Help! Horizontal Asymptote!

Once again, I'm coming to Brilliant with another question. This time, it's on the limits of probabilities.

I was investigating this problem:

You are playing a computer game with \(n-1\) of your friends (a total of \(n\) people), and each of you get a chest with a \(\frac{1}{n}\) chance of giving you a prize. What is the probability that none of you get a prize?

I realized that each party member has a \(\frac{n-1}{n}\) chance of not getting a prize. This approaches 100% as you have more and more party members. The party probability is easy to find once you find the individual probability. Since the probability for each party member is \(\frac{n-1}{n}\), you just need to raise it to the \(n^{\text{th}}\) power, since there are \(n\) people in the party.

Then, I got curious about the infinite limit of the function: \[\displaystyle\lim_{x\rightarrow \infty} f(x)=\left(\frac{x-1}{x}\right)^x\] After I graphed it, I realized that there is a horizontal asymptote, and it cannot be calculated through traditional methods. Could someone explain how to get the solution (not just through WolframAlpha...)? I know it is approximately \(y=.368\) through direct observation.

It is actually very interesting that, if there is an arbitrarily large amount of people, the probability that no one will get a prize is more than a third! Talk about bad odds!

Note by Blan Morrison
2 weeks, 5 days ago

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Well, since the probability of a person getting a prize depends on the number of people themselves, the result is not so surprising.

How did you stumble upon this problem?

Agnishom Chattopadhyay - 2 weeks, 4 days ago

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Actually, this problem is based off of an actual scenario that I was involved in. I just got curious and decided to generalize it.

Blan Morrison - 2 weeks, 4 days ago

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Did you realise that the number you produced is actually \(\frac{1}{e}\)? You have a real scenario? That is interesting. Can you tell me about it?

Agnishom Chattopadhyay - 2 weeks, 4 days ago

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@Agnishom Chattopadhyay Oh my goodness; you're correct! That's crazy. Did you actually work it out, or were you just messing around? If you did calculate it, how so?


Scenario: I play this mobile game called Clash Royale. They recently added a feature where everyone in your group, called a clan, gets a chest after winning in an event. The winning chest has a 1 in 10 chance of giving you a special prize. Now, my clan has 30 members, so I knew that I should expect 3 people in my clan to get the prize. Then, I started getting curious about the probability of no one getting a prize. You can infer what happened from there: I calculated it, decided to generalize for a clan of \(n\) members with a 1 in \(n\) chance of getting the prize, and then I got here.

Blan Morrison - 2 weeks, 3 days ago

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@Blan Morrison Well, it is an well known fact that \[ e^x = \lim_{n \to \infty} ( 1 + \frac{x}{n})^n \]


I am not sure I am following. Why does the chance of someone getting a prize in the clan reduce when there are more people?

Agnishom Chattopadhyay - 2 weeks, 3 days ago

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@Agnishom Chattopadhyay Ah, I see. Replacing \(x\) with -1 gives you that same function, which makes the limit equal to \(e^{-1}\). What is the name of this famous limit? I've never seen it.


In the original scenario, the probability of getting the prize was independent of the number of members in the clan. I modified that scenario, giving us the one mentioned in the note.

Blan Morrison - 2 weeks, 3 days ago

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@Blan Morrison I don't really know what it is called. I learnt it in high school.

Cool. Maybe you can post this as a problem?

Agnishom Chattopadhyay - 2 weeks, 3 days ago

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@Agnishom Chattopadhyay Here it is! Thanks for the help!

Blan Morrison - 2 weeks, 1 day ago

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