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Danu

11:08 PM

Awwwyis @ACuriousMind

Loving the new color on you

@rob Congrats :D

rob

@Danu Thanks

Danu

Interesting results on the votes :P

You went from 4/5 to 1/5 as the stats progressed :D

So wait the results came in just an hour ago @rob?

ACuriousMind

@Danu More like three hours

rob

Yeah, I've been trying to decipher that. Looking at the raw votes it seems like I was the most common second or third choice. But it was close.

Danu

Oh well, what does it matter :) You are our new janitor, welcome :D

rob

11:14 PM

Probably close enough that a different vote-counting strategy would have changed the outcome.

0celo7

I DID A QUESTION RIGHT

Obliv

how would setting $4ac - b^2 = 0$ in $w = \frac{1}{4a}\left[ 4a^2(x+\frac{b}{2a}y)^2 + (4ac-b^2)y^2\right]$ make this function equivalent to $z = x^2$

wanna share @0celo7

0celo7

@Obliv An urn contains a white chip and a black chip

If the white chip is drawn, the game ends

if a black chip is drawn, the chip is replaced and one more black chip is added

What is the average number of chips that need to be drawn to end the game?

Obliv

.5 @0celo7

or equivalently setting $4ac - b^2 > 0$ means $z = x^2 + y^2$ in that expression

0celo7

Next problem is nontrival

Derive the expectation value of a hypergeometric distribution

Obliv

11:24 PM

@0celo7 so you add another black chip if you draw a black chip

0celo7

yes

Obliv

so it's the product $\Pi \frac{1}{n}$ where $n$ is iterated in the interval $\mathbb{N}$ @0celo7

starting from n = 2

0celo7

wtf is that

Obliv

$\Pi \frac{1}{n} = \frac{1}{2}\frac{1}{3}\frac{1}{4} ...$

0celo7

what is that supposed to be

Obliv

11:28 PM

that's the odds of the game being ended for $n$ draws

0celo7

proof?

Obliv

no nvm you were only asking about expectation value

0celo7

I can explain it do you

but I won't!

haha

Obliv

u no what i don't care. @0celo7 explain to me

0celo7

Let $X$ be the random variable given by "number of chips drawn to get a white chip"

We need to compute $p_X(k)$, namely the probability for $X=k$.

Clearly $p_X(1)=\frac{1}{2}$.

Then $p_X(2)=\frac{1}{2}\cdot\frac{1}{3}$.

This is where it gets tricky

$p_X(3)=\frac{1}{3}\cdot\frac{1}{4}$

In general, $p_X(k)=\frac{1}{k(k+1)}$.

So $$E(X)=\sum_{k=1}^\infty kp_X(k)=\sum_{k=1}^\infty\frac{1}{k+1}$$

This diverges since it is a harmonic series.

So the average is infinite.

Obliv

11:39 PM

but..say you drew 3 times. the first time the probability was 1/2 then the second time was 1/3 and the third time should be 1/4 so $\frac{1}{2}\frac{1}{3}\frac{1}{4}$ ?

0celo7

If you draw three times

this is what happens

let $B$ denote a black chip being drawn

$W$ a white chip

the only way you get 3 is by drawing $BBW$

The first black is $1/2$ probability

the second is $2/3$

Obliv

oh my god..

0celo7

then the white is $1/4$