Science, Tech, Math › Math Expected Value for Chuck-a-Luck Share Flipboard Email Print 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 March 06, 2017 Chuck-a-Luck is a game of chance. Three dice are rolled, sometimes in a wire frame. Due to this frame, this game is also called birdcage. This game is more often seen in carnivals rather than casinos. However, due to the use of random dice, we can use probability to analyze this game. More specifically we can calculate the expected value of this game. Wagers There are several types of wagers that are possible to bet on. We will only consider the single number wager. On this wager we simply choose a specific number from one to six. Then we roll the dice. Consider the possibilities. All of the dice, two of them, one of them or none could show the number that we have chosen. Suppose that this game will pay the following: $3 if all three dice match the number chosen.$2 if exactly two dice match the number chosen.$1 if exactly one of the dice matches the number chosen. If none of the dice matches the number chosen, then we must pay $1. What is the expected value of this game? In other words, in the long run how much on average would we expect to win or lose if we played this game repeatedly? Probabilities In order to find the expected value of this game we need to determine four probabilities. These probabilities correspond to the four possible outcomes. We note that each die is independent of the others. Due to this independence, we use the multiplication rule. This will help us in determining the number of outcomes. We also assume that the dice are fair. Each of the six sides on each of the three dice is equally likely to be rolled. There are 6 x 6 x 6 = 216 possible outcomes from rolling these three dice. This number will be the denominator for all of our probabilities. There is one way to match all three dice with the number chosen. There are five ways for a single die to not match our chosen number. This means that there are 5 x 5 x 5 = 125 ways for none of our dice to match the number that was chosen. If we consider exactly two of the dice matching, then we have one die that does not match. There are 1 x 1 x 5 = 5 ways for the first two dice to match our number and the third to be different.There are 1 x 5 x 1 = 5 ways for the first and third dice to match, with the second be different.There are 5 x 1 x 1 = 5 ways for the first die to be different and for the second and third to match. This means that there is a total of 15 ways for exactly two dice to match. We now have calculated the number of ways to obtain all but one of our outcomes. There are 216 rolls possible. We have accounted for 1 + 15 + 125 = 141 of them. This means that there are 216 -141 = 75 remaining. We collect all of the above information and see: The probability our number matches all three dice is 1/216.The probability our number matches exactly two dice is 15/216.The probability our number matches exactly one die is 75/216.The probability our number matches none of the dice is 125/216. Expected Value We are now ready to calculate the expected value of this situation. The formula for expected value requires us to multiply the probability of each event by the net gain or loss if the event occurs. We then add all of these products together. The calculation of the expected value is as follows: (3)(1/216) + (2)(15/216) +(1)(75/216) +(-1)(125/216) = 3/216 +30/216 +75/216 -125/216 = -17/216 This is approximately -$0.08. The interpretation is that if we were to play this game repeatedly, on average we would lose 8 cents each time that we played.