![]() ⋅# ways to assign three 1's ⋅ # ways to assign two 2's # full houses = # varieties of full houses In other words, there is only oneway to pick the dice the 2's are assigned to.Ĭombining resultsby multiplyingwe have the total numberof full houses. Once we choose the set of dice for the 1's, the set of dicefor the 2's has to be the remaining dice. All that matters is the set of dice wepick. Conversely, we could assume we work from left to rightwhen assigning the dice that are 1's. Although we might assign them in aparticular order, only the set of diceassigned 1's matters. Starting with the three 1's, we are interested only inwhich three dice are assigned 1's. Thus, we need only determine how many ways there are toassign numbers to dice for one particular full house, say three 1's and two2's. The varieties of full houses form a partition of allpossible full houses.īy symmetry, each of the 30 different full houses has thesame probability. Thus, permutingthe assignment of numbers to dice cannot cause one variety of full house tolook like another. Since the full houseaccounts for all five dice, any uncertainty that might result from numbers onextra dice, (which would apply if we were considering only the probability ofgetting a certain 3-of-a-kind, for example), is eliminated. This will correspond to exactly one of the 30 possibleways to choose the numbers for the 3-of-a-kind and pair. In the present case, if a roll is a full house,there will be a unique number showing for the 3-of-a-kind and a unique numbershowing for the pair. In other words, we consider whether the way we assign thenumbers in a full house to the different dice can result in one full house givingthe same outcome as another. Note that we would get the same result if we chose thenumber for the pair first, (6 choices), and the 3-of-a-kind second, (5choices).īefore proceeding, we consider whether all the full housesare distinct. Thus, we have 6⋅5 = 30varieties of full houses. Once we choose this number, we must choose a different number for the pair. We have six choices for the number for the 3-of-a-kind. # varieties = # 3-of-a-kind numbers ⋅ # pair numbers Referring tothe possibilities as "varieties" of full houses, the number ofvarieties is the product of the number of numbers for the 3-of-a-kind and thenumber of numbers for the pair. To correctly count the number of full houses, we firstconsider what the numbers are in the 3-of-a-kind and in the pair. ![]() Note that the three-of-a-kind may occur on non-adjacent dice.Īnother roll that is counted as a separate outcome in the correctapproach but would have been counted as the same outcome in the two incorrectapproaches discussed earlier is shown below. Turning to the number of full-house outcomes, theillustration below shows an example of a full house achieved by rolling thedice in order. By extension,we have a total of 6 5 = 7776possible outcomes for all five dice. Thus, we have6⋅6 = 36 possible outcomes for the last two dice. For each of these sixvalues, the second-to-last die may have six values. Thenumbers are the same as counting in base 6. Wemay imagine making a diagram of all the possible outcomes, as shown below. We calculate the total number of all possible outcomes bycounting all the combinations of numbers that can appear on the five dice. Sincethe number rolled on each die is independent of numbers on other dice, rollingthe dice in color-order gives the same probabilities of outcomes as rollingthem all at once. We may also assume that the dice are rolled in a specific color order. For simplicity, we may even assume that thelabels are colors and we may imagine that the dice actually have those colors. Since the dice are physically distinct objects, this has noeffect on the probabilities. One correct approach begins by identifying each die with aunique label. This distorts probabilities because a set ofnumbers such as five 1's requires every die to show a 1, whereas a set ofnumbers such as 1, 2, 3, 4, 5 allows us to exchange numbers on any two dice andstill have the same outcome. This distorts probabilities because it is clearly muchmore likely that we will have a pair than that we will have four-of-a-kind.Īnother incorrect approach, for example, is to equateoutcomes with the unique sets of numbers on the dice when the numbers arelisted from smallest to largest. ![]() One incorrect approach, for example, is to equate outcomeswith categories such as one pair, two pairs, three-of-a-kind, full house,four-of-a-kind, etc. The primary difficulty here is deciding how to define anoutcome so we can avoid distorting probabilities. Note that the number showing on the pair must be different from the numbershowing on the three of a kind, as otherwise we would have five-of-a-kind. Find the probability that exactly three diceshow the same number, (i.e., three of a kind), and the remaining two dice showthe same number, (i.e., a pair).
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