Probability that any outcome of a dice roll happens more than X times out of Y trials

I'm trying to determine the probability that a person experiences a "lucky number" when rolling a single, fair, 6-sided dice over a set of rolls in a single trial. A "lucky number" in this case is any face of the die that occurs visibly more common than one would normally expect. If you roll a six-sided die 100 times, you expect the outcome to occur with ~16.6 results of 1, 2, 3, 4, 5, and 6 ea, on average.

For example, you roll a six-sided dice in 100 independent trials, what is the probability that the occurrence of rolling any side of the dice happens at least 33 times over the course of the 100 independent trials? It doesn't matter if the roll was 1, 2, 3, 4, 5, or 6, just that the same result happened at least 33 times out of the 100 trials.

How would I calculate this?

Thanks.

Answer

The chance that $1$ comes up exactly $33$ times in $100$ comes from the binomial distribution. The chance of success is $\frac 16$ and failure is $\frac 56$ so it is ${100 \choose 33}(\frac 16)^{33}(\frac 56)^{67}\approx 0.00003$ If we sum from $33$ to $100$ we get the chance of at least $33\ 1$s, which is about $0.0005$ per Alpha. You can multiply these by $6$ to get the chance for any number, as it is very unlikely we doublecount by having at least $33$ of two different numbers. So the chance of a "lucky number" happening by chance is about $0.0003$ or one in $3300$. Pretty unlikely, but rarer things happen all the time.

limits - Discuss the differentiability of $e^{-|x|}$?

Discuss the differentiability of $e^{-|x|}$ ?

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I tried something like making $2$ domains, $x\ge 0$ and $x<0$

$e^{-|x|} = e^{-x} | x\ge 0 $

and

$e^{-|x|} = e^{x} | x<0 $

By using definition of differentiability,

$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^+}\frac{e^{-x}-1}{x} = -1$$

and Similarly

$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^-}\frac{e^{x}-1}{x} =+1$$

Hence, I can say that it is not differentiable at $x=0$.

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Is my understanding right or Am I missing something ?

real analysis - Bijection f: [0, 1] to (0, 1)?

My solution for this function is let 0 map to 1/2, 1 to 1/3, and then for any other values map them to that value + 1/2. Is this the correct solution for this mapping?

real analysis - dedekind cut for the square root

Can somebody give me the hints to solve it ?

1. What is the Dedekind cut $(A, B)$ for $\sqrt 2$ ?



2. What is the Dedekind cut $(C, D)$ for
   $\sqrt 2 + \sqrt 3$ ?


3. In $\mathbb R[x]/(x^2 + 1)$, what is the value of $[x^3 + x^2 + x + 1]$?
   (Just simplify $x^3 + x^2 + x + 1$)
   where $[p]$ means the equivalence class that has $p$ as an element;
   $[p]$ is the set of polynomials $q$ in $\mathbb R[x]$ such that
   $x^2 + 1$ divides $p - q$.
   $\mathbb R[x]$ is the set of polynomials whose coefficients are all real numbers.

divisibility - Numbers divisible by all of their digits: Why don't 4's show up in 6- or 7- digit numbers?

For reasons I'll explain below the question if you're interested, I stumbled across a peculiar phenomenon involving numbers divisible by their digits.

I'm concerned with numbers that are divisible by all of its digits, and do not have any zeros or repeated digits.

Ex: 175, 9867312, 1

Not: 111, 105

There are 548 such numbers: 105 of them have 6 or more digits. (There can't be any with more than 7 digits, for reasons I'll leave you to discover.*) For some reason, though, no six- or seven- digit numbers have any 4's in their digits.
Why?

I know why there can't be any 8 digit numbers: 1+2+3+4+6+7+8+9 = 40, which is not divisible by 3, so 9, 6, and 3 won't divide. But why don't 4's show up past 6 digits?

---

Explanation for why I have this question:

*About two years ago I was in a TI-Basic programming competition which required me to write the following program:

A number is said to be "digisible" if it meets the following three conditions:
- It has no 0;
- All digits which compose it are different from each other;
- It is divisible by each of the digits that compose it.

You will have to make a program that asks a positive integer greater than or equal to 10 and displays 1 if it is digisible, 0 if it is not.

This was a fun challenge to create and super-optimize in TI-Basic. TI-Basic is really slow, so it wasn't possible to check all numbers for "digisible"-ness. However, in the past year I learned Java, which is speedy-fast. So I returned to the problem and made a program to list out ALL of the digisible numbers.

Looking at all of the Digisibles, I noticed some cool things, some of which makes sense, others which I couldn't find an explanation for. This question is one that I couldn't find an explanation for, but highly suspect one exists.

Hint: 5's do not show up past 5 digit numbers.

Answer

As you point out, a 7-digit number cannot have a 5 in it (otherwise, it would have to end in 5, but would have to have one of 2, 4, 6, or 8 in it, so must be even). Thus a 7-digit number $x$ contains 7 of 1, 2, 3, 4, 6, 7, 8, 9. We need only identify the missing digit. If it is not 9, the resulting number must be divisible by 9, so the missing digit must be 4 ($40-4 = 36$). If the missing digit is 9, then the digits are 1, 2, 3, 4, 6, 7, 8. But the sum here is $40-9=31$, which is not divisible by 3. So this cannot happen.

Thus any 7-digit number of this type must consist of the digits 1, 2, 3, 6, 7, 8, 9.

group theory - Is there an operational isomorphism from $(mathbb{Z},+)$ to $(mathbb{Q}^{+},cdot)$?

Let $\left(\mathbb{Z},+\right)$ and $\left(\mathbb{Q}^{+},\cdot\right)$ be groups (let the integers be a group with respect to addition and the positive rationals be a group with respect to multiplication). Is there a function $\phi\colon\mathbb{Z}\mapsto\mathbb{Q}^{+}$ such that:

• $\phi(a)=\phi(b) \implies a=b$ (injection)


• $\forall p\in\mathbb{Q}^{+} : \exists a\in\mathbb{Z} : \phi(a)=p$ (surjection)


• $\phi(a+b) = \phi(a)\cdot\phi(b)$ (homomorphism)

? If so, provide an example. If not, disprove.

Answer

If such an isomorphism existed it would of course be onto so there would exist some $n \in \mathbb{Z}$ such that $\phi(n) = \frac{1}{2}$ for example. But then, $$\phi(n) = \phi(1+\cdots + 1) = \phi(1)\cdots \phi(1) = \phi(1)^n = \frac{1}{2}.$$ This implies $n=1$ since otherwise $\left(\frac{1}{2}\right)^{1/n} \notin \mathbb{Q}$. So, $\phi(1) = \frac{1}{2}$. Thus, for any $n \in \mathbb{Z}$, $\phi(n) = \phi(1)^n = \frac{1}{2^n}$, so clearly $\phi$ is not onto since we only achieve powers of two in the image. So such an isomorphism cannot exist.

decimal expansion - I'm puzzled with 0.99999

After reading all the kind answers for this previous question question of mine,
I wonder... How do we get a fraction whose decimal expansion is the simple $0.\overline{9}$?

I don't mean to look like kidding or joking (of course, one can teach math with fun so it becomes more interesting), but this series has really raised a flag here, because $\frac{9}{9}$ won't solve this case, although it solves for all other digits (e.g. $0.\overline{8}=\frac{8}{9}$ and so on).

Thanks!
Beco.

Answer

The number $0.9999\cdots$ is in fact equal to $1$, which is why you get $\frac{9}{9}$. See this previous question.

To see it is equal to $1$, you can use any number of ideas:

1. The hand-wavy but convincing one: Let $x=0.999\cdots$. Then $10x = 9.999\cdots = 9 + x$. So $9x = 9$, hence $x=1$.




2. The formal one. The decimal expansion describes an infinite series. Here we have that
   
   $$ x = \sum_{n=1}^{\infty}\frac{9}{10^n}.$$
   This is a geometric series with common ration $\frac{1}{10}$ and initial term $\frac{9}{10}$, so
   $$x = \sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{\quad\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\quad\frac{9}{10}\quad} = 1.$$

In general, a number whose decimal expansion terminates (has a "tail of 0s") always has two decimal expansions, one with a tail of 9s. So:
$$\begin{align*}
1.0000\cdots &= 0.9999\cdots\\
2.480000\cdots &= 2.4799999\cdots\\
1938.01936180000\cdots &= 1938.019361799999\cdots

\end{align*}$$
etc.