Problem 74 - A First Course in Probability

Question

\(A\) and \(B\) alternate rolling a pair of dice, stopping either when \(A\) rolls the sum 9 or when \(B\) rolls the sum \(6 .\) Assuming that \(A\) rolls first, find the probability that the final roll is made by \(A.\)

Step-by-Step Solution

Answer

The probability that the final roll is made by \(A\) is \(\boxed{\frac{6}{5}}\).

Step 1: Calculate the probability of \(A\) rolling a sum of \(9.\)

We will now calculate the probability that \(A\) rolls a sum of \(9.\) There are a total of 36 possible outcomes for rolling two dice. We need to find the number of outcomes with a sum of \(9.\) The pairs that give a sum of \(9\) are: \((2,7),(3,6),(4,5),(5,4),(6,3),\) and \((7,2).\) That is a total of 6 pairs that sum to 9. The probability of rolling a sum of \(9\) in one turn for A is: P(A rolls 9) = \(\frac{6}{36} = \frac{1}{6}.\)

Step 2: Calculate the probability of \(B\) rolling a sum of \(6.\)

Now, let's calculate the probability that \(B\) rolls a sum of \(6.\) The pairs that give a sum of \(6\) are: \((1,5),(2,4),(3,3),(4,2),(5,1).\) That is a total of 5 pairs that sum to 6. The probability of rolling a sum of \(6\) in one turn for B is: P(B rolls 6) = \(\frac{5}{36}.\)

Step 3: Derive the conditional probabilities

Now, we must consider the situations where the final roll is made by \(A.\) We need to calculate the probability of \(A\) rolling a sum of \(9\) during his turn, while \(B\) did not roll a sum of \(6\) during her previous turn. To do this, we will compute the probability of B not rolling a 6: P(B does not roll 6) = 1 - P(B rolls 6) = 1 - \(\frac{5}{36} = \frac{31}{36}\) Notice that this problem has a geometric sequence because the probability of reaching player A's nth round is multiplied by the probability of not reaching A's win condition nor B's win condition during previous rounds.

Step 4: Calculate the probability of the final roll is made by \(A.\)

Finally, let's calculate the probability that the final roll is made by \(A.\) We will add the probabilities of each round, considering the probability that A wins on that round, and that neither A nor B had won during their previous turns. P(final roll by A) = P(A rolls 9) + P(B does not roll 6) * P(A rolls 9) + (P(B does not roll 6))^2 * P(A rolls 9) + ... This is a geometric series with an infinite number of terms: \[ P(final roll by A) = \frac{1}{6}\left( 1 + \frac{31}{36} + \left(\frac{31}{36}\right)^2 + \left(\frac{31}{36}\right)^3 + ... \right) \] The sum of an infinite geometric series with common ratio r is given by: S = \(\frac{a}{1-r}\), where a is the first term. The probability that the final roll is made by \(A\) can now be calculated using the formula for the sum of an infinite geometric series: \[ P(final roll by A) = \frac{1/6}{1 - (31/36)} = \frac{\frac{1}{6}}{\frac{5}{36}} = \boxed{\frac{6}{5}} \]

Key Concepts

Conditional Probability

Conditional probability is a central concept in probability theory that describes the likelihood of an event occurring given that another event has already occurred. In our dice-rolling exercise, we use conditional probability to determine the likelihood of a certain player's final roll under specific conditions.
For example, we found the probability that A rolls a sum of 9, given that B did not roll a sum of 6 in their previous turn. We say the event "A rolls 9" is conditional on "B does not roll 6".

• This requires calculating the probability for each turn, taking into account past events.
• It applies the formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] where \( P(A|B) \) is the probability of A given B.

This step-by-step application helps in evaluating each round played, considering the result of the previous turns.

Infinite Geometric Series

An infinite geometric series is a series with an infinite number of terms that follow a geometric progression. In our problem, it represents the sum of probabilities over all possible rounds as long as a particular condition holds.
Consider the following:

• Each term in the series is formed by multiplying the probability that neither player reached their win condition previously with the probability that A rolls their needed number in this turn.
• The series takes the form \[ a + ar + ar^2 + ar^3 + ... \]

To find its sum, the formula is used: \[ S = \frac{a}{1-r} \] where \( a \) is the initial term, and \( r \) is the common ratio between consecutive terms.
This mathematical concept helps us solve for the probability that A makes the final roll when the series has an infinite scope.

Dice Rolling Probability

Dice rolling probability is a fundamental part of probability theory dealing with the likelihood of obtaining a particular sum from a roll or series of rolls of dice. In our problem, we calculated the probabilities for specific sums.
Here's an example:

• There are 36 possible outcomes for a pair of six-sided dice.
• We specifically calculated the probabilities for A rolling a 9 and B rolling a 6 from these outcomes.

These probabilities were computed by identifying all possible combinations that produce the desired sums and then dividing by the total possible outcomes. For instance, for a total of 9, the probability was calculated as: \[ P(\text{A rolls 9}) = \frac{6}{36} = \frac{1}{6} \]
Such calculations provide the necessary foundation for using geometric distributions and series in resolving the given problem.

Geometric Distribution

A geometric distribution models the number of trials needed for a success to occur for the first time in a sequence of independent and identically distributed Bernoulli trials. In the context of our exercise, it describes the distribution for A rolling a 9 before B rolls a 6.
Key features of the geometric distribution include:

• Each outcome is independent of others.
• Probability remains constant for each trial.

We used a geometric distribution to formulate the probability of reaching a scenario where A would ultimately get to roll again without B having achieved their goal. The series considered: \[ P(n) = (1-p)^{n-1}p \] where \( n \) is the number of trials, and \( p \) is the probability of success (here, A rolling a 9) on each trial.
This helps understand how long or how many turns it might take for A to win first under the game's premises.