Science, Tech, Math › Math Probabilities and Liar's Dice Share Flipboard Email Print Riou/Photographer's Choice RF/Getty Images Math Statistics Probability & Games Statistics Tutorials Formulas Descriptive Statistics Inferential Statistics Applications Of Statistics Math Tutorials Geometry Arithmetic Pre Algebra & Algebra Exponential Decay Functions Worksheets By Grade Resources View More By Courtney Taylor Professor of Mathematics Ph.D., Mathematics, Purdue University M.S., Mathematics, Purdue University B.A., Mathematics, Physics, and Chemistry, Anderson University Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra." our editorial process Courtney Taylor Updated January 27, 2019 Many games of chance can be analyzed using the mathematics of probability. In this article, we will examine various aspects of the game called Liar’s Dice. After describing this game, we will calculate probabilities related to it. A Brief Description of Liar’s Dice The game of Liar’s Dice is actually a family of games involving bluffing and deception. There are a number of variants of this game, and it goes by several different names such as Pirate’s Dice, Deception, and Dudo. A version of this game was featured in the movie Pirates of the Caribbean: Dead Man’s Chest. In the version of the game that we will examine, each player has a cup and a set of the same number of dice. The dice are standard, six-sided dice that are numbered from one to six. Everyone rolls their dice, keeping them covered by the cup. At the appropriate time, a player looks at his set of dice, keeping them hidden from everyone else. The game is designed so that each player has perfect knowledge of his own set of dice, but has no knowledge about the other dice that have been rolled. After everyone has had an opportunity to look at their dice that were rolled, bidding commences. On each turn a player has two choices: make a higher bid or call the previous bid a lie. Bids can be made higher by bidding a higher dice value from one to six, or by bidding a greater number of the same dice value. For example, a bid of “Three twos” could be increased by stating “Four twos.” It could also be increased by saying “Three threes.” In general, neither the number of dice nor the values of the dice can decrease. Since most of the dice are hidden from view, it is important to know how to calculate some probabilities. By knowing this is it easier to see what bids are likely to be true, and what ones are likely to be lies. Expected Value The first consideration is to ask, “How many dice of the same kind would we expect?” For example, if we roll five dice, how many of these would we expect to be a two? The answer to this question uses the idea of expected value. The expected value of a random variable is the probability of a particular value, multiplied by this value. The probability that the first die is a two is 1/6. Since the dice are independent of one another, the probability that any of them is a two is 1/6. This means that the expected number of twos rolled is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6. Of course, there is nothing special about the result of two. Neither is there anything special about the number of dice that we considered. If we rolled n dice, then the expected number of any of the six possible outcomes is n/6. This number is good to know because it gives us a baseline to use when questioning bids made by others. For example, if we are playing liar's dice with six dice, the expected value of any of the values 1 through 6 is 6/6 = 1. This means that we should be skeptical if someone bids more than one of any value. In the long run, we would average one of each of the possible values. Example of Rolling Exactly Suppose that we roll five dice and we want to find the probability of rolling two threes. The probability that a die is a three is 1/6. The probability that a die is not three is 5/6. Rolls of these dice are independent events, and so we multiply the probabilities together using the multiplication rule. The probability that the first two dice are threes and the other dice are not threes is given by the following product: (1/6) x (1/6) x (5/6) x (5/6) x (5/6) The first two dice being threes is is just one possibility. The dice that are threes could be any two of the five dice that we roll. We denote a die that is not a three by a *. The following are possible ways to have two threes out of five rolls: 3, 3, * , * ,*3, * , 3, * ,*3, * , * ,3 ,*3, * , * , *, 3*, 3, 3, * , **, 3, *, 3, **, 3, * , *, 3*, *, 3, 3, **, *, 3, *, 3*, *, *, 3, 3 We see that there are ten ways to roll exactly two threes out of five dice. We now multiply our probability above by the 10 ways that we can have this configuration of dice. The result is 10 x(1/6) x (1/6) x (5/6) x (5/6) x (5/6) = 1250/7776. This is approximately 16%. General Case We now generalize the above example. We consider the probability of rolling n dice and obtaining exactly k that are of a certain value. Just as before, the probability of rolling the number that we want is 1/6. The probability of not rolling this number is given by the complement rule as 5/6. We want k of our dice to be the selected number. This means that n - k are a number other than the one we want. The probability of the first k dice being a certain number with the other dice, not this number is: (1/6)k(5/6)n - k It would be tedious, not to mention time-consuming, to list all possible ways to roll a particular configuration of dice. That is why it is better to use our counting principles. Through these strategies, we see that we are counting combinations. There are C(n, k) ways to roll k of a certain kind of dice out of n dice. This number is given by the formula n!/(k!(n - k)!) Putting everything together, we see that when we roll n dice, the probability that exactly k of them are a particular number is given by the formula: [n!/(k!(n - k)!)] (1/6)k(5/6)n - k There is another way to consider this type of problem. This involves the binomial distribution with probability of success given by p = 1/6. The formula for exactly k of these dice being a certain number is known as the probability mass function for the binomial distribution. Probability of at Least Another situation that we should consider is the probability of rolling at least a certain number of a particular value. For example, when we roll five dice what is the probability of rolling at least three ones? We could roll three ones, four ones or five ones. To determine the probability we want to find, we add together three probabilities. Table of Probabilities Below we have a table of probabilities for obtaining exactly k of a certain value when we roll five dice. Number of Dice k Probability of Rolling Exactly k Dice of a Particular Number 0 0.401877572 1 0.401877572 2 0.160751029 3 0.032150206 4 0.003215021 5 0.000128601 Next, we consider the following table. It gives the probability of rolling at least a certain number of a value when we roll a total of five dice. We see that although it is very likely to roll at least one 2, it is not as likely to roll at least four 2's. Number of Dice k Probability of Rolling at Least k Dice of a Particular Number 0 1 1 0.598122428 2 0.196244856 3 0.035493827 4 0.00334362 5 0.000128601 Do You Know the Probabilities for Rolling Two Dice? What Are the Probability Outcomes for Rolling Three Dice? The Probability of Rolling a Yahtzee What's the Probability of Going to Jail in Monopoly? Expected Value for Chuck-a-Luck Learn How to Play Liar's Dice, a Chinese New Year's Tradition See How to Calculate Backgammon Probabilities How to Calculate the Probability of a Small Straight in Yahtzee What's the Probability of a Full House in Yahtzee in a Single Roll? What Is the Probability of the Union of 3 or More Sets? What's the Probability of a Large Straight in Yahtzee in One Roll? Probability simplified. Binomial Table for n=7, n=8 and n=9 Binomial Table for n =10 and n= 11 Example Chi-Square Test for a Multinomial Experiment What Is the Factorial (!) in Mathematics and Statistics?