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In a game of bridge, West has no aces. What is the probability of his partner's having (a) no aces and (b) 2 or more aces? (c) What would the probabilities be if West had exactly 1 ace?

Short Answer

Expert verified
The probability of East having (a) no aces is \(\frac{C(27, 13)}{C(39, 13)}\), (b) 2 or more aces is \(1 - \left(\frac{C(27, 13)+C(12, 1)*C(27, 12)}{C(39, 13)}\right)\), and when West has exactly 1 ace, the probabilities are (c) no aces: \(\frac{C(27, 13)}{C(38, 13)}\), and 2 or more aces: \(1 - \left(\frac{C(27, 13)+C(11, 1)*C(27, 12)}{C(38, 13)}\right)\).

Step by step solution

01

Calculate the total number of card combinations for East

Since there are 52 cards in total and West has 13 of them without any aces, this leaves 39 cards for East to receive from (12 of them being aces). As each player receives 13 cards, we must find all possible combinations (C) of 13 cards that East can get from the remaining 39: Total combinations for East = C(39, 13)
02

Calculate the probability of East having no aces

In this case, East would receive all his 13 cards from the remaining 27 non-ace cards. Thus, we have to find: Combinations without aces = C(27, 13) Now, we divide this by the total combinations calculated in Step 1 to find the probability: Probability of East having no aces = \(\frac{C(27, 13)}{C(39, 13)}\)
03

Calculate the probability of East having 2 or more aces

We can find this probability by getting the complement of having 0 or 1 aces: Probability of having 2 or more aces = 1 - (Probability of having no aces + Probability of having 1 ace) We have already found the probability of having no aces. Now, let's find the probability of having 1 ace, which involves East having exactly 1 of the 12 aces and the remaining 12 cards coming from the 27 non-ace cards: Combinations with 1 ace = C(12, 1) * C(27, 12) Probability of East having 1 ace = \(\frac{C(12, 1) * C(27, 12)}{C(39, 13)}\) Now, we proceed to calculate the probability of having 2 or more aces: Probability of East having 2 or more aces = \(1 - \left(\frac{C(27, 13)+C(12, 1)*C(27, 12)}{C(39, 13)}\right)\)
04

Conditional probability with West having 1 ace

Now that West has one of the aces, East can only receive from the remaining 38 cards (11 of them being aces). In this case, total combinations for East become: Total combinations with West having 1 ace = C(38, 13) Now, we repeat the calculations from the previous steps accordingly: - Probability of East having no aces: \(P(\text{no aces})=\frac{C(27, 13)}{C(38, 13)}\) - Probability of East having 2 or more aces: We calculate the complementary probability again: \(P(\text{2 or more aces}) = 1 - \left(\frac{C(27, 13)+C(11, 1)*C(27, 12)}{C(38, 13)}\right)\) So, the probability of East having (a) no aces is \(\frac{C(27, 13)}{C(39, 13)}\), the probability of having (b) 2 or more aces is \(1 - \left(\frac{C(27, 13)+C(12, 1)*C(27, 12)}{C(39, 13)}\right)\), and the probabilities when West has exactly 1 ace are (c) for no aces: \(\frac{C(27, 13)}{C(38, 13)}\), and for 2 or more aces: \(1 - \left(\frac{C(27, 13)+C(11, 1)*C(27, 12)}{C(38, 13)}\right)\).

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Key Concepts

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

Combinatorics
Combinatorics is a field of mathematics that focuses on counting, arranging, and finding the structure within distinct patterns. In card games like bridge, combinatorics allows us to calculate the different ways cards can be distributed amongst players.

For example, considering a standard deck of 52 cards, if you want to determine how many different 13-card hands there are, you would use a combinatory formula called 'combinations'. The formula for combinations is represented as \( C(n, k) = \frac{n!}{k!(n - k)!} \), where \( n \) is the total number of items to choose from, \( k \) is the number of items to choose, and \( ! \) denotes a factorial, the product of an integer and all the integers below it.

This principle is crucial in calculating the probabilities in card games, as knowing how many combinations are possible underpins the entire calculation of chances.
Conditional Probability
Conditional probability is the likelihood of an event occurring, given that another event has already happened. In card games, calculating conditional probabilities can significantly affect strategic decisions.

For instance, if you know West has no aces in a game of bridge, you adjust the probability calculations for East's hand considering this information. This is different from calculating the chances of East receiving no aces from a full untouched deck.

The formula for conditional probability is represented as \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of event \( A \) occurring given that \( B \) has happened, and \( P(A \cap B) \) is the probability of both events happening. In our context, this could involve calculating the probability of East getting two or more aces, knowing West's lack of them.
Probability Calculations
Probability calculations involve determining the chance of a particular event happening. They are fundamental to understanding games of chance and skill, such as card games.

To calculate a probability, you divide the number of favorable outcomes by the total number of possible outcomes. In card games, favorable outcomes refer to the specific card distributions of interest, while total outcomes are the combination of all possible ways the cards could be dealt.

For instance, if you need to know the likelihood of East getting exactly two aces in bridge, given West's hand, you find all the ways East can receive two aces and divide by all the possible ways East can receive any 13 cards. It's important to subtract probabilities of other specific outcomes when calculating the probability of 'at least' scenarios, such as having 'at least two aces,' which was done in the step-by-step solution by using the complement rule.

Practical application of these calculations in actual gameplay can enhance a player's strategic decisions and their understanding of the odds they are facing.

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Most popular questions from this chapter

A person tried by a 3-judge panel is declared guilty if at least 2 judges cast votes of guilty. Suppose that when the defendant is, in fact, guilty, each judge will independently vote guilty with probability. 7, whereas when the defendant is, in fact, innocent, this probability drops to \(.2\). If 70 percent of defendants are guilty, compute the conditional probability that judge number 3 votes guilty given that (a) judges 1 and 2 vote guilty; (b) judges 1 and 2 cast 1 guilty and 1 not guilty vote; (c) judges 1 and 2 both cast not guilty votes. Let \(E_{i}, i=1,2,3\) denote the event that judge \(i\) casts a guilty vote. Are these events independent. Are they conditionally independent? Explain.

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die \(A\); if it lands tails, then die \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}\).

Suppose that a nonmathematical but philosophically minded friend of yours claims that Laplace's rule of succession must be incorrect because it can lead to ridiculous conclusions. "For instance," says he, "if a boy is 10 years old, the rule states that having lived 10 years, the boy has probability \(\frac{15}{12}\) of living another year. On the other hand, if the boy has an 80 -year-old grandfather, then by Laplace's rule the grandfather has probability \(\frac{81}{82}\) of surviving another year. However, this is ridiculous. Clearly, the boy is more likely to survive an additional year than is the grandfather." How would you answer your friend?

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