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Chapter 4: Problem 2

The following problem is called the two aces problem. This problem, dating back to 1936 , has been attributed to the English mathematician J. H. C. Whitehead (see Gridgeman \(^{24}\) ). This problem was also submitted to Marilyn vos Savant by the master of mathematical puzzles Martin Gardner, who remarks that it is one of his favorites. A bridge hand has been dealt, i. e. thirteen cards are dealt to each player. Given that your partner has at least one ace, what is the probability that he has at least two aces? Given that your partner has the ace of hearts, what is the probability that he has at least two aces? Answer these questions for a version of bridge in which there are eight cards, namely four aces and four kings, and each player is dealt two cards. (The reader may wish to solve the problem with a 52 -card deck.)

Short Answer

Expert verified
The probability of at least two aces given one ace is \( \frac{3}{11} \); given an ace of hearts, it's \( \frac{3}{7} \).

Step by step solution

01

Define Total Outcomes

For this version of bridge, there are eight cards total 鈥� four aces and four kings. Each player is dealt two cards. First, we calculate the total possible combinations of two cards from eight, which can be calculated as \( \binom{8}{2} \).
02

Confirm Total Outcomes Calculation

The number of possible combinations of dealing two cards from eight is \( \binom{8}{2} = 28 \). Therefore, there are 28 equally likely ways to deal two cards to a player.
03

Calculate Outcomes for At Least One Ace

To determine the probability of having at least one ace, consider that there are two scenarios: having one ace with one king or having two aces. Calculate distinct pairings for each scenario.
04

Count Outcomes for One Ace and One King

There are 4 possible choices for one ace and 4 possible choices for a king, giving us \( 4 \times 4 = 16 \) combinations where the two cards consist of one ace and one king.
05

Count Outcomes for Two Aces

There are \( \binom{4}{2} = 6 \) combinations to select two cards that are both aces.
06

Total Outcomes for At Least One Ace

Combine the outcomes of having one ace and one king with those of having two aces. The total number of outcomes with at least one ace is \( 16 + 6 = 22 \).
07

Probability of Having Two Aces Given At Least One

Out of 22 outcomes where at least one ace is present, 6 outcomes include two aces. The conditional probability is \( \frac{6}{22} = \frac{3}{11} \).
08

Calculate Outcomes for One Heart Given

Now focus on the scenario where the ace of hearts is included. Consider the combinations including the ace of hearts and any other card.
09

Count All Combinations with the Ace of Hearts

There are 3 remaining aces (since hearts is one) and 4 kings. Thus, there are \( 3 + 4 = 7 \) valid combinations when one card is the ace of hearts.
10

Outcomes Having at Least Two Aces Given Ace of Hearts

With the ace of hearts included, the potential combinations that include at least one other ace are \( 3 \) since any of the three remaining aces can pair with the ace of hearts.
11

Probability of Two Aces Given Ace of Hearts

Out of 7 possible combinations where the ace of hearts is present, 3 include at least two aces. The conditional probability is \( \frac{3}{7} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics that studies the counting, arrangement, and combination of objects. In problems like the two aces problem, combinatorics helps determine the total number of ways to choose cards. This forms the basis for calculating probabilities. Here, we're interested in the number of possible combinations of two cards from a deck of eight cards (four aces and four kings). This can be mathematically expressed using a combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Where:
  • \(n\) is the total number of items to choose from
  • \(r\) is the number of items to choose
In this exercise, we calculate \(\binom{8}{2}\), which equals 28. This result reflects the total possible ways to deal two cards out of the eight-card deck.
Bridge Card Game
Bridge is a popular card game involving strategic decision-making, often played with a 52-card deck. However, in this exercise, a simplified version is played with only eight cards 鈥� four aces and four kings. Each player receives two cards. This scenario frames a unique environment for applying probability theory, where attention is directed toward smaller combinations of cards. By reducing the deck size, the problem becomes a more focused examination of outcomes, especially concerning the appearance of aces. Players seek to leverage these deterministic possibilities to improve their understanding of the broader, more complex standard settings.
Probability Theory
Probability theory is the mathematical framework that quantifies the likelihood of events occurring. It is crucial for making predictions in situations with uncertain outcomes. In this problem, conditional probability is employed to find the probability of an event occurring, given that another event has already occurred.The concept of conditional probability is represented by: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]where:
  • \(A\) is the event we are calculating the probability for (e.g., having two aces)
  • \(B\) is the given condition (e.g., having at least one ace)
In this problem, knowing that your partner has at least one ace forms the condition, and we calculate the probability of them having two aces based on this condition. The calculated probabilities are \(\frac{3}{11}\) when having at least one ace, and \(\frac{3}{7}\) when specifically having the ace of hearts.
Pairing Outcomes
Pairing outcomes are essential in determining the probability of drawing specific combinations of cards. Not only does it help calculate how many combinations meet a particular criterion, but it also helps compare them against the total number of possible outcomes.In our exercise:
  • For drawing one ace and one king, there are 16 possible pairings (4 ways to choose the ace multiplied by 4 ways to choose the king).
  • For drawing two aces, there are 6 combinations available due to four aces and choosing any two at a time, calculated by \(\binom{4}{2}\).
By understanding these pairings, we can effectively determine the odds of specific favorable outcomes, such as possessing at least two aces under given conditions.

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Most popular questions from this chapter

A coin is in one of \(n\) boxes. The probability that it is in the \(i\) th box is \(p_{i}\). If you search in the \(i\) th box and it is there, you find it with probability \(a_{i}\). Show that the probability \(p\) that the coin is in the \(j\) th box, given that you have looked in the \(i\) th box and not found it, is $$ p=\left\\{\begin{array}{cc} p_{j} /\left(1-a_{i} p_{i}\right), & \text { if } j \neq i \\ \left(1-a_{i}\right) p_{i} /\left(1-a_{i} p_{i}\right), & \text { if } j=i \end{array}\right. $$

Probability theory was used in a famous court case: People v. Collins. \(^{10}\) In this case a purse was snatched from an elderly person in a Los Angeles suburb. A couple seen running from the scene were described as a black man with a beard and a mustache and a blond girl with hair in a ponytail. Witnesses said they drove off in a partly yellow car. Malcolm and Janet Collins were arrested. He was black and though clean shaven when arrested had evidence of recently having had a beard and a mustache. She was blond and usually wore her hair in a ponytail. They drove a partly yellow Lincoln. The prosecution called a professor of mathematics as a witness who suggested that a conservative set of probabilities for the characteristics noted by the witnesses would be as shown in Table 4.5 . The prosecution then argued that the probability that all of these characteristics are met by a randomly chosen couple is the product of the probabilities or \(1 / 12,000,000\), which is very small. He claimed this was proof beyond a reasonable doubt that the defendants were guilty. The jury agreed and handed down a verdict of guilty of second-degree robbery. $$ \begin{aligned} &\begin{array}{lc} \text { man with mustache } & 1 / 4 \\ \text { girl with blond hair } & 1 / 3 \\ \text { girl with ponytail } & 1 / 10 \\ \text { black man with beard } & 1 / 10 \\ \text { interracial couple in a car } & 1 / 1000 \\ \text { partly yellow car } & 1 / 10 \end{array}\\\ &\text { Table 4.5: Collins case probabilities. } \end{aligned} $$ If you were the lawyer for the Collins couple how would you have countered the above argument? (The appeal of this case is discussed in Exercise 5.1.34.)

Two cards are drawn from a bridge deck. What is the probability that the second card drawn is red?

Let \(x\) and \(y\) be chosen at random from the interval [0,1] . Which pairs of the following events are independent? (a) \(x>1 / 3\) (b) \(y>2 / 3\). (c) \(x>y\) (d) \(x+y<1\).

Let \(\Omega=\\{a, b, c, d, e, f\\} .\) Assume that \(m(a)=m(b)=1 / 8\) and \(m(c)=\) \(m(d)=m(e)=m(f)=3 / 16 .\) Let \(A, B,\) and \(C\) be the events \(A=\\{d, e, a\\}\) \(B=\\{c, e, a\\}, C=\\{c, d, a\\} .\) Show that \(P(A \cap B \cap C)=P(A) P(B) P(C)\) but no two of these events are independent.
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