Hilbert's Casino (pt. 1)

The probability is $\left(1 - \frac 1 n\right)^n.$ In general, $\lim_{n\to\infty} \left(1 + \frac x n\right)^n = e^x;$ in this case, $x = -1$ so that the probability is $e^{-1} \approx \boxed{0.368}.$

Relevant wiki: Probability - By Complement

Clearly, the probability for a finite number $n$ of players to all lose is $\left(1 - \dfrac{1}{n} \right)^n$ , and we need to find its limit as $n \to \infty$ .

We know that the very similar expression $\left(1 + \dfrac{1}{n} \right)^n$ tends to $e$ as $n \to \infty$ , so we manipulate our limit:

$\begin{aligned} \lim_{n \to \infty} \left(1 - \dfrac{1}{n} \right)^n &= \lim_{n \to \infty} \left( \dfrac{n - 1}{n} \right)^n\\ \\ &= \lim_{n \to \infty} \dfrac{1}{\left( \dfrac{n}{n - 1} \right)^n} \\ \\ &= \lim_{n \to \infty} \dfrac{1}{\left( 1 + \dfrac{1}{n - 1} \right)^n} \\ \\ &= \dfrac{1}{\displaystyle\lim_{n \to \infty} \left( 1 + \dfrac{1}{n - 1} \right)^{n-1} } \cdot \dfrac{1}{\displaystyle\lim_{n \to \infty} \left( 1 + \dfrac{1}{n - 1} \right)} \\ \\ &= \dfrac{1}{e} \cdot 1 = \boxed{\dfrac{1}{e}} \end{aligned}$

Relevant wiki: Probability - By Complement

Is a really really beautiful problem. I'm going to cry.

The answer is e Yes, the calculus constant. Yes, it is not a lie Follow me

Probability that guest 1 doesn't win

$(1-1/n)$

Probability that neither guest 1 nor 2 win

$(1-1/n)(1-1/n) = (1-1/n)^2$

Probability that neither guest 1, guest 2 nor guest 3 win

$(1-1/n)(1-1/n)(1-1/n)...=(1-1/n)^3$

Probability that neither guest 1, guest 2, guest 3... guest n-1 nor guest n win.

$(1-1/n)^n$

If you take the limit of that as n approaches infinity (for infinite guests) the result is 1/e

How is this possible? Follow me

Set $A = (1-1/n)^n$

$\ln(A) =\ln((1-1/n)^n)$

$\ln(A)=n*\ln(1-1/n)$

$\ln(A)=\ln(1-1/n)/n^{-1}$

Now, use L'Hopital to take the limit as n approaches infinity for ln(A) Differentiate the numerator and denominator of $\frac{\ln(1-1/n)}{n^{-1}}$ Then take the limit again

Besides all the more calculus-like approaches I've read, there is a more statistically approach I'd like to share:

Since we have the probability of success, the number of trials and the constraint that there is only one shot to empty the machine, this problem can be analized as a series of Bernoulli trials, or most commonly known as binomial distribution since the outcome is dicotomical, I mean, you win or you loose.

Parameters of binomial distributions are number of trials (n) and probability of success (p), the problem states that n and p are related since "p" will be the inverse number of "n". With this data is easy to make some trials with different increasing "n" numbers and see that the binomial probability starts to point out to 0.3678 as you go higher in "n" hence lowering "p" as I show here:

Binomial probability calculation: P(n,p) P(10,1/10) = 0.387420... P(100,1/100) = 0.369729... P(1000,1/1000) = 0.368063... P(10000,1/10000) = 0.367897... P(100000,1/100000) = 0.367881... ... So... after a few calculations you can see the probability is approaching to 0.3678... as guests number approaches to infinity.

I know this solution is not as formal or elegant as the 1/e, but yet is another way to solve this riddle.

Hope you like it

Disclaimer: A cheap, mathematically impure solution follows. It's your fault if you read it.

Use a scientific calculator, permitted in Lemma 1, to evaluate the probability of no one winning, found in Lemma 2, for n = 100,000. By Lemma 3, to two decimal places, report this value as your answer. Q.E.D.

Lemma 1: This item is scientific calculator-permitted.

Because the answer must be a rounded decimal rather than a mathematical expression, this item is calculator permitted. (I don't think we're expected to have the reciprocal of e memorized.) Specifically, a scientific calculator is permitted because four-function calculators generally don't have an e button.

Lemma 2: The probability of no one winning when there are n guests is $\left(1 - \frac 1 n\right)^n.$

For no guest to win, all guests have to lose. The probability of the first guest winning, and thus the probability of any one guest winning given all previous guests lost, is $\frac{1}{n}$ . Thus, the probability of any given guest losing is $\frac{n - 1}{n}$ . This has to occur for each guest, so the loss probability is raised to the nth power.

Lemma 3: To two decimal places, the probability of no one winning out of ∞ guests is equal to the probability of no one winning out of 100,000 guests.

This is provable, but we are given three opportunities to answer correctly, so it is reasonable to assume this lemma for the first attempt.

We're looking for the limit of $\left(1 - \displaystyle\frac{1}{n}\right)^n$ as $n \rightarrow \infty$ .

$\begin{aligned} \lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n &= \lim_{n \rightarrow \infty} \text{e}^{\displaystyle n\ln\left(1 - \frac{1}{n}\right)} \\ & = \lim_{x \rightarrow 0} \text{e}^{\displaystyle \frac{\ln\left(1 - x\right)}{x}} \\ & = \lim_{x \rightarrow 0} \text{e}^{\displaystyle -\frac{\ln\left(1 - x\right)}{-x}} \\ \end{aligned}$

Now, knowing $\lim_{x \rightarrow 0} \displaystyle\frac{\ln\left(1 + x\right)}{x} = 1,$ we get:

$\boxed{ \displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \text{e}^{-1}}.$

Could you lot at Brilliant.org please program this, so as to allow for symbol input? Ie $e$ ^( $-1$ ) or $\exp(-1)$ or $1/e$ , etc.? You would need to connect an interpreter (eg wolfram), but that should not be too problematic. Numerical approximations, where exact irrationals are demanded are unsatisfactory.

I took a more pragmatic approach to summing an infinite sequence.

I expanded (1 - 1/n)^n using the first few binomial coefficients and ignoring anything with n in the denominator (which disappears as n becomes large) to get the infinite sum 1 – 1 + 1/2! – 1/3! + 1/4! – 1/5! + 1/6! ...

Then I took out my calculator and calculated each term, either adding or subtracting it from memory, like this: 1 / 2 M+ / 3 M– / 4 M+ / 5 M– / 6 M+ By the time I got to / 8 M+ the terms were small enough that I felt confident in rounding off 'MR' to 4 decimal places and submitting the answer.

Sure, don't specify the number of digits you want in your answer.