| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To compute the joint entropy of the sequence of colors \(X_1, X_2, \ldots, X_{52}\), we need to consider the distribution of the sequence of colors when the deck is shuffled. Since there are 26 red cards and 26 black cards, the total number of possible sequences is \(\binom{52}{26}\), which is the number of ways to choose 26 positions for the red cards out of 52 total positions. The joint probability of any particular sequence of colors is \(\frac{1}{\binom{52}{26}}\), because each sequence is equally likely when the deck is shuffled. The joint entropy \(H(X_1, X_2, \ldots, X_{52})\) is given by the formula: \[H(X_1, X_2, \ldots, X_{52}) = -\sum_{x_1, x_2, \ldots, x_{52}} p(x_1, x_2, \ldots, x_{52}) \log_2 p(x_1, x_2, \ldots, x_{52})\] where the sum is over all possible sequences of colors \(x_1, x_2, \ldots, x_{52}\), and \(p(x_1, x_2, \ldots, x_{52})\) is the probability of the sequence. Substituting the probability of any sequence \(\frac{1}{\binom{52}{26}}\) into the entropy formula, we get: \[H(X_1, X_2, \ldots, X_{52}) = -\binom{52}{26} \cdot \frac{1}{\binom{52}{26}} \cdot \log_2 \frac{1}{\binom{52}{26}}\] \[H(X_1, X_2, \ldots, X_{52}) = -\log_2 \frac{1}{\binom{52}{26}}\] Using the fact that \(\binom{52}{26} = \frac{52!}{26! \cdot 26!}\), we can simplify the expression for the entropy: \[H(X_1, X_2, \ldots, X_{52}) = \log_2 \binom{52}{26}\] Using a calculator or computer to evaluate \(\binom{52}{26}\), we find that \(\binom{52}{26} \approx 5.36447 \times 10^{28}\). Therefore, the joint entropy of the sequence of colors is: \[H(X_1, X_2, \ldots, X_{52}) \approx \log_2 (5.36447 \times 10^{28}) \approx 96.57 \text{ bits}\] So, the joint entropy of the sequence of colors when a deck of cards is shuffled and dealt out one card at a time without replacement is approximately 96.57 bits. |
48.8 |
| deepseekr1 | To compute the entropy \( H(X_1, X_2, \ldots, X_{52}) \) for the sequence of card colors drawn from a shuffled deck with 26 red and 26 black cards, we consider the following: 1. **Entropy of a Uniform Distribution**: The entropy of a sequence of random variables is maximized when all outcomes are equally likely. For a shuffled deck, each permutation of the cards is equally likely. However, since we are only concerned with the color sequence, we need to consider the number of distinct color sequences. 2. **Number of Distinct Color Sequences**: The number of distinct color sequences is given by the binomial coefficient \( \binom{52}{26} \), which represents the number of ways to arrange 26 red and 26 black cards in a deck of 52 cards. 3. **Entropy Calculation**: The entropy of a uniform distribution over all possible outcomes is the logarithm of the number of possible outcomes. Therefore, the entropy \( H(X_1, X_2, \ldots, X_{52}) \) is the base-2 logarithm of the binomial coefficient \( \binom{52}{26} \). Thus, the entropy \( H(X_1, X_2, \ldots, X_{52}) \) is: \[ \boxed{\log_2 \dbinom{52}{26}} \] bits. |
48.8 |
| gpto1 | To compute the entropy \( H(X_1, X_2, \ldots, X_{52}) \) in bits, we need to determine the total uncertainty associated with the sequence of colors as the cards are dealt from the shuffled deck. **Understanding the Problem:** - An ordinary deck has **52 cards**: **26 red** and **26 black**. - The deck is thoroughly shuffled, so all permutations of the deck are equally likely. - We are interested in the entropy of the sequence of colors \( (X_1, X_2, \ldots, X_{52}) \) as cards are dealt one by one without replacement. **Key Concepts:** - **Entropy (in bits)**: For a discrete random variable with equally likely outcomes, the entropy \( H \) is the logarithm (base 2) of the number of possible outcomes. \[ H = \log_2 (\text{Number of Possible Outcomes}) \] - **Number of Possible Sequences:** - The total number of ways to arrange 26 red and 26 black cards in a sequence is given by the binomial coefficient: \[ \text{Number of Sequences} = \binom{52}{26} \] This represents the number of ways to choose 26 positions out of 52 for the red cards (the black cards fill the remaining positions). **Calculating the Entropy:** 1. **Compute the Total Number of Sequences:** \[ \text{Number of Sequences} = \binom{52}{26} = \frac{52!}{26! \times 26!} \] 2. **Compute the Entropy:** \[ H = \log_2 \left( \binom{52}{26} \right) = \log_2 \left( \frac{52!}{26! \times 26!} \right) \] This expression gives the entropy in bits for the sequence \( (X_1, X_2, \ldots, X_{52}) \). It represents the total uncertainty in the color sequence of the shuffled deck. **Answer:** An explicit expression: H(X₁,…,X₅₂) = log₂ [52! ⁄ (26!·26!)] bits. |
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| GPT-4o | To solve the problem of computing \( H(X_1, X_2, \ldots, X_{52}) \), where \( X_i \) represents the color of the \( i \)-th card in a shuffled deck of 52 cards (26 red and 26 black), we need to determine the entropy of the sequence of colors. Entropy \( H(X_1, X_2, \ldots, X_{52}) \) measures the uncertainty of the entire sequence of colors. To find this, we can use the property of entropy for a deterministic process and the fact that the total number of possible sequences is fixed by the combinatorics of the problem. 1. **Number of Possible Sequences:** The total number of possible sequences of colors when dealing all 52 cards is given by the combination \[ \binom{52}{26} \] because we are choosing 26 positions out of 52 for the red cards (or equivalently for the black cards). 2. **Combinatorial Entropy:** The entropy of a system where each sequence is equally likely is given by the logarithm of the number of possible sequences. Thus, \[ H(X_1, X_2, \ldots, X_{52}) = \log_2 \binom{52}{26} \] 3. **Calculating \( \binom{52}{26} \):** \[ \binom{52}{26} = \frac{52!}{26! \cdot 26!} \] Although calculating the factorials directly can be cumbersome, we can use approximations or computational tools to find the exact value. However, for entropy purposes, we will focus on the logarithmic form. 4. **Using Stirling's Approximation:** Stirling's approximation for factorial \( n! \) is: \[ n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \] Applying this approximation to our combination, we get: \[ \binom{52}{26} \approx \frac{\sqrt{2 \pi \cdot 52} \left(\frac{52}{e}\right)^{52}}{\left(\sqrt{2 \pi \cdot 26} \left(\frac{26}{e}\right)^{26}\right)^2} = \frac{\sqrt{2 \pi \cdot 52} \left(\frac{52}{e}\right)^{52}}{2 \pi \cdot 26 \left(\frac{26}{e}\right)^{52}} \] Simplifying, we get: \[ \binom{52}{26} \approx \frac{\sqrt{52}}{2 \pi \cdot 26} \cdot \left(2\right)^{52} \] 5. **Entropy Calculation:** \[ H(X_1, X_2, \ldots, X_{52}) \approx \log_2 \left( \frac{\sqrt{52}}{2 \pi \cdot 26} \cdot 2^{52} \right) \] Simplifying further, \[ H(X_1, X_2, \ldots, X_{52}) \approx \log_2 \left(2^{52}\right) + \log_2 \left(\frac{\sqrt{52}}{2 \pi \cdot 26} \right) \] \[ H(X_1, X_2, \ldots, X_{52}) \approx 52 + \log_2 \left(\frac{\sqrt{52}}{2 \pi \cdot 26} \right) \] Since \(\log_2 \left(\frac{\sqrt{52}}{2 \pi \cdot 26}\right)\) is a small correction term, the dominant term is \(52\). 6. **Final Answer:** Therefore, the entropy \( H(X_1, X_2, \ldots, X_{52}) \) in bits is approximately: \[ H(X_1, X_2, \ldots, X_{52}) \approx 52 \text{ bits} \] This result aligns with the fact that the entropy of a perfectly shuffled deck of cards (considering only the color sequence) mainly comes from the combinatorial possibilities of arranging 26 red and 26 black cards, leading to an entropy close to 52 bits. |
48.8 |
| llama3.1-70B-Instruct | The entropy of the sequence of cards is equal to the entropy of a single card, since the cards are dealt without replacement and the sequence is completely determined by the first card. The entropy of a single card is $$H(X_1) = -\sum_{x\in\{R,B\}} p(x)\log_2 p(x) = -\frac{1}{2}\log_2\frac{1}{2} - \frac{1}{2}\log_2\frac{1}{2} = 1$$ Therefore, the entropy of the sequence of cards is $H(X_1,X_2,\ldots,X_{52}) = 1$ bit. Note: The entropy of a single card is 1 bit because there are two possible outcomes (red or black) with equal probability, and the entropy of a binary random variable with equal probabilities is 1 bit. |
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| llama3.1-405B-Instruct | Since the cards are dealt out without replacement, the probability of each card being red or black is not independent. However, since there are an equal number of red and black cards, the probability of each card being red or black is still 1/2. Since there are 52 cards in total, the probability distribution of the sequence $X_1, X_2, \ldots, X_{52}$ is uniform over all possible sequences of 52 bits with 26 ones and 26 zeros. The number of such sequences is given by the binomial coefficient: $$\binom{52}{26} = \frac{52!}{26!26!}$$ The entropy of the sequence is therefore: $$H(X_1, X_2, \ldots, X_{52}) = \log_2 \binom{52}{26}$$ Using Stirling's approximation, we can approximate the binomial coefficient as: $$\binom{52}{26} \approx \frac{2^{52}}{\sqrt{52\pi}}$$ Taking the logarithm base 2, we get: $$H(X_1, X_2, \ldots, X_{52}) \approx 52 - \frac{1}{2} \log_2 (52\pi)$$ Simplifying, we get: $$H(X_1, X_2, \ldots, X_{52}) \approx 52 - 5.35$$ Therefore, the entropy of the sequence is approximately: $$H(X_1, X_2, \ldots, X_{52}) \approx 46.65$$ So the entropy of the sequence is approximately 46.65 bits. |
48.8 |
| DeepSeek-Coder-V2 | To solve the problem of computing the joint entropy \( H(X_1, X_2, \ldots, X_{52}) \) in bits, we need to follow these steps: 1. **Understand the Entropy Formula**: The entropy \( H(X_1, X_2, \ldots, X_{52}) \) for a sequence of random variables is given by: \[ H(X_1, X_2, \ldots, X_{52}) = -\sum_{x_1, x_2, \ldots, x_{52}} P(x_1, x_2, \ldots, x_{52}) \log_2 P(x_1, x_2, \ldots, x_{52}) \] where \( P(x_1, x_2, \ldots, x_{52}) \) is the probability of a specific sequence of 52 cards. 2. **Determine the Total Number of Sequences**: Since the deck is shuffled randomly, each possible sequence of 52 cards is equally likely. The total number of possible sequences of 52 cards from a deck of 52 cards is: \[ 52! \] 3. **Calculate the Probability of Each Sequence**: Each specific sequence of 52 cards has the same probability, which is: \[ P(x_1, x_2, \ldots, x_{52}) = \frac{1}{52!} \] 4. **Compute the Entropy**: Substituting the probability into the entropy formula, we get: \[ H(X_1, X_2, \ldots, X_{52}) = -\sum_{x_1, x_2, \ldots, x_{52}} \frac{1}{52!} \log_2 \left( \frac{1}{52!} \right) \] Since there are \( 52! \) terms in the sum, and each term is the same, the sum simplifies to: \[ H(X_1, X_2, \ldots, X_{52}) = -52! \cdot \frac{1}{52!} \log_2 \left( \frac{1}{52!} \right) \] This further simplifies to: \[ H(X_1, X_2, \ldots, X_{52}) = -\log_2 \left( \frac{1}{52!} \right) \] Using the properties of logarithms, we can rewrite this as: \[ H(X_1, X_2, \ldots, X_{52}) = \log_2 (52!) \] 5. **Conclusion**: Therefore, the joint entropy \( H(X_1, X_2, \ldots, X_{52}) \) in bits is: \[ H(X_1, X_2, \ldots, X_{52}) = \log_2 (52!) \] Thus, the final answer is: \[ \boxed{\log_2 (52!)} \] |
48.8 |
| Qwen2-72B-Instruct | To compute the joint entropy of the sequence of colors $X_1, X_2, \ldots, X_{52}$, we need to consider the distribution of the sequence of colors when the deck is shuffled. Since there are 26 red cards and 26 black cards, the total number of possible sequences is $\binom{52}{26}$, which is the number of ways to choose 26 positions for the red cards out of 52 total positions. The joint probability of any particular sequence of colors is $\frac{1}{\binom{52}{26}}$, because each sequence is equally likely when the deck is shuffled. The joint entropy $H(X_1, X_2, \ldots, X_{52})$ is given by the formula: $$H(X_1, X_2, \ldots, X_{52}) = -\sum_{x_1, x_2, \ldots, x_{52}} p(x_1, x_2, \ldots, x_{52}) \log_2 p(x_1, x_2, \ldots, x_{52})$$ where the sum is over all possible sequences of colors $x_1, x_2, \ldots, x_{52}$, and $p(x_1, x_2, \ldots, x_{52})$ is the probability of the sequence. Substituting the probability of any sequence $\frac{1}{\binom{52}{26}}$ into the entropy formula, we get: $$H(X_1, X_2, \ldots, X_{52}) = -\binom{52}{26} \cdot \frac{1}{\binom{52}{26}} \cdot \log_2 \frac{1}{\binom{52}{26}}$$ $$H(X_1, X_2, \ldots, X_{52}) = -\log_2 \frac{1}{\binom{52}{26}}$$ Using the fact that $\binom{52}{26} = \frac{52!}{26! \cdot 26!}$, we can simplify the expression for the entropy: $$H(X_1, X_2, \ldots, X_{52}) = \log_2 \binom{52}{26}$$ Using a calculator or computer to evaluate $\binom{52}{26}$, we find that $\binom{52}{26} \approx 5.36447 \times 10^{28}$. Therefore, the joint entropy of the sequence of colors is: $$H(X_1, X_2, \ldots, X_{52}) \approx \log_2 (5.36447 \times 10^{28}) \approx 96.57 \text{ bits}$$ So, the joint entropy of the sequence of colors when a deck of cards is shuffled and dealt out one card at a time without replacement is approximately 96.57 bits. |
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| baichuan4 | To compute the entropy $H(X_1,X_2,\ldots,X_{52})$ in bits, we need to consider the probability distribution of the sequence of colors for the 52 cards. Since the deck is shuffled randomly, each possible ordering of the cards is equally likely. There are $2^{52}$ possible sequences of 52 cards, each with a probability of $\frac{1}{2^{52}}$. However, not all sequences are distinguishable in terms of color patterns because there are only 26 red and 26 black cards. The number of distinguishable color patterns is given by the binomial coefficient $\binom{52}{26}$, which is the number of ways to choose 26 positions for the red cards out of 52 positions. Now, let's calculate the entropy: $$ H(X_1,X_2,\ldots,X_{52}) = -\sum_{i=1}^{2^{52}} P(x_i) \log_2 P(x_i) $$ Since each distinguishable color pattern has the same probability, we can simplify this to: $$ H(X_1,X_2,\ldots,X_{52}) = -\binom{52}{26} \cdot \frac{1}{2^{52}} \cdot \log_2 \left(\frac{1}{2^{52}}\right) $$ Substituting the value of $\binom{52}{26}$ and simplifying: $$ H(X_1,X_2,\ldots,X_{52}) = -\frac{52!}{26!26!} \cdot \frac{1}{2^{52}} \cdot (-52) $$ $$ H(X_1,X_2,\ldots,X_{52}) = 52 \cdot \frac{52!}{26!26!} \cdot \frac{1}{2^{52}} $$ This is the entropy of the sequence of 52 card colors in bits. |
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| doubao | Each card has two possible colors (red or black), so for each $X_i$, there are 2 possible outcomes. The entropy of a single card is $H(X_i) = -\log_2(\frac{1}{2}) = 1$ bit. For 52 cards, the total entropy is $H(X_1,X_2,\ldots,X_{52}) = 52 \times 1 = 52$ bits. |
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| minimax | To compute the joint entropy $H(X_1, X_2, \ldots, X_{52})$ of the random variables representing the colors of the cards in a deck, we need to understand that the entropy of a set of random variables is the expected value of the sum of the individual entropies of each random variable, given the values of the other random variables. However, in this case, the cards are dealt one at a time without replacement, and the color of each card is independent of the order in which the cards are dealt. This means that the probability of drawing a red card or a black card remains constant at each draw. Specifically, the probability of drawing a red card is always 26/52 (or 1/2), and the probability of drawing a black card is also 26/52 (or 1/2). Since the cards are independent and identically distributed, the joint entropy of the sequence of cards is simply the sum of the entropies of each draw. The entropy of a single draw is given by the Shannon entropy formula: $$ H(X) = -\sum_{i=1}^{n} p(x_i) \log_2 p(x_i) $$ For a single card, where $p(\text{red}) = p(\text{black}) = 1/2$, the entropy is: $$ H(X_i) = -\left( \frac{1}{2} \log_2 \frac{1}{2} + \frac{1}{2} \log_2 \frac{1}{2} \right) $$ $$ H(X_i) = -\left( \frac{1}{2} \cdot (-1) + \frac{1}{2} \cdot (-1) \right) $$ $$ H(X_i) = 1 \text{ bit} $$ Since each card's draw is independent, the joint entropy for all 52 cards is simply 52 times the entropy of a single draw: $$ H(X_1, X_2, \ldots, X_{52}) = 52 \cdot H(X_i) $$ $$ H(X_1, X_2, \ldots, X_{52}) = 52 \cdot 1 \text{ bit} $$ $$ H(X_1, X_2, \ldots, X_{52}) = 52 \text{ bits} $$ Therefore, the joint entropy of the colors of the cards in a shuffled deck dealt out one at a time without replacement is 52 bits. |
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| kimi | To compute the entropy \( H(X_1, X_2, \ldots, X_{52}) \) of the sequence of 52 cards, we first need to understand what entropy is and how it is calculated. Entropy, in the context of information theory, is a measure of the uncertainty or randomness in a set of outcomes. In the case of a deck of cards, each card has two possible outcomes: red or black. For a single card \( X_i \), the probability of drawing a red card \( P(X_i = \text{red}) \) is \( \frac{26}{52} = \frac{1}{2} \), and the probability of drawing a black card \( P(X_i = \text{black}) \) is also \( \frac{1}{2} \). The entropy \( H(X_i) \) for a single card is given by the formula: \[ H(X_i) = -\sum_{k} p_k \log_2(p_k) \] where \( p_k \) is the probability of the \( k \)-th outcome. For a single card, this becomes: \[ H(X_i) = -\left( \frac{1}{2} \log_2 \frac{1}{2} + \frac{1}{2} \log_2 \frac{1}{2} \right) \] \[ H(X_i) = -\left( \frac{1}{2} \times (-1) + \frac{1}{2} \times (-1) \right) \] \[ H(X_i) = 1 \text{ bit} \] Since each card is dealt without replacement, the entropy of the entire sequence \( H(X_1, X_2, \ldots, X_{52}) \) is simply the sum of the entropies of each individual card because the cards are independent events. \[ H(X_1, X_2, \ldots, X_{52}) = H(X_1) + H(X_2) + \ldots + H(X_{52}) \] Since \( H(X_i) = 1 \) bit for each card, and there are 52 cards: \[ H(X_1, X_2, \ldots, X_{52}) = 52 \times 1 \text{ bit} \] \[ H(X_1, X_2, \ldots, X_{52}) = 52 \text{ bits} \] So, the entropy of the sequence of 52 cards is 52 bits. |
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| wenxinyiyan | 要计算$H(X_1,X_2,\ldots,X_{52})$，我们首先需要理解这个问题。这是一个关于信息论中熵的计算问题，其中$H$表示熵，$X_i$表示第$i$张卡片的颜色。 首先，我们考虑第一张卡片$X_1$。由于牌堆中红色和黑色卡片各有26张，所以$P(X_1=\text{red}) = P(X_1=\text{black}) = \frac{1}{2}$。因此，$H(X_1) = -(\frac{1}{2}\log_2\frac{1}{2} + \frac{1}{2}\log_2\frac{1}{2}) = 1$ bit。 接下来，我们考虑第二张卡片$X_2$。在已知$X_1$的情况下，$X_2$的概率分布会发生变化。例如，如果$X_1=\text{red}$，那么剩下的卡片中有25张红色和26张黑色，所以$P(X_2=\text{red}|X_1=\text{red}) = \frac{25}{51}$，$P(X_2=\text{black}|X_1=\text{red}) = \frac{26}{51}$。类似地，我们可以计算$P(X_2=\text{red}|X_1=\text{black})$和$P(X_2=\text{black}|X_1=\text{black})$。然后，我们可以使用这些条件概率来计算$H(X_2|X_1)$。 然而，由于这个问题的特殊性，我们可以观察到，在已知前$i-1$张卡片的情况下，第$i$张卡片的颜色只有两种可能，且这两种可能的概率相等（因为红色和黑色卡片的数量总是相等的）。因此，对于所有的$i$，我们都有$H(X_i|X_{i-1},\ldots,X_1) = 1$ bit。 最后，由于熵的链式法则，我们有 $H(X_1,X_2,\ldots,X_{52}) = H(X_1) + H(X_2|X_1) + H(X_3|X_2,X_1) + \cdots + H(X_{52}|X_{51},\ldots,X_1) = 52 \times 1 = 52$ bits。 因此，$H(X_1,X_2,\ldots,X_{52}) = 52$ bits。 |
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| zhipu | To solve this problem, we need to compute the entropy $H(X_1, X_2, \ldots, X_{52})$ of the joint distribution of the colors of the 52 cards dealt from the deck. Entropy in information theory measures the uncertainty or randomness of a random variable or a set of random variables. ### Step-by-Step Solution: 1. **Understanding the Variables:** - Each $X_i$ can take on two values: Red (R) or Black (B). - There are 26 red cards and 26 black cards in the deck. 2. **Joint Distribution:** - The joint distribution of $(X_1, X_2, \ldots, X_{52})$ describes the probability of every possible sequence of 52 cards. - Since the deck is shuffled, every permutation of the 52 cards is equally likely. 3. **Total Number of Permutations:** - The total number of ways to arrange 52 cards is $52!$. - The number of ways to arrange 26 red cards and 26 black cards in a specific sequence is given by the multinomial coefficient: \[ \frac{52!}{26!26!} \] - Each specific sequence (permutation) has a probability of: \[ P((X_1, X_2, \ldots, X_{52}) = (x_1, x_2, \ldots, x_{52})) = \frac{1}{\frac{52!}{26!26!}} \] 4. **Entropy Calculation:** - The entropy $H(X_1, X_2, \ldots, X_{52})$ is given by: \[ H(X_1, X_2, \ldots, X_{52}) = -\sum_{(x_1, x_2, \ldots, x_{52})} P((X_1, X_2, \ldots, X_{52}) = (x_1, x_2, \ldots, x_{52})) \log_2 P((X_1, X_2, \ldots, X_{52}) = (x_1, x_2, \ldots, x_{52})) \] - Since all permutations are equally likely, we can simplify this to: \[ H(X_1, X_2, \ldots, X_{52}) = - \left( \frac{52!}{26!26!} \right) \left( \frac{1}{\frac{52!}{26!26!}} \log_2 \frac{1}{\frac{52!}{26!26!}} \right) \] - This simplifies to: \[ H(X_1, X_2, \ldots, X_{52}) = \log_2 \left( \frac{52!}{26!26!} \right) \] 5. **Computing the Logarithm:** - We need to compute $\log_2 \left( \frac{52!}{26!26!} \right)$. - Using Stirling's approximation for factorials, $n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$, we get: \[ \log_2 (n!) \approx \frac{1}{2} \log_2 (2 \pi n) + n \log_2 \left( \frac{n}{e} \right) \] - Applying this to $52!$, $26!$, and $26!$: \[ \log_2 (52!) \approx \frac{1}{2} \log_2 (2 \pi \cdot 52) + 52 \log_2 \left( \frac{52}{e} \right) \] \[ \log_2 (26!) \approx \frac{1}{2} \log_2 (2 \pi \cdot 26) + 26 \log_2 \left( \frac{26}{e} \right) \] - Therefore: \[ \log_2 \left( \frac{52!}{26!26!} \right) \approx \left( \frac{1}{2} \log_2 (2 \pi \cdot 52) + 52 \log_2 \left( \frac{52}{e} \right) \right) - 2 \left( \frac{1}{2} \log_2 (2 \pi \cdot 26) + 26 \log_2 \left( \frac{26}{e} \right) \right) \] 6. **Simplifying the Expression:** - After simplifying the above expression, we get an approximate value for the entropy. ### Exact Calculation: For exact calculation, we can use logarithmic properties and compute directly: \[ \log_2 \left( \frac{52!}{26!26!} \right) = \log_2 (52!) - 2 \log_2 (26!) \] Using a calculator or software to compute the exact values: \[ \log_2 (52!) \approx 225.58 \] \[ \log_2 (26!) \approx 107.99 \] \[ \log_2 \left( \frac{52!}{26!26!} \right) = 225.58 - 2 \times 107.99 = 225.58 - 215.98 = 9.60 \] ### Final Answer: \[ H(X_1, X_2, \ldots, X_{52}) \approx 9.60 \text{ bits} \] Thus, the entropy $H(X_1, X_2, \ldots, X_{52})$ in bits is approximately 9.60 bits. |
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| gpto1 | To compute the entropy \( H(X_1, X_2, \ldots, X_{52}) \) in bits, we need to determine the total uncertainty associated with the sequence of colors as the cards are dealt from the shuffled deck. **Understanding the Problem:** - An ordinary deck has **52 cards**: **26 red** and **26 black**. - The deck is thoroughly shuffled, so all permutations of the deck are equally likely. - We are interested in the entropy of the sequence of colors \( (X_1, X_2, \ldots, X_{52}) \) as cards are dealt one by one without replacement. **Key Concepts:** - **Entropy (in bits)**: For a discrete random variable with equally likely outcomes, the entropy \( H \) is the logarithm (base 2) of the number of possible outcomes. \[ H = \log_2 (\text{Number of Possible Outcomes}) \] - **Number of Possible Sequences:** - The total number of ways to arrange 26 red and 26 black cards in a sequence is given by the binomial coefficient: \[ \text{Number of Sequences} = \binom{52}{26} \] This represents the number of ways to choose 26 positions out of 52 for the red cards (the black cards fill the remaining positions). **Calculating the Entropy:** 1. **Compute the Total Number of Sequences:** \[ \text{Number of Sequences} = \binom{52}{26} = \frac{52!}{26! \times 26!} \] 2. **Compute the Entropy:** \[ H = \log_2 \left( \binom{52}{26} \right) = \log_2 \left( \frac{52!}{26! \times 26!} \right) \] This expression gives the entropy in bits for the sequence \( (X_1, X_2, \ldots, X_{52}) \). It represents the total uncertainty in the color sequence of the shuffled deck. **Answer:** An explicit expression: H(X₁,…,X₅₂) = log₂ [52! ⁄ (26!·26!)] bits. |
48.8 |
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