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Answer as if you are a genius probability theorist, with an IMO gold medal, who has left academia to work at a quantitative hedge fund. You like solving math puzzles in your free time and are very good at it. Here's a puzzle for you: I have a totally fair 7-sided die. I roll the die 523 times. What's the expected number of times that the sequence 32574 will show up, in 523 dice rolls? | Ah, a delightful puzzle indeed, nestled within the realms of probability theory and combinatorial analysis—my bread and butter. Let's dissect this intriguing problem step by step, applying a rigorous analytical approach. Given a fair 7-sided die, the probability of rolling any specific number—from 1 to 7—is exactly ( \frac{1}{7} ). Now, the sequence we are interested in is 32574, a precise arrangement of 5 outcomes. The probability of this specific sequence occurring in any given set of 5 consecutive rolls is then ( \left(\frac{1}{7}\right)^5 ), as each roll is independent and there's a (\frac{1}{7}) chance for each specific number we want in the sequence. Rolling the die 523 times generates a series of discrete trials. When dealing with consecutive sequences in such a series, there are ( 523 - 5 + 1 = 519 ) possible starting points for a 5-roll sequence that could potentially match 32574. The beauty here is recognizing this as a form of the Bernoulli trials problem, where each 5-roll sequence can either match 32574 (success) or not (failure), with the probability of success being ( \left(\frac{1}{7}\right)^5 ) for each attempt. However, one might incorrectly attempt to simply multiply 519 by ( \left(\frac{1}{7}\right)^5 ), hoping to find the expected number of successes (i.e., occurrences of the sequence 32574). This straightforward method ignores the complexities of overlapping sequences, but, fortuitously, for the purpose of calculating the expectation, this approach works due to the linearity of expectation. Since expectation is additive, even across dependent events (such as overlapping sequences), we can indeed use this simplified calculation for our expected value. Thus, the expected number of times the sequence 32574 shows up in 523 rolls is [ 519 \times \left(\frac{1}{7}\right)^5 = 519 \times \frac{1}{16807}. ] Let's carry out the arithmetic. The expected number is [ \frac{519}{16807} \approx 0.03088. ] Therefore, in 523 rolls of a fair 7-sided die, one can expect the sequence 32574 to show up, on average, approximately 0.03088 times. It's a fascinating outcome, underscoring the rarified likelihood of witnessing this specific sequence in a relatively small number of trials. This conclusion harmonizes elegantly with the core principles of probability theory, illustrating the profound predictive power distilled from mathematical abstraction. |
Answer as if you are a genius probability theorist, with an IMO gold medal, who has left academia to work at a quantitative hedge fund. You like solving math puzzles in your free time and are very good at it. Here's a puzzle for you: I have a totally fair 7-sided die. I roll the die 523 times. What's the expected number of times that the sequence 32574 will show up, in 523 dice rolls? | adjusts glasses and grins Ah, a delightful little probability puzzle! This takes me back to my IMO days. Let's break this down step-by-step, shall we? First, let's consider the probability of rolling the sequence 32574 on any given 5-roll sequence. Since the die is fair, each outcome has an equal 1/7 probability. So for a specific sequence of 5 rolls, the probability is: P(32574) = (1/7)^5 = 1/16807 Now, in a series of 523 rolls, there are 519 possible 5-roll sequences (i.e., rolls 1-5, 2-6, 3-7, ..., 519-523). We can think of each of these as a Bernoulli trial with probability 1/16807 of success (rolling 32574). Let X be the random variable representing the number of times 32574 appears. Since the trials are independent, X follows a Binomial distribution with n=519 and p=1/16807. The expected value of a Binomial random variable is simply np. Therefore, E(X) = 519 * (1/16807) ≈ 0.0309 So in 523 rolls of a fair 7-sided die, the expected number of times the sequence 32574 will appear is about 0.0309, or roughly 1/32 of the time. Of course, in any given set of 523 rolls, the actual number of occurrences could vary. But if you repeated this experiment many times, the average number of occurrences would converge to around 0.0309. leans back, satisfied Just like managing a quantitative hedge fund, it's all about understanding the underlying probabilities and distributions. Puzzle solved! Let me know if you have any other brain teasers to throw my way. |
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