Science, Tech, Math › Math The Probability of a Full House in Yahtzee in a Single Roll Share Flipboard Email Print Virginia State Parks staff [CC BY 2.0 (https://creativecommons.org/licenses/by/2.0)], via Wikimedia Commons 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 31, 2019 The game of Yahtzee involves the use of five standard dice. On each turn, players are given three rolls. After each roll, any number of dice may be kept with the goal being to obtain particular combinations of these dice. Every different kind of combination is worth a different amount of points. One of these types of combinations is called a full house. Like a full house in the game of poker, this combination includes three of a certain number along with a pair of a different number. Since Yahtzee involves the random rolling of dice, this game can be analyzed by using probability to determine how likely it is to roll a full house in a single roll. Assumptions We will begin by stating our assumptions. We assume that the dice used are fair and independent of one another. This means that we have a uniform sample space consisting of all possible rolls of the five dice. Although the game of Yahtzee allows three rolls, we will only consider the case that we obtain a full house in a single roll. Sample Space Since we are working with a uniform sample space, the calculation of our probability becomes a calculation of a couple of counting problems. The probability of a full house is the number of ways to roll a full house, divided by the number of outcomes in the sample space. The number of outcomes in the sample space is straightforward. Since there are five dice and each of these dice can have one of six different outcomes, the number of outcomes in the sample space is 6 x 6 x 6 x 6 x 6 = 65 = 7776. Number of Full Houses Next, we calculate the number of ways to roll a full house. This is a more difficult problem. In order to have a full house, we need three of one kind of dice, followed by a pair of a different type of dice. We will split this problem into two parts: What is the number of different types of full houses that could be rolled?What is the number of ways that a particular type of full house could be rolled? Once we know the number to each of these, we can multiply them together to give us the total number of full houses that can be rolled. We begin by looking at the number of different types of full houses that can be rolled. Any of the numbers 1, 2, 3, 4, 5 or 6 could be used for the three of a kind. There are five remaining numbers for the pair. Thus there are 6 x 5 = 30 different types of full house combinations that can be rolled. For example, we could have 5, 5, 5, 2, 2 as one type of full house. Another type of full house would be 4, 4, 4, 1, 1. Another yet would be 1, 1, 4, 4, 4, which is different than the preceding full house because the roles of the fours and ones have been switched. Now we determine the different number of ways to roll a particular full house. For example, each of the following gives us the same full house of three fours and two ones: 4, 4, 4, 1, 14, 1, 4, 1, 41, 1, 4, 4, 41, 4, 4, 4, 14, 1, 4, 4, 1 We see that there are at least five ways to roll a particular full house. Are there others? Even if we keep listing other possibilities, how do we know that we have found all of them? The key to answering these questions is to realize that we are dealing with a counting problem and to determine what type of counting problem we are working with. There are five positions, and three of these must be filled with a four. The order in which we place our fours does not matter as long as the exact positions are filled. Once the position of the fours has been determined, the placement of the ones is automatic. For these reasons, we need to consider the combination of five positions taken three at a time. We use the combination formula to obtain C(5, 3 ) = 5!/(3!2!) = (5 x 4) / 2 = 10. This means that there are 10 different ways to roll a given full house. Putting all of this together, we have our number of full houses. There are 10 x 30 = 300 ways to obtain a full house in one roll. Probability Now the probability of a full house is a simple division calculation. Since there are 300 ways to roll a full house in a single roll and there are 7776 rolls of five dice possible, the probability of rolling a full house is 300/7776, which is close to 1/26 and 3.85%. This is 50 times more likely than rolling a Yahtzee in a single roll. Of course, it is very likely that the first roll is not a full house. If this is the case, then we are allowed two more rolls making a full house much more likely. The probability of this is much more complicated to determine because of all of the possible situations that would need to be considered.