For all problems of this type, there are some important points that apply, which are expanded upon in the article linked in the Reference section. First of all, the problem will probably refer to a "fair" die. This just means that no tricks are involved. It's not weighted.
We also assume a normal situation. Don't question whether or not the die could come to rest on a corner or up against the wall, or anything like that. It will land on one of the six sides fairly. The problems typically refer to a standard six-sided die, unless you are told otherwise. Note that some textbooks may refer to a die as a "standard number cube," so that the publisher will not be accused of promoting gambling among children.
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Here is a typical question: "A fair die lands on 1 five times in a row. What are the chances that it will land on 1 on the next roll?" The answer to the question is 1/6. That is it. Any other answer is wrong. Each roll of a die is totally independent. See the article in the Resource section for expanded information on that important concept.
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Another typical question is, "What are the chances that a fair die will land on 7?" The answer is 0%. Since there is no 7 on a standard die, the outcome cannot occur. Similarly, if you are asked what are the chances that the die will land on 1, 2, 3, 4, 5, or 6, the answer is 100%. There are 6 possible ways for success out of 6 ways total. From that standpoint, we could write our answer as 6/6, which simplifies to 1.
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Note that all probabilities are a number between 0 and 1, which we often convert to a percent by converting the fraction to a decimal, and then moving the decimal two places to the right. For example, 0 becomes 0%, 1/2 becomes 50%, 1/6 becomes 16.67% (rounded), and 1 becomes 100%.
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Another possible question is: "What are the chances that a die will land on 2 or 3 when rolled?" We are dealing with an "or" condition, which means we must add the possible outcomes together (since either is OK). In this case there are two possible means of success, out of 6 total. Our answer is 2/6. Interestingly, in probability we would usually leave the answer just like that, rather than reduce it to 1/3 which would de-emphasize the denominator of 6 total possible outcomes.
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Another possible question is: "A fair standard number cube is rolled twice. What are the chances that it will land on 2 the first time, and an odd number the second time?" In this case we have an "and" condition, separating two independent events. Each die roll knows nothing of the other. We use multiplication for this. For the first roll, the chances of success are 1/6. For the second roll it is 3/6, since there are three odd values on a die (1, 3, 5). Multiply 1/6 times 3/6 to get 3/36. In this case, we would probably reduce our answer to 1/12.
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The final type of question discussed is: "A fair die is to be rolled 10 times. What are the chances that it will land on 3 every time? Express your answer using an exponent." The chances of a 3 on each roll is 1/6. We're dealing an "and" condition. Each roll has to be a 3. We must compute 1/6 times 1/6 times 1/6, repeated a total of 10 times. The simplest way of representing this is shown at left. It is (1/6) raised to the 10th power. The exponent is applied to both the numerator and the denominator. Since 1 to the power of 10 is just 1, we could also just write our answer as 1 divided by (6 to the 10th power).
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It is interesting to note that the actual odds of the above happening are about one in a 60 million. While it is unlikely that any one particular person will experience this, if you were to ask every single American to conduct this experiment honestly and accurately, there is a good chance that a few people would report success.
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Students should make sure that they are comfortable working with the basic probability concepts discussed in this article since they come up quite frequently.
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