Mathematics - 2021-09-02 (page 2 of 2)

« first day (4047 days earlier) ← previous day next day → last day (1270 days later) »

Ted Shifrin

20:14

@shintuku Of course you can. If $f$ is monotone decreasing.

shintuku

oh!

i got it now

thanks a lot!

copper.hat

@robjohn nice! i'm still sad that apple dropped dylan.

SAJW

Is this a dilemma: you need ("potentially" untrusted) software in order to write your own software, but how can you be sure that the software you write is not affected? (potentially as in: not 100% excluded)

I'm not asking for a solution to the problem, just if it is a dilemma.

Or would that better be in philosophy SE?

Avra

20:44

Given a fair die, $p(x \ge 2) = \frac{5}{6}$?

SAJW

@Avra if its a die6 yes

Avra

@SAJW. Really! I thought it's $2/6$

SAJW

How you came to that result?

Avra

How you calculate it please? $ p(x≥2)= 2/6+3/6+4/6+5/6+6/6$

This is how I understand it honestly

SAJW

no, assuming it's a fair(like you do): 1/6+1/6+1/6+1/6+1/6

because each side has the probability 1/6

and you only look for p(x!=1)

so 1-p(x!=1)

Avra

20:51

@robjohn told me that $E[x] = \Sigma(i \times p(X=i))$ is same as $E[x] = \Sigma(p(X \ge i))$

SAJW

roll a dice 12 times as random as you can and confirm it

Avra

So, I concluded that $p(x≥2)=p(x=2) \times p(x=3) \times p(x=4) \times p(x=5) \times p(x=6)$

SAJW

What's the probability to flip heads on a coin?

Avra

1/2

SAJW

what you did do is evaluate 2,3,4,5,6 in exactly this order

but you only want to evaluate one throw

Avra

20:55

Since each event is independent, then why it's not please

$ p(x≥2)=p(x=2)×p(x=3)×p(x=4)×p(x=5)×p(x=6)$

SAJW

because multiplying comes only if you throw multiple times

(in the case of a dice)

Have you a dice at hand?

Avra

Oh! Thank you, so p(x>=2) literaly means the probability to get at least 2 face is same as saying getting 2 is 1/6, getting 3 is 1/6, etc.?

we sum all of them together

SAJW

yes

Avra

This is sum rule not product rule

SAJW

can only account that to a typo

Avra

20:59

and we have one task not separate task

I remember that if we have separate tasks (multiple throws), then we go for product rule otherwise use sum rule

This is what I remember

SAJW

if you threw a dice 2 times then, what do you think is the probability that both times it's 2 or higher?

Avra

Since we throw it 2 times, we have separate tasks, so we go for product rule. Though each throw itself is answered using sum rule as you did above

so, 5/6 x 5/6

SAJW

yes!

Avra

Sorry! It has been 10 years

SAJW

I think you got it now :)

Avra

21:01

I forgot the content

I am doing my best to recover what I took

Thank you

SAJW

@Avra or expressed in another way $(\frac{5}{6})^2$

Avra

Yes! Thank you

@SAJW. I see now why if we have a hash table with $m$ slots and $n$ elements such that $n<m$, then $E[probes] = \Sigma(Pr[at ~least~ one ~probe] \times Pr[at ~least~ 2 ~probe]) \times \cdots = n/m \times (n-1)/(m-1) \times \cdots$

Sorry. I meant $\times$ not $+$

PM 2Ring

@SAJW Back in the 1980s, Ken Thompson, one of the co-creators of Unix, gave a famous talk on this topic: Reflections On Trusting Trust. See softwareengineering.stackexchange.com/q/184874

Avra

@PM2Ring. Have you wrote a compiler before please?

PM 2Ring

21:21

@Avra No, I've never written a compiler. Or an interpreter.

But I have written numerous programs that create programs as their output.

SAJW

@PM2Ring thanks for digging that up!

PM 2Ring

No worries.

SAJW

So in essence: my chain of trust has to start somewhere and can't be only myself without huge timecost?

like building a computer with transistors alone in some bunker without internet

PM 2Ring

If you don't want to trust anybody, you have to design & manufacture your own CPU. :)

I guess you don't need to actually fabricate chips. You could build a CPU with discrete transistors, but it would be huge, expensive, error-prone, and consume a lot of energy.

robjohn

@user27286 Mathematician who programs.

Avra

21:37

@PM2Ring. The only factory in EU that produce CPUs is the British one I guess?

@PM2Ring. How Germany can not have their own CPU?

robjohn

@Avra $P[x\ge2]=\overset{=2}{\frac16}+\overset{=3}{\frac16}+\overset{=4}{\frac16}+\overset{=5}{\frac16}+\overset{=6}{\frac16}=\overset{\ge2}{\frac56}$

Avra

@PM2Ring. Taiwan is #1 in the world now in chips industry

Saxony-Valley in Germany produces 75-80% of all mircochips for Europe

@robjohn. Thank you

Avra

22:04

@robjohn. Can you look this by any chance if you have time please:

Q: Prove that $ Pr[X \ge i] = \frac{n}{m}\times \frac{n-1}{m-1} \times \cdots \frac{n-i+2}{m-i+2}$

Question: Assume uniform hashing. Let us define the random variable $X$ to be the number of probes made in an unsuccessful search, and let us also define the event $A_i$ , for $i = 1, 2,\cdots,$ to be the event that there is an ith probe and it is to an occupied slot. $X$ is defined in terms of i...