ENT

CasinoGames

Okay, so I need to figure out how to solve this problem about the expected number of people who have to be asked before finding someone with a birthday on a specific day, like January 1st. Let me start by understanding the problem.


First, the question is about probability and expectation. We have a group of people, and we want to know the expected number of people we need to ask until we find one whose birthday is on January 1st. Assuming birthdays are uniformly distributed throughout the year, ignoring leap years, there are 365 days. So the probability that any one person has a birthday on January 1st is 1/365.


Hmm, this seems similar to the concept of geometric distribution. In a geometric distribution, we're interested in the number of trials needed to get the first success, where each trial has a probability p of success. The expected value for a geometric distribution is 1/p. So in this case, p is 1/365, so the expected number would be 365. But wait, let me make sure I'm not missing anything here.


Wait, the geometric distribution models the number of trials until the first success, right? So if each person we ask has a 1/365 chance of having a birthday on January 1st, then the expected number of people we need to ask is indeed 365. But let me think again to verify.


Alternatively, maybe I can approach this using expectation directly. Let's denote E as the expected number of people we need to ask. When we ask the first person, there's a 1/365 chance they have the birthday, and a 364/365 chance they don't. If they don't, then we've already asked one person, and we're back in the same situation, expecting to ask E more people. So the equation would be:


E = (1/365)1 + (364/365)(1 + E)


Let me explain this equation. With probability 1/365, the first person has the birthday, so we only needed to ask 1 person. With probability 364/365, the first person doesn't have the birthday, so we've spent 1 person and still expect to need E more. So solving this equation:


E = (1/365) + (364/365)*(1 + E)


Multiply out the right side:


E = 1/365 + (364/365) + (364/365)E


Combine the constants:


1/365 + 364/365 = 365/365 = 1


So now we have:


E = 1 + (364/365)E


Subtract (364/365)E from both sides:


E - (364/365)E = 1


(1 - 364/365)E = 1


(1/365)E = 1


Multiply both sides by 365:


E = 365


Okay, that confirms it. So the expected number is 365. That makes sense because if each trial has a 1/n chance, the expectation is n. So in this case, n is 365. Therefore, the answer should be 365.


Wait, but let me check if there's a different interpretation. For example, sometimes people might think about the probability of at least one person having a birthday on a specific day in a group of n people, but this question is different. It's about the expectation of the number of people to ask until finding the first one with that birthday. So yes, geometric distribution applies here.


Another way to think about it: the probability that the first person has the birthday is 1/365. The probability that the second person is the first one is (364/365)(1/365), the third is (364/365)^2(1/365), and so on. The expectation E is the sum over k=1 to infinity of k * (364/365)^(k-1) * (1/365). This is the standard expectation formula for a geometric distribution, which is indeed 1/p = 365.


So all methods point to the same answer. Therefore, I'm confident that the expected number of people to ask is 365.




The expected number of people you need to ask before finding someone with a birthday on January 1st is 365.


Step-by-Step Explanation:





Problem Setup:

We want the expected number of trials (people asked) until the first success (finding someone with a birthday on January 1st). Each trial has a success probability ( p = \frac{1}{365} ).



Geometric Distribution:

This scenario follows a geometric distribution, where the expected value ( E ) is given by ( E = \frac{1}{p} ). Substituting ( p = \frac{1}{365} ):
[
E = \frac{1}{\frac{1}{365}} = 365.
]



Direct Calculation via Expectation:

Let ( E ) be the expected number. If the first person succeeds (prob ( \frac{1}{365} )), we’re done in 1 trial. If not (prob ( \frac{364}{365} )), we’ve spent 1 trial and expect ( E ) more:
[
E = \left(\frac{1}{365}\right)(1) + \left(\frac{364}{365}\right)(1 + E).
]
Solving:
[
E = \frac{1}{365} + \frac{364}{365} + \frac{364}{365}E \implies E - \frac{364}{365}E = 1 \implies \frac{1}{365}E = 1 \implies E = 365.
]




Answer:

\boxed{365}