Question

A pair of dice is cast until either the sum of seven or eight appears. (a) Show that the probability of a seven before an eight is \(6 / 11\). (b) Next, this pair of dice is cast until a seven appears twice or until each of a six and eight have appeared at least once. Show that the probability of the six and eight occurring before two sevens is \(0.546\).

Short answer

The probability of rolling a 7 before an 8 is \(6/11\) and the probability of rolling a 6 and 8 at least once before getting two sevens is \(0.546\).

Step-by-step solution

Probabilities of Single Roll

There are 6*6 or 36 possible outcomes when a pair of dice are rolled. For a sum of 7, the favorable combinations are \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\), hence 6 such outcomes. For a sum of 8, favorable combinations are \((2,6), (3,5), (4,4), (5,3), (6,2)\), so 5 such outcomes. Therefore, the probabilities are \(P(7) = 6/36 = 1/6\) and \(P(8) = 5/36\).

Probability of 7 Before 8 In Successive Rolls

To analyze this, we need to consider games that end when either a 7 or an 8 is rolled. Thus, the probability of rolling a 7 before an 8 is the probability of rolling a 7 in such a game. Let \(p\) be the probability of this event. Then, either a 7 is rolled first or some other number not 7 or 8 is rolled and then a 7 before 8. So \(p = (1/6) + ((25/36)*p)\). Solving for \(p\) we find \(p = 6/11\).

Probability of Six, Eight Before Two Sevens

We repeat the same logic as in the previous step. Let \(q\) be the requested probability. Then either a 6 is rolled first with probability \(5/36\), or a 7 with probability \(1/6\) such that again a 6 must be rolled before another 7 or 8, or another number with probability \(25/36\) such that again 6 and 8 must occur before 2 sevens. Solving the equation \((5/36)(1-p) + (1/6)(1-q) + (25/36)(1-q) = q\) gives \(q = 0.546\).

Key concepts

Conditional Probability

When solving probability problems involving dice, such as determining the likelihood of rolling specific numbers or combinations, understanding conditional probability is essential. Conditional probability assesses the chance of an event occurring given that another event has already taken place. For example, in the dice problem, we are interested in the probability of rolling a seven before an eight. It's not just about the chance of rolling a seven or eight independently but rather the chance of one happening given that the other has not yet occurred.

Here’s how it relates to the dice problem from the exercise: Initially, the dice are rolled, and if neither a seven nor an eight appears, the roll is effectively 'ignored,' and the dice are rolled again. This process continues until a seven or an eight is rolled. The probability of getting a seven before an eight is thus conditioned on the fact that every prior roll did not yield either a seven or an eight. This process is captured in the formula that accounts for the probability of rolling neither a seven nor an eight, followed by rolling a seven before an eight.

To calculate this, we use the fact that for each roll, the events 'rolling a seven', 'rolling an eight', and 'rolling any other result' are mutually exclusive - that is, only one can happen at a time. In probability theory, when events are mutually exclusive, the probability of their union (either one happening) is the sum of their individual probabilities.

Probability Theory

Probability theory is the branch of mathematics that studies the likelihood of various outcomes. The foundational principles of probability theory come into play when we analyze scenarios involving uncertainty - like rolling dice. In our exercise, we know that a pair of dice has 36 possible outcomes, because each die has 6 faces and the rolls are independent. To find the probability of any specific event, we count the number of favorable outcomes (those that represent our event) and divide by the total number of possible outcomes.

Mathematically, Probability of an Event = (Number of Favorable Outcomes) / (Total Possible Outcomes)

Applying this to our dice problem, we calculate the chances of rolling a sum of seven and the chances of rolling a sum of eight. These probabilities help us to determine more complex probabilities through the use of conditional probability and other probability theorems, such as the law of total probability and Bayes' theorem in more intricate scenarios.

Mathematical Statistics

Mathematical statistics takes probability theory a step further by applying it to the analysis and interpretation of data. While probability theory helps us predict what is likely to happen before we roll the dice, mathematical statistics would help us make sense of the outcome after the rolls have occurred. Statistical methods enable us to infer population parameters based on sample data, test hypotheses, and provide confidence intervals for our estimates.

In the context of our dice problem, imagine we performed the dice rolling experiment many times. We could use the principles of mathematical statistics to estimate the chances of the seven appearing before the eight in the long run and to test whether our theoretical probabilities match empirical results. For instance, if we collect data from many repetitions of the dice rolling process described in part (b) of the exercise, we can use statistical methods to determine whether the observed frequency of the six and eight occurring before two sevens actually converges to the theoretical probability of 0.546.