In the Basic Concepts of Probability, we were considering a Monopoly game where, if your sister rolled a sum of 4, 5, or 7 with 2 standard dice, you would win the game. What is the probability of this event? Use tables to determine your answer.

We should think of this experiment as occurring in two stages: (1) one die roll, then (2) another die roll. Even though these two stages will usually occur simultaneously in practice, since they’re independent, it’s okay to treat them separately.

Step 1: Since we have two independent stages, let’s create a table (Figure 7.27), which is probably the most efficient method for determining the sample space.

Now, each of the 36 ordered pairs in the table represent an equally likely outcome.

Step 2: To make our analysis easier, let’s replace each ordered pair with the sum (Figure 7.28).

Step 3: Since the event we’re interested in is the one consisting of rolls of 4, 5, or 7. Let’s shade those in (Figure 7.29).

Our event contains 13 outcomes, so the probability that your sister rolls a losing number is 13 36 13 36 .

Your Turn 7.18

Example 7.19, using tree diagrams to compute theoretical probability.

If you flip a fair coin 3 times, what is the probability of each event? Use a tree diagram to determine your answer

• You flip exactly 2 heads.
• You flip 2 consecutive heads at some point in the 3 flips.
• All 3 flips show the same result.

Let’s build a tree to identify the sample space (Figure 7.30).

The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which has 8 elements.

• Flipping exactly 2 heads occurs three times (HHT, HTH, THH), so the probability is 3 8 3 8 .
• Flipping 2 consecutive heads at some point in the experiment happens 3 times: HHH, HHT, THH. So, the probability is 3 8 3 8 .
• There are 2 outcomes that all show the same result: HHH and TTT. So, the probability is 2 8 = 1 4 2 8 = 1 4 .

Your Turn 7.19

People in mathematics, gerolamo cardano.

The first known text that provided a systematic approach to probabilities was written in 1564 by Gerolamo Cardano (1501–1576). Cardano was a physician whose illegitimate birth closed many doors that would have otherwise been open to someone with a medical degree in 16th-century Italy. As a result, Cardano often turned to gambling to help ends meet. He was a remarkable mathematician, and he used his knowledge to gain an edge when playing at cards or dice. His 1564 work, titled Liber de ludo aleae (which translates as Book on Games of Chance ), summarized everything he knew about probability. Of course, if that book fell into the hands of those he played against, his advantage would disappear. That’s why he never allowed it to be published in his lifetime (it was eventually published in 1663). Cardano made other contributions to mathematics; he was the first person to publish the third degree analogue of the Quadratic Formula (though he didn’t discover it himself), and he popularized the use of negative numbers.

Method 2: Empirical Probability

Theoretical probabilities are precise, but they can’t be found in every situation. If the outcomes in the sample space are not equally likely, then we’re out of luck. Suppose you’re watching a baseball game, and your favorite player is about to step up to the plate. What is the probability that he will get a hit?

In this case, the sample space is {hit, not a hit}. That doesn’t mean that the probability of a hit is 1 2 1 2 , since those outcomes aren’t equally likely. The theoretical method simply can’t be used in this situation. Instead, we might look at the player’s statistics up to this point in the season, and see that he has 122 hits in 531 opportunities. So, we might think that the probability of a hit in the next plate appearance would be about 122 531 ≈ 0.23 122 531 ≈ 0.23 . When we use the outcomes of previous replications of an experiment to assign a probability to the next replication, we’re defining an empirical probability . Empirical probability is assigned using the outcomes of previous replications of an experiment by finding the ratio of the number of times in the previous replications the event occurred to the total number of previous replications.

Empirical probabilities aren’t exact, but when the number of previous replications is large, we expect them to be close. Also, if the previous runs of the experiment are not conducted under the exact set of circumstances as the one we’re interested in, the empirical probability is less reliable. For instance, in the case of our favorite baseball player, we might try to get a better estimate of the probability of a hit by looking only at his history against left- or right-handed pitchers (depending on the handedness of the pitcher he’s about to face).

Probability and Statistics

One of the broad uses of statistics is called statistical inference, where statisticians use collected data to make a guess (or inference) about the population the data were collected from. Nearly every tool that statisticians use for inference is based on probability. Not only is the method we just described for finding empirical probabilities one type of statistical inference, but some more advanced techniques in the field will give us an idea of how close that empirical probability might be to the actual probability!

Example 7.20

Finding empirical probabilities.

Assign an empirical probability to the following events:

• Jose is on the basketball court practicing his shots from the free throw line. He made 47 out of his last 80 attempts. What is the probability he makes his next shot?
• Amy is about to begin her morning commute. Over her last 60 commutes, she arrived at work 12 times in under half an hour. What is the probability that she arrives at work in 30 minutes or less?
• Felix is playing Yahtzee with his sister. Felix won 14 of the last 20 games he played against her. How likely is he to win this game?
• Since Jose made 47 out of his last 80 attempts, assign this event an empirical probability of 47 80 ≈ 59 % 47 80 ≈ 59 % .
• Amy completed the commute in under 30 minutes in 12 of the last 60 commutes, so we can estimate her probability of making it in under 30 minutes this time at 12 60 = 20 % 12 60 = 20 % .
• Since Felix has won 14 of the last 20 games, assign a probability for a win this time of 14 20 = 70 % 14 20 = 70 % .

Your Turn 7.20

Work it out, buffon’s needle.

A famous early question about probability (posed by Georges-Louis Leclerc, Comte de Buffon in the 18th century) had to do with the probability that a needle dropped on a floor finished with wooden slats would lay across one of the seams. If the distance between the slats is exactly the same length as the needle, then it can be shown using calculus that the probability that the needle crosses a seam is 2 π 2 π . Using toothpicks or matchsticks (or other uniformly long and narrow objects), assign an empirical probability to this experiment by drawing parallel lines on a large sheet of paper where the distance between the lines is equal to the length of your dropping object, then repeatedly dropping the objects and noting whether the object touches one of the lines. Once you have your empirical probability, take its reciprocal and multiply by 2. Is the result close to π π ?

Method 3: Subjective Probability

In cases where theoretical probability can’t be used and we don’t have prior experience to inform an empirical probability, we’re left with one option: using our instincts to guess at a subjective probability . A subjective probability is an assignment of a probability to an event using only one’s instincts.

Subjective probabilities are used in cases where an experiment can only be run once, or it hasn’t been run before. Because subjective probabilities may vary widely from person to person and they’re not based on any mathematical theory, we won’t give any examples. However, it’s important that we be able to identify a subjective probability when we see it; they will in general be far less accurate than empirical or theoretical probabilities.

Example 7.21

Distinguishing among theoretical, empirical, and subjective probabilities.

Classify each of the following probabilities as theoretical, empirical, or subjective.

• An eccentric billionaire is testing a brand new rocket system. He says there is a 15% chance of failure.
• With 4 seconds to go in a close basketball playoff game, the home team need 3 points to tie up the game and send it to overtime. A TV commentator says that team captain should take the final 3-point shot, because he has a 38% chance of making it (greater than every other player on the team).
• Felix is losing his Yahtzee game against his sister. He has one more chance to roll 2 dice; he’ll win the game if they both come up 4. The probability of this is about 2.8%.
• This experiment has never been run before, so the given probability is subjective.
• Presumably, the commentator has access to each player’s performance statistics over the entire season. So, the given probability is likely empirical.
• Rolling 2 dice results in a sample space with equally likely outcomes. This probability is theoretical. (We’ll learn how to calculate that probability later in this chapter.)

Your Turn 7.21

Benford’s law.

In 1938, Frank Benford published a paper (“The law of anomalous numbers,” in Proceedings of the American Philosophical Society ) with a surprising result about probabilities. If you have a list of numbers that spans at least a couple of orders of magnitude (meaning that if you divide the largest by the smallest, the result is at least 100), then the digits 1–9 are not equally likely to appear as the first digit of those numbers, as you might expect. Benford arrived at this conclusion using empirical probabilities; he found that 1 was about 6 times as likely to be the initial digit as 9 was!

New Probabilities from Old: Complements

One of the goals of the rest of this chapter is learning how to break down complicated probability calculations into easier probability calculations. We’ll look at the first of the tools we can use to accomplish this goal in this section; the rest will come later.

Given an event E E , the complement of E E (denoted E ′ E ′ ) is the collection of all of the outcomes that are not in E E . (This is language that is taken from set theory, which you can learn more about elsewhere in this text.) Since every outcome in the sample space either is or is not in E E , it follows that n ( E ) + n ( E ′ ) = n ( S ) n ( E ) + n ( E ′ ) = n ( S ) . So, if the outcomes in S S are equally likely, we can compute theoretical probabilities P ( E ) = n ( E ) n ( S ) P ( E ) = n ( E ) n ( S ) and P ( E ′ ) = n ( E ′ ) n ( S ) P ( E ′ ) = n ( E ′ ) n ( S ) . Then, adding these last two equations, we get

P ( E ) + P ( E ′ ) = n ( E ) n ( S ) + n ( E ′ ) n ( S ) = n ( E ) + n ( E ′ ) n ( S ) = n ( S ) n ( S ) = 1 P ( E ) + P ( E ′ ) = n ( E ) n ( S ) + n ( E ′ ) n ( S ) = n ( E ) + n ( E ′ ) n ( S ) = n ( S ) n ( S ) = 1

Thus, if we subtract P ( E ′ ) P ( E ′ ) from both sides, we can conclude that P ( E ) = 1 − P ( E ′ ) P ( E ) = 1 − P ( E ′ ) . Though we performed this calculation under the assumption that the outcomes in S S are all equally likely, the last equation is true in every situation.

P ( E ) = 1 − P ( E ′ ) P ( E ) = 1 − P ( E ′ )

How is this helpful? Sometimes it is easier to compute the probability that an event won’t happen than it is to compute the probability that it will . To apply this principle, it’s helpful to review some tricks for dealing with inequalities. If an event is defined in terms of an inequality, the complement will be defined in terms of the opposite inequality: Both the direction and the inclusivity will be reversed, as shown in the table below.

Example 7.22

Using the formula for complements to compute probabilities.

• If you roll a standard 6-sided die, what is the probability that the result will be a number greater than one?
• If you roll two standard 6-sided dice, what is the probability that the sum will be 10 or less?
• If you flip a fair coin 3 times, what is the probability that at least one flip will come up tails?
• Here, the sample space is {1, 2, 3, 4, 5, 6}. It’s easy enough to see that the probability in question is 5 6 5 6 , because there are 5 outcomes that fall into the event “roll a number greater than 1.” Let’s also apply our new formula to find that probability. Since E E is defined using the inequality roll > 1 roll > 1 , then E ′ E ′ is defined using roll ≤ 1 roll ≤ 1 . Since there’s only one outcome (1) in E ′ E ′ , we have P ( E ′ ) = 1 6 P ( E ′ ) = 1 6 . Thus, P ( E ) = 1 − P ( E ′ ) = 5 6 P ( E ) = 1 − P ( E ′ ) = 5 6 .

Here, the event E E is defined by the inequality sum ≤ 10 sum ≤ 10 . Thus, E ′ E ′ is defined by sum > 10 sum > 10 . There are three outcomes in E ′ E ′ : two 11s and one 12. Thus, P ( E ) = 1 − P ( E ′ ) = 1 − 3 36 = 11 12 P ( E ) = 1 − P ( E ′ ) = 1 − 3 36 = 11 12 .

• In Example 7.15, we found the sample space for this experiment consisted of these equally likely outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Our event E E is defined by T ≥ 1 T ≥ 1 , so E ′ E ′ is defined by T < 1 T < 1 . The only outcome in E ′ E ′ is the first one on the list, where zero tails are flipped. So, P ( E ) = 1 − P ( E ′ ) = 1 − 1 8 = 7 8 P ( E ) = 1 − P ( E ′ ) = 1 − 1 8 = 7 8 .

Your Turn 7.22

Check your understanding, section 7.5 exercises.

For the following exercises, use the following table of the top 15 players by number of plate appearances (PA) in the 2019 Major League Baseball season to assign empirical probabilities to the given events. A plate appearance is a batter’s opportunity to try to get a hit. The other columns are runs scored (R), hits (H), doubles (2B), triples (3B), home runs (HR), walks (BB), and strike outs (SO).

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 7.

• Intro to theoretical probability
• Experimental versus theoretical probability simulation

Theoretical and experimental probability: Coin flips and die rolls

• Random number list to run experiment
• Random numbers for experimental probability
• Interpret results of simulations

Part 1: Flipping a coin

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
• (Choice A)   Results from an experiment don't always match the theoretical results, but they should be close after a large number of trials. A Results from an experiment don't always match the theoretical results, but they should be close after a large number of trials.
• (Choice B)   Dave's coin is obviously unfair. B Dave's coin is obviously unfair.
• (Choice A)   The experimental probability got closer to the theoretical probability after more flips. A The experimental probability got closer to the theoretical probability after more flips.
• (Choice B)   The experimental probability got farther away from the theoretical probability after more flips. B The experimental probability got farther away from the theoretical probability after more flips.

Part 2: Rolling a die

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4.30 Loaded dice. There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 (which is opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face. Because the spot is solid plastic, this works even with transparent dice. If a die is loaded so that 6 comes up with probability 0.24 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of probabilities to the six faces?

Dice Probability Calculator

Use this dice odds calculator to easily calculate any type of dice roll probability: sum of two dice, sum of multiple dice, getting a value greater than or less than on a given throw of N dice, and so on. Different types of dice are supported: from four-sided, six-sided, all the way to 20-sided (D4, D6, D8, D10, D12, and D20) so that success probabilities and dice odds in popular dice games can be computed.

Related calculators

• Using the dice probability calculator
• Two (6-sided) dice probabilities

Sample space of the two dice problem

Calculating two (6-sided) dice probability, two dice probability chart.

• Dice probability formula
• Probabilities with a single die roll
• Probability of rolling a certain number with n dice throws
• Probability of rolling a certain number on all n dice

What are the odds of throwing more than 9 at craps?

What are the odds of rolling 38 or more in d&d,     using the dice probability calculator.

The tool can be used to compute dice probabilities for any type of game of chance or probability problem as used in teaching basic statistical concepts such as sample space and p-values. It supports the classic scenario of computing probabilities of the sum of two six-sided dice, but also supports 4-sided, 8-sided, 10-sided, 12-sided, and 20-sided dice. In the classic problem two dice are thrown, but with this dice calculator you can also explore it with three or more dice.

The calculator can output, depending on the choice of probability calculation, the probability of:

• throwing an exact sum of two or more dice
• a sum less than or greater than or equal to a given number
• at least one die having a face equal to a given number
• all dice having rolling a value equal to a given number
• at least one die rolling a value less than or equal, or greater than or equal to a number
• all dice rolling a value less than or equal, or greater than or equal to a number

With such versatility you can calculate dice probabilities for most games of chance such as Craps, Backgammon, Dungeons & Dragons (D&D), Balut, Dice 10000, Diceball!, Dudo, Elder Sign, Kismet, Yahtzee, Bunco, and many more.

    Two (6-sided) dice probabilities

The classic case of exploring dice throw probabilities (dice rolling odds) is to estimate the chance of landing a given sum on the faces of two six-sided dice. In this example, two dice are thrown together and one records their face values, and computes their sum. Each die is a cube with six faces with the numbers from one to six printed on each side. One throws two dice and the value of a roll is whatever number faces up once the die settles in place.

The image above shows six such dice each with a different face up. The question is: what is the probability of rolling a given sum with two six-sided dice? . The dice are assumed to be fair (unbiased), meaning each side has equal probability of turning up. Also, the outcome of each roll is independent of rolls preceding or succeeding it. A series of fair dice rolls can be modelled as independent events .

In essence, one needs to first estimate the size of the set of all possible outcomes of the dice throw known as the sample space , and then figure out how many of these result in the desired sum. To solve this problem, one needs to calculate every possibility which might turn up which amounts to estimating how many possible permutations there are in total.

For two 6-sided dice there are exactly 6^2 = 6 · 6 = 36 possible permutations with repetition (selecting two numbers from a set of six).

Knowing the sample space means now we need only compute how many possible ways there are for the dice to result in the sum of interest. For example, if the sum of interest is 12, there is just a single dice permutation which results in such a sum - it only happens if both dice thrown roll a six (double sixes, a.k.a. 'boxcars'). The chance to roll a twelve is then 1 out of 36 = 1 / 36 = 0.02777 or 2.77%. It is also the chance of rolling double ones (a.k.a. 'snake eyes').

If the question is what is the chance of throwing a seven with two dice, then one should consider all possible permutations in which a seven turns up. A seven can be the sum of (6 and 1), (5 and 2), (4 and 3) for a total of three possible permutations, and then we need to consider them in the other possible way, namely (1 and 6), (2 and 5), (3 and 4), for a total of six permutations. So we have six chances out of thirty six, meaning the probability of throwing exactly seven is 6/36 = 1/6 = 0.1666 = 16.66%.

Doing the calculations for all possible outcomes between two and twelve for the sum of two dice rolls one arrives at the following chart of dice probabilities and dice odds. It is for two dice rolled simultaneously or one after another (classic 6-sided dice):

If two dice are thrown together, the odds of getting a seven are the highest at 6/36, followed by six and eight with equal odds of 5/36 (13.89%), then five and nine with odds of 4/36 (11.11%), and so on. Here is the same information in table form:

    Dice probability formula

The probability ( P ) of rolling a given sum ( p ) by throwing any number of dice ( n ) with a given number of sides or faces ( s ) can be calculated using the general formula as provided by Uspensky 1937 [1] :

The brackets ⌊ ... ⌋ in the upper limit of the sigma operator denote the floor mathematical function. The round brackets with two numbers one on top of the other denote combinations without repetition. It is read as "n choose k" and its computation uses the factorial function. This makes it somewhat difficult to compute the equation by hand so a dice probability calculator comes in super handy even with a small number of dice.

The formula can be used to produce dice probability distribution charts for any type and number of dice, and dice rolls. With a larger number of dice the distribution converges to the normal distribution in accordance with the Central Limit Theorem [2] . Calculating the probability of rolling greater than or less than a certain sum simply entails summing the probabilities of all possible outcomes greater than the sum of interest.

    Probabilities with a single die roll

The probability of rolling any given number with a single die on a single roll is equal to one divided by the number of die faces. For example, with a 20-sided die, the odds of rolling 20 is 1/20 or a 5% chance. With a classic six-sided die the probability of seeing any particular face is 16.667% or chance odds of 1/6. As you can see, single die roll probabilities require only simple division to calculate.

    Probability of rolling a certain number with n dice throws

If there is more than one throw possible, e.g. if throwing two dice simultaneously or one after another, the probability of rolling a given number with at least one of the throws increases. However, it never reaches 100%! For example, if one needs a six and has six throws of a D6 die, the probability of rolling a six is not 16.667% · 6 = 100%.

A clever way to easily calculate it is to first raise the probability of not getting a six on each throw to the power of the number of throws possible. The probability of not rolling a six is (6-1) / 6 = 5/6 or 0.8333. With six dice, we have 5/6^6 = 5/6 · 5/6 · 5/6 · 5/6 · 5/6 · 5/6 = 0.3349. This means that the probability of throwing at least one six with six tries is 1 - 0.3349 = 0.6651 or 66.51% giving odds of roughly 2/3. In effect, the multiplication in the denominator calculates the sample space whereas the multiplication in the numerator computes the number of unfavourable outcomes in the sample space. We have 5^5 = 15,625 possibilities for not seeing a single six out of 6^6 = 46,656 total possibilities.

Here is a chart of the probabilities of getting at least one six with increasing number of dice rolls :

While the probability increases slowly with each subsequent dice throw, it never actually reaches 100%.

    Probability of rolling a certain number on all n dice

In contrast to the probability of throwing a certain number by having n tries (either throwing n dice together or throwing a die n times), the probability of rolling the same face value on a number of dice decreases substantially the more dice there are. The probability never reaches zero, but it can be considered practically nil after a certain number of dice. Here is a graph with these probabilities:

With just ten dice throws, the probability of rolling a six on all is a mere 0.000002%, and the chance only decreases further when more dice throws are added.

    Examples

Here are a few more examples for the odds in different games with dice.

Craps is played with two six-sided dice. To calculate the odds of rolling 9 or more we need to use the dice probability formula above and compute the probabilities for all possible outcomes of throwing the two dice: 9, 10, 11, and 12, then sum them up. Using the table above we can see the odds are 4/36, 3/36, 2/36, and 1/36 respectively. Summing them up (you can use our fractions calculator for this task) results in odds of 10/36 (10 to 36), or a probability of 0.2778 (27.78%).

Dice play an important role in the tabletop game Dungeons and Dragons (DnD). Sometimes one needs to roll a high number to have a chance at performing a given action such as successfully casting a spell. What is the probability of rolling 38 or more given that one can throw two 20-sided dice? Using the equation for the sum of n dice above, we can compute the probability of getting exactly 38, 39, and 40 to be 0.75%, 0.5%, and 0.25%. Summing these up, we get that the chance to roll 38 or higher in D&D, is 0.75% + 0.5% + 0.25% = 1.5% (or odds of 1 out of 66.7).

    References

1 Uspensky, J. V. (1937) "Introduction to Mathematical Probability", New York: McGraw-Hill , pp. 23-24

2 Spanos A. (2019) "Probability Theory and Statistical Inference", Cambridge University Press , pp. 396-404, doi: 10.1017/9781316882825

Cite this calculator & page

If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Dice Probability Calculator" , [online] Available at: https://www.gigacalculator.com/calculators/dice-probability-calculator.php URL [Accessed Date: 18 Apr, 2024].

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     Statistical calculators

Dice Probability Calculator

Polyhedral dice, how to calculate dice roll probability, when to use dice probability calculator.

• Should I play or should I pass? – Let's play a game!

The dice probability calculator is a great tool if you want to estimate the dice roll probability over numerous variants . There are many different polyhedral dice included, so you can explore the likelihood of a 20-sided die as well as that of a regular cubic die.

So, just evaluate the odds, and play a game! You'll also find short descriptions of each option in the text.

🔎 Don't you have any physical dice? No problem - try our dice roller calculator !

Everybody knows what a regular 6-sided die is, and, most likely, many of you have already played thousands of games where you used one (or more). But, did you know that there are different types of die ? Out of the countless possibilities, the most popular dice are included in the Dungeons & Dragons dice set , which contains seven different polyhedral dice:

• 4-sided dice , also known as a tetrahedron — each face is an equilateral triangle;
• 6-sided dice , a classic cube — each face is a square;
• 8-sided dice , also known as an octahedron — each face is an equilateral triangle;
• 10-sided dice , also known as a pentagonal trapezohedron — each face is a kite;
• 10-sided dice counting from 00 to 90 in increments of 10 — used for percentile rolls in combination with the other 10-sided dice;
• 12-sided dice , also known as a dodecahedron — each face is a regular pentagon; and
• 20-sided dice , also known as an icosahedron — each face is an equilateral triangle.

💡 You can master your D&D strategy using Omni's point buy calculator 5e .

Don't worry, we take each of these dice into account in our dice probability calculator. You can choose whichever you like, and e.g., pretend to roll five 20-sided dice at once!

Well, the question is more complex than it seems at first glance, but you'll soon see that the answer isn't that scary! It's all about maths and statistics.

First of all, we have to determine what kind of dice roll probability we want to find . We can distinguish a few, which you can see in this dice probability calculator.

Before we make any calculations, let's define some variables which we'll use in the formulas. n – the number of dice, s – the number of individual die faces, p – the probability of rolling any value from a die, and P – the overall probability for the problem. There is a simple relationship – p = 1/s , so the probability of getting 7 on a 10-sided die is twice that of a 20-sided die.

The probability of rolling the same value on each die – while the chance of getting a particular value on a single die is p , we only need to multiply this probability by itself as many times as the number of dice. In other words, the probability P equals p to the power n , or P = p n = (1/s) n . If we consider three 20-sided dice, the chance of rolling 15 on each of them is: P = (1/20) 3 = 0.000125 (or P = 1.25·10 -4 in scientific notation). And if you are interested in rolling the set of any identical values — not just three 15s, but three of any number — you simply multiply the result by the total die faces: P = 0.000125 · 20 = 0.0025 .

The probability of rolling all the values equal to or higher than y – the problem is similar to the previous one, but this time p is 1/s multiplied by all the possibilities which satisfy the initial condition. For example, let's say we have a regular die and y = 3 . We want to rolled value to be either 6 , 5 , 4 , or 3 . The variable p is then 4 · 1/6 = 2/3 , and the final probability is P = (2/3) n .

The probability of rolling all the values equal to or lower than y – this option is almost the same as the previous one, but this time we are interested only in numbers that are equal to or lower than our target. If we take identical conditions ( s=6 , y=3 ) and apply them in this example, we can see that the values 1 , 2 , & 3 satisfy the rules, and the probability is: P = (3 · 1/6) n = (1/2) n .

The probability of rolling exactly X same values (equal to y ) out of the set — imagine you have a set of seven 12-sided dice, and you want to know the chance of getting exactly two 9s . It's somehow different than previously because only a part of the whole set has to match the conditions . This is where the binomial probability comes in handy. The binomial probability formula is:

P(X=k) = nCk · p k · (1-p) n-k ,

where k is the number of elements we want to select, and nCk is the number of combinations that fulfil this condition (also known as " n choose k ", or, alternatively as " n choose r ").

In our example we have n = 7 , p = 1/12 , k = 2 , nCk = 21 , so the final result is: P(X=2) = 21 · (1/12) 2 · (11/12) 5 = 0.09439 , or P(X=2) = 9.439% as a percentage.

🙋 You can find more information about this topic in our binomial distribution calculator .

The probability of rolling at least X same values (equal to y ) out of the set — the problem is very similar to the prior one, but this time the outcome is the sum of the probabilities for X=2, 3, 4, 5, 6, 7 . Moving to the numbers, we have: P = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 0.11006 = 11.006% . As you may expect, the result is a little higher. Sometimes the precise wording of the problem will increase your chances of success.

The probability of rolling an exact sum r out of the set of n s -sided dice — the general formula is pretty complex:

However, we can also try to evaluate this problem by hand. One approach is to find the total number of possible sums. With a pair of regular dice, we can have 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 , but these results are not equivalent !

Take a look; there is only one way you can obtain 2 : 1+1 , but for 4 , there are three different possibilities: 1+3 , 2+2 , 3+1 , and for 12 there is, once again, only one variant: 6+6 . It turns out that 7 is the most likely result with six possibilities: 1+6 , 2+5 , 3+4 , 4+3 , 5+2 , and 6+1 . The number of permutations with repetitions in this set is 36 . Our permutation calculator may be handy for finding permutations for other dice types. We can estimate the probabilities as the ratio of favorable outcomes to all possible outcomes: P(2) = 1/36 , P(4) = 3/36 = 1/12 , P(12) = 1/36 , P(7) = 6/36 = 1/6 .

The higher the number of dice, the closer the distribution function of sums gets to the normal distribution. As you may expect, as the number of dice and faces increases, more time is consumed evaluating the outcome on a sheet of paper. Luckily, this isn't the case for our dice probability calculator!

The probability of rolling a sum out of the set, not lower than X — like the previous problem, we have to find all results which match the initial condition and divide them by the number of all possibilities. Taking into account a set of three 10-sided dice, we want to obtain a sum at least equal to 27 . As we can see, we have to add all permutations for 27 , 28 , 29 , and 30 , which are 10, 6, 3, and 1, respectively. In total, there are 20 good outcomes in 1,000 possibilities, so the final probability is: P(X ≥ 27) = 20 / 1,000 = 0.02 .

The probability of rolling a sum out of the set, not higher than X — the procedure is precisely the same as for the prior task, but we have to add only sums below or equal to the target. Having the same set of dice as above, what is the chance of rolling at most 26 ? If you were to do it step by step, it would take ages to obtain the result (to sum all 26 sums). But, if you think about it, we have just worked out the complementary event in the previous problem. The total probability of complementary events is exactly 1 , so the probability here is: P(X ≤ 26) = 1 − 0.02 = 0.98 .

There are a lot of board games where you take turns to roll a die (or dice), and the results may be used in numerous contexts. Let's say you're playing Dungeons & Dragons and attacking. Your opponent's armor class is 17 . You roll a 20-sided dice, hoping for a result of at least 15 — with your modifier of +2. That should be enough. With these conditions, the probability of a successful attack is 0.30 . If you know the odds of a successful attack, you can choose whether you want to attack this target or pick another with better odds.

Or maybe you're playing The Settlers of Catan , and you hope to roll the sum of exactly 8 with two 6-sided dice, as this result will yield you precious resources. Just use our dice probability calculator, and you'll see the chance is around 0.14 — you'd better get lucky this turn!

Should I play or should I pass? – Let's play a game!

There are various types of games, like lotteries, where your task is to make a bet depending on the odds. Rolling dice is one of them. Although taking some risks is inevitable, you can choose the most favorable option and maximize your chances of a win. Take a look at this example.

Imagine you are playing a game where you have one of three options to choose from , which are:

• The sum of five 10-sided dice is at least 30 ;
• The sum of five 12-sided dice is at most 28 ;
• The sum of five 20-sided dice is at least 59 .

You only win if the option you pick comes up. You can also pass if you feel none of these will happen. Intuitively, it's difficult to estimate the most likely success, but with our dice probability calculator, it takes only a blink of an eye to evaluate all the probabilities.

The resulting values are:

• P 1 = 0.38125 for 10-sided dice;
• P 2 = 0.3072 for 12-sided dice; and
• P 3 = 0.3256 for 20-sided dice.

The probability for a pass to be successful is the product of the complementary events of the remaining options:

• P 4 = (1-P 1 ) · (1-P 2 ) · (1-P 3 ) = 0.61875 · 0.6928 · 0.6744 = 0.2891 .

We can see that the most favorable option is the first one, while passing is the least likely event to happen. We cannot assure you'll win all the time, but we strongly recommend that you pick the 10-sided dice set to play.

What is a probability?

Probability determines how likely certain events are to occur . The simple formula for probability is the number of desired outcomes/number of possible outcomes . In board games or gambling, dice probability is used to determine the chance of throwing a certain number , e.g., what is the possibility of getting a specific number with one die?

How many possible outcomes are there from rolling two dice?

There are 36 outcomes when you throw two dice . For a single die , there are six faces, and for any roll, there are six possible outcomes . For two dice , you should multiply the number of possible outcomes together to get 6 × 6 = 36 . With subsequent dice, simply multiply the result by 6 . If you use dice of a different shape, enter the number of their sides instead of 6 .

When rolling 2 dice, what is the probability of 7?

It's 1/6 or 0.1666667 . Let's assume that a total of 7 occurs at least once . For 2 dice, there are 6 ways to throw the sum of 7 — (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) . The total number of combinations for a pair of cube dice is 36 . So the probability of summing up to 7 is 6/36 = 1/6 = 0.1666667 .

How many times do I throw 5 on a pair of dice?

20 . Let's assume that a pair of dice is rolled 180 times . You have 4 ways to get a sum of 5 — (1,4), (2,3), (3,2), and (4,1) . The probability of throwing a sum of 5 is 4/36 = 1/9 . The expected number in 180 throws is 180 × (1/9) = 20 . You will throw a sum of 5 at least 20 times .

Can I always roll a 6 on a dice?

No , it is always a matter of luck, but you can increase your chances by using some tricks:

• Place your index finger and thumb on the numbers that are on opposite sides revealing 1 and 6 .
• Throw the dice slowly and hope the dice rolls straight.

The trick is that you let the dice roll with the remaining numbers, creating a higher probability of landing on the number 6 .

Binomial distribution

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Live updates, outrage as high school student is suspended just for using the term ‘illegal alien’ in class discussion.

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A 16-year-old North Carolina high school student says he was suspended just for saying “illegal alien” while discussing word meaning in English class — possibly ruining his chances of landing a college sports scholarship.

Christian McGhee, a student at Central Davidson High School in Lexington, received a three-day suspension last week after he used the term in English class, the Carolina Journal reported .

His mother, Leah McGhee, said his teacher had given an assignment that used the word “alien,” and Christian asked: “Like space aliens or illegal aliens without green cards?”

Another student reportedly took offense and threatened to fight Christian, so the teacher took the matter to the assistant principal, according to the Carolina Journal.

Eventually, his words were determined to be offensive and disrespectful to Hispanic classmates, so he was suspended.

 “I didn’t make a statement directed towards anyone — I asked a question,” Christian told the outlet.

“I wasn’t speaking of Hispanics because everyone from other countries needs green cards, and the term ‘illegal alien’ is an actual term that I hear on the news and can find in the dictionary,” he added.

The suspension may also affect the student-athlete’s prospects of securing a college sports scholarship, the Journal noted.

“Because of his question, our son was disciplined and given THREE days OUT of school suspension for ‘racism,’” Leah wrote in an email describing the incident.

“He is devastated and concerned that the racism label on his school record will harm his future goal of receiving a track scholarship. We are concerned that he will fall behind in his classes due to being absent for three consecutive days,” she added in the message, which was shared with the outlet.

The irate mom said the assistant principal has refused to remove the suspension from the boy’s record, so the family has hired an attorney.

On Tuesday, Leah appeared on “The Pete Kaliner Show,” which airs on radio station WBT, and said her family had once lived in England, and Christian mentioned how Britons also need green cards to live in the US, Newsweek reported . She said she and her husband told the assistant principal that “illegal alien” is a term their son can look up in a dictionary. “It is a term used as federal code, and it is a term that is heard frequently on many news broadcasts,” Leah said on the show. “I feel that if this was handled properly in the classroom, it could have easily been used as a teachable moment for everyone.”

Republican state Sen. Steve Jarvis said he has contacted the school district superintendent about the matter — but he has not yet taken a stance on what should be done.

“I do not see that that would be an offensive statement, just in getting clarification,” Jarvis told the Journal. “But there again, I don’t know. I don’t know the situation of this particular incident.”

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The popular X account Libs of TikTok also weighed in by saying Christian’s record could be “damaged” by the brouhaha over political correctness.

“Please support this based student by helping to raise awareness to his story!” the conservative account wrote in the post, which has received more than 4 million views.

Among those to respond was X owner Elon Musk, who wrote: “This is absurd.”

Conservative personality Ian Miles Chong called it “insane.”

“How does one get suspended for using the term illegal alien?” he asked.

Libs of TikTok added: “Hopefully North Carolina officials can step in and ensure his record isn’t tarnished in any way because he’s trying to secure an athletic scholarship for college.

“He should not be persecuted for using the correct term just because the left is trying to change our entire language,” the account added.

A staffer at Central Davidson High School told Newsweek that they could not comment about a specific student due to federal protections.

“Please know that Davidson County Schools administrators take all discipline incidents seriously and investigate each one thoroughly,” the rep told the mag. “Any violation of the code of conduct is handled appropriately by administrators.”

The student handbook says that “schools may place restrictions on a student’s right to free speech when the speech is obscene, abusive, promoting illegal drug use, or is reasonably expected to cause a substantial disruption to the school day,” the Carolina Journal reported.

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2024 NBA playoffs bracket: Postseason matchups as 76ers advance to face Knicks, Bulls eliminate Hawks

Here's the up-to-date nba playoff bracket.

The 2024 NBA postseason kicked off this week, and two teams have already been eliminated from the bracket. The Warriors and Hawks have seen their seasons come to a close in the Play-In Tournament, while the Lakers and 76ers have advanced to the playoffs as the No. 7 seeds in their respective conferences. The Bulls were the most recent postseason victor, ousting the Hawks on Wednesday night to advance to face the Heat for the East's No. 8 seed on Friday. Coby White, the potential Most Improved Player , scored a game-high 42 points to end the Hawks' season.

The Kings -- who knocked out the Warriors -- will face the Zion Williamson-less Pelicans on Friday in the final Play-In game with the West's No. 1 seed up for grabs. A notable injury hangs over the Heat-Bulls game on Friday as well, as Miami star Jimmy Butler is dealing with a knee issue.

With the Lakers and 76ers advancing, six of the eight first-round playoff matchups are set. It will be Knicks-76ers, Bucks-Pacers and Cavaliers-Magic in the East, and Nuggets-Lakers, Timberwolves-Suns and Clippers-Mavericks in the West.

Here's a look at the complete playoff bracket with all the seeds set. You can see the full playoff schedule here .

2024 NBA playoff bracket

West (1) Thunder vs. (8) Pelicans/Kings (2) Nuggets vs. (7) Lakers (3) Timberwolves vs. (6) Suns (4) Clippers vs. (5) Mavericks

East (1) Celtics vs. (8) Heat/Bulls (2) Knicks vs. (7) 76ers (3) Bucks vs. (6) Pacers (4) Cavaliers vs. (5) Magic

Final Eastern Conference standings

• Boston Celtics - 64-18
• New York Knicks - 50-32
• Milwaukee Bucks - 49-33
• Cleveland Cavaliers - 48-34
• Orlando Magic - 47-35
• Indiana Pacers - 47-35
• Philadelphia 76ers - 47-35
• Miami Heat - 46-36
• Chicago Bulls - 39-43
• Atlanta Hawks - 36-46

Final Western Conference standings

• Oklahoma City Thunder - 57-25
• Denver Nuggets - 57-25
• Minnesota Timberwolves - 56-26
• Los Angeles Clippers - 51-31
• Dallas Mavericks - 50-32
• Phoenix Suns - 49-33
• New Orleans Pelicans - 49-33
• Los Angeles Lakers - 47-35
• Sacramento Kings - 46-36
• Golden State Warriors - 46-36

Play-In Tournament schedule, scores

(All times Eastern)

Tuesday, April 16 No. 8 Lakers 110, No. 7 Pelicans 106 No. 9  Kings  118, No. 10 Warriors 94

Wednesday, April 17 No. 7 76ers 105, No. 8 Heat 104 No. 9 Bulls 131, No. 10 Hawks 116

Friday, April 19 Heat vs. Bulls, 7 p.m. ESPN/ fubo Pelicans vs. Kings, 9:30 p.m., TNT

2024 NBA playoffs schedule: First weekend

Saturday, April 20 Game 1:  Cavaliers  vs. Magic, 1 p.m., ESPN/ fubo Game 1:  Timberwolves  vs. Suns, 3:30, ESPN/ fubo Game 1: Knicks vs. 76ers, 6 p.m., ESPN/ fubo Game 1:  Nuggets  vs. Lakers, 8:30 p.m., ABC/ fubo

Sunday, April 21 Game 1:  Celtics  vs. East No. 8, 1 p.m., ABC/ fubo Game 1:  Clippers  vs.  Mavericks , 3:30 p.m., ABC/ fubo Game 1: Bucks vs.  Pacers , 7 p.m., TNT Game 1: Thunder vs. West No. 8, 9:30 p.m., TNT

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