/sci/ - Science & Math

[Anonymous No.15750877]

/g/ is too retarded to solve this

[Barkun No.15750886]

35%

[Anonymous No.15750890]

50%

[Barkun No.15750895]

>>15750886
Actually. 33.75%

You have to consider, you might get 2 chests, but then again you might not, so the odds aren't incremtative.

[Barkun No.15750898]

>>15750895
33.9%

Final answer

[Barkun No.15750902]

>>15750898
No 42.5%

Final answer. Promise.

[Anonymous No.15750926]

32%

[Anonymous No.15750954]

>>15750877
43.6%

[Anonymous No.15750959]

this is very similar to slot machine with free spins logic, redoing my old formulas for that

[Barkun No.15750967]

>>15750954
This. I was wrong.

[Anonymous No.15750976]

>>15750954
[math]\frac{\sqrt{15}-3}{2}[/math]

[Anonymous No.15750987]

>>15750877
While you say the two chests are identically distributed as the former, you do not say that they are independently distributed.
It is therefore unclear how to solve exercise.

[Anonymous No.15750989]

0.5 * loss
0.3 * win
0.2 * 2 free spins

loss probability is x
x = 0.5 + 0.2x2
squared because you have to lose both for it to be a loss
5x = 2.5 + x2
x2 - 5x + 2.5 = 0
x = 4.436, 0.563
x = 0.563
win is 1 - loss
win = 1 - x
win = 1 - 0.563
win = 0.427

[Anonymous No.15751002]

>>15750976
confirm

[Anonymous No.15751015]

>>15750987
>>15750989
[math]p = 0.3 + 0.2\left(1 - (1-p)^2\right) \implies 2p^2 + 6p -3 = 0\\
p = \dfrac{\sqrt{15} - 3}{2} = 0.4365[/math]

[Barkun No.15751021]

>>15751015
You forgot no negate your working out

[Anonymous No.15751179]

>>15751002
How do you get the equation on second row? Wouldn't the problem imply a sort of infinite recursion of opening pairs of boxes inside pairs of boxes?

Picrel is me trying to think this problem

[Anonymous No.15751212]

2.5-sqrt(3.75)

[Anonymous No.15751220]

>>15751179
The answer for any box is p, just use that for the bonus chests in the equation.

[Anonymous No.15751747]

50 %
it either does or does not

[Anonymous No.15752287]

>>15751747
kek
always gets me

[Anonymous No.15753397]

>>15751179

The probability 1 box has metal is P(M), the probability it does not is therefore 1-P(M)
The probability 2 boxes do not is therefore (1-P(M))^2

The probability 2 boxes have at least 1 is therefore 1 - (1-P(M))^2

[Anonymous No.15753407]

>>15750877
Quite easy if you have ever studied probabilities.

[Anonymous No.15753434]

>>15750877
>>15751179
Unless you wrongly assume the gold bars get smaller (which the problem does not state) there exists one box in the population where the recursion does not converge, and many boxes where it converges at different numbers of bars, negating there ever forming a conserved population from which to even define the original percentages that define the possible different contents of the boxes. Therefore, the question cannot be answered.

The reason this was originally a /g problem was to get morons to create infinite loop programs that would crash, or would not compile, driving them nuts.

[Anonymous No.15753463]

>>15750877
35,71%

It's not harder than +30% which leaves 2/7 and 5/7
so 20% for two which is 2/7 x 2 which is 0,571 and then just adjust it to correct decimal position

[Anonymous No.15753474]

>>15753463
or actually you have to account for it to every decimal so
35,71 + 0,571 + 0,0571 + 0,00571...
So 36,343 %

[Anonymous No.15753989]

>>15751015
This is the correct answer, verified with a simulation of 100 million runs.

[Anonymous No.15753994]

A better question would be, how many bars of metal do we expect to obtain on average from a single chest?

[Anonymous No.15754010]

>>15753994
actually it's not interesting at all, it's 3/8 = 0.375

[Anonymous No.15754013]

>>15754010
im retarded, its 0.5

[Anonymous No.15754340]

>>15750877
P(M) = P(Ag) + P(Au) + (P(Chests) * P(M in at least one))

P(M) = 0.2 + 0.1 + 0.2 * (1 - (1 - P(M))^2)

It then follows that

P(M)^2 - P(M) = 0.30
P(M) * (P(M) - 1) = 0.3

P(M) = 0.30 because P(M) cannot be 1.

[Anonymous No.15754902]

>>15750877
>open chest
>see no gold
ok but what if grug
>open chest
>open chest
>open chest
>open chest etc
then what. grug get bored and angry. touch lady by force. grug now happy but lady mad. use chest 3 to kill lady. hide lady in chest 5. grug hungry. no gold for food. grug now use chest 2 to threaten rich grug. rich grug give grug silver. grug now have silver and chest. grug happy. grug now has silver, chests, and lady.

[Anonymous No.15754909]

>>15751747
this is the correct answer though that's the wrong reason.

[Anonymous No.15755000]

>>15750877
It's a simple GP so it's 3/10(1/(1-2/10)) = 3/10 * 10/8 = 3/8 = 0.375. How is this even an inconvinience is beyond me. /g/ is also just a forum for posers though so makes sense. Anyone actually good at computer science or maths browses /sci/

[Anonymous No.15755003]

>>15755000
wait okay i messed up, it's two chests. Ill do it again

[Anonymous No.15755028]

>>15755003
It's 0.43 approx. More specifically it's (-6 + sqrt(60))/4

My points about /g/ tards still stands

[Anonymous No.15755030]

>>15755028
0.4364916731 on a calculator

[Anonymous No.15755031]

>>15755000
>Anyone actually good at computer science or maths browses /sci/
No one on /sci/ is good at anything STEM-related. You're the proof.

[Anonymous No.15755034]

>>15754902
tfw no gold

[Anonymous No.15755035]

>>15755031
Im haven't slept and it's 4 am, everyone makes mistakes tard. I realized and rectified it immediately

[Anonymous No.15755038]

>>15755035
Congratulations on finding the answer only 16 hours after it was already posted in the thread.

[Anonymous No.15755039]

>>15755038
Congrats to me finding the answer only 20 minutes after i saw it while struggling to keep my eyes open and jerking off to your faggot dad's ass

[Anonymous No.15755043]

>>15755039
Congrats to you being a brainlet who took 20 minutes to solve a kindergarten-level problem and got it wrong.

[Anonymous No.15755045]

>>15755039
Oh, you're an engineer. Now it all makes sense.

[Anonymous No.15755047]

>>15755043
>brainlet
Actually agreed on that one

>took 20 minutes to solve a kindergarten-level problem and got it wrong.
typed the first answer instantantly upon seeing the question. Missed the two chests part because sleep deprived and realized that 3 minutes later, posted reply acknowledging mistake and corrected in 10 minutes.

[Anonymous No.15755050]

>>15755047
You will never be intelligent is the bottom line. How hilariously defensive you're getting and how hell-bent you are to prove yourself with a kindergarten problem proves this. You were obviously so elated over being able to figure out this triviality. It must've been crushing to realize you got it wrong and no one is impressed. lol

[Anonymous No.15755055]

>>15755050
Lmao, i know i am not intelligent, i am fucking retarded. But i love how you, being a midwit, are so much enjoying shitting on some random retard on the internet because he got a simple question wrong once. So affirming to your 'intelligence'.

My point is that i am absolutely autistic but /g/ even more than me

[Anonymous No.15755061]

>>15755055
>i know i am not intelligent, i am fucking retarded.
Ok.

[Anonymous No.15755063]

>>15755061
Ok.

[Anonymous No.15755138]

>>15750877
What was diificult about this?

#include<stdio.h>
#include<time.h>
#include<stdlib.h>

int trial()
{
int total=0,success,conclude;
while(1)
{
int roll = rand()%10;
conclude = (roll<8)?1:0;
if(roll<3)return 1;
if(conclude)return 0;
if(trial())return 1;
if(trial())return 1;
return 0;
}

}

void main()
{
int total,success;time_t t;
srand((unsigned)time(&t));
while(total<2000000)
{
success += trial();
total+=1;
}
printf("Probability of getting a metal is %f, with %d successes on %d total",(float)success/total,success,total);
}

In C
Results:
>Probability of getting a metal is 0.436291, with 872583 successes on 2000000 total
>Probability of getting a metal is 0.437417, with 874834 successes on 2000000 total
>Probability of getting a metal is 0.436357, with 872714 successes on 2000000 total
>Probability of getting a metal is 0.436322, with 872644 successes on 2000000 total

[Anonymous No.15755220]

>>15755028
>>15755030
>>15755000
Nope

[Barkun No.15755247]

>>15755220
42.9

[Barkun No.15755249]

42.93

[Anonymous No.15755271]

>>15755138
Do this:
int roll = rand() * 10ull / (RAND_MAX + 1ull);
instead of this:
int roll = rand()%10;
The later is slower and favors lower bits of rand() that are usually low quality.

[Anonymous No.15755286]

>>15755271
Huh thanks. Could you explain what it's doing to me? It just looks odd and confusing with a lot of alien terms

Also, Is my solution correct otherwise?

[Anonymous No.15755328]

>>15755286
If you the undefined behavior that brainlet's code invokes through signed integer overflow, it's the equivalent of scaling the output of rand down down to a range of 0-1, multiplying by 10 and truncating after the floating point.

[Anonymous No.15755331]

>>15755328
If you ignore*

[Anonymous No.15755393]

>>15755328
How does running a modulus on Rnd invoke a signed integer overflow? i don't know much about it, but doesn't it generate a value between 0 and RAND_MAX which is 2^31-1 (highest positive value) by default? so how can a signed integer overflow be evoked here?

[Anonymous No.15755400]

>>15755393
>How does running a modulus on Rnd invoke a signed integer overflow?
It doesn't. I meant the code from the other poster who was trying to "enlighten" you. Multiplying the output of rand() by 10 can easily overflow a signed int.

[Anonymous No.15755405]

>>15750877
>anime girl
Im tired of yuo people

[Anonymous No.15755410]

>>15755393
>>15755400
Actually I was wrong. It's not undefined behavior. Arithmetic involving signed and unsigned values promotes the signed values into unsigned, so the result is just implementation-defined and not UB.

[Anonymous No.15755412]

>>15755400
Oh i see. Yeah i guess that's true, but that's why he multiplied with a ull i think. Multiplying an int and an ull int will give a ull int which can go upto ~19 digits, so multiplying by 10 will not actually cause an overflow.

I assume the gist of what the poster was trying to say is that multiplying and dividing like that is faster than using a modulus on a large integer? Also i think this is the same method to generate random numbers that was used in BASIC programs on old computers right?

[Anonymous No.15755422]

>>15755410
Happens to the best of us ig. Thanks for explaining though.

[Anonymous No.15755474]

>>15755220
Enlighten me

[Anonymous No.15755606]

ITT: low IQ retards who are programmed to apply some formula to every problem instead of using their own low IQ brain to solve this low IQ problem
the answer is 50%

[Anonymous No.15755806]

>>15755286
The result of rand() is an integer in the range [0 .. RAND_MAX] and RAND_MAX + 1ull is a power of two, the integer gets promoted to unsigned long long that is at least 64 bits. In rand() * 10ull the result of rand() is also promoted to unsigned long long. The result is an unsigned between 0 and 9 inclusive. The division by an unsigned power of two is equivalent of displacement to the right, a simple and fast operation.

[Anonymous No.15756046]

>>15755474
Not my job.

[Anonymous No.15757806]

>>15754013
Can confirm, it is 1/2.

Let R be reward:

E[R] = 1/2*0 + 3/10*1 + 1/5 * 2*E[R]

E[R] = 3/10 + 2/5*E[R]

E[R] = 1/2

[Anonymous No.15757989]

>>15750877
100% with enough privateers

[Anonymous No.15759499]

37.5 tho

[Anonymous No.15759686]

>>15750877
x = 0
x = 0.3 + (x*0.2)

[Anonymous No.15759788]

>>15759686
hold on I got the second part wrong
x = 0.3 + (x*0.4)

[Anonymous No.15759836]

/g/ is gaymers who gloat about their gpu and coomers who think diffusion "ai" is skynet.

[Anonymous No.15759845]

>>15759836
Well yeah, it wasn't too long about that it was getting called "tech support for /v/"

[Barkun No.15759951]

The answer is 1.2-3 in secret language. Don't ask why. Just trust me. In your language it's 42.93

[Anonymous No.15760243]

>>15759836
generally higher quality than /sci/, it's a /pol/ infested shithole here

[Anonymous No.15760445]

>>15759788
>>15755138
>>15755000
>>15753463
>>15751015
>>15751002
>>15750989
jfc why is everyone retarded? 10% chance of gold and 20% chance of silver is NOT the same as 30% chance of metal.
The base odds of NOT getting silver AND NOT getting gold is 0.8*0.9 = 0.72, so the base odds of getting silver OR gold is only 28%, not 30%

>>15750877
OP is retarded too. if the odds of silver, gold, and two more chests are 20%, 10%, and 20% respectively, the odds of there being nothing in the chest is 0.8*0.9*0.8=0.576 = 57.6%, not 50%.

the answer is 36.6% from 10M simulated runs

[Anonymous No.15760468]

>>15760445
I figured out why everyone is retarded. They think a chest gets only one roll that determines its contents (i.e., roll between 1-10; 1-2=silver, 3=gold, 4-5=two more chests, 6-10=nothing). In that case there actually is a 50% chance of nothing. but the post also says “odds of AT LEAST one bar” which is retarded because in this scenario it’s not possible to get more than 1 bar

if each item is rolled for independently, you have a higher chance to get nothing and thus lower odds of finding metal, but you do have odds to get multiple bars per chest AND additional chests

[Anonymous No.15760790]

>>15760445
semantics
the retards are right because logically we expect one item (or two bonus chests) on each roll - it's meant to represent a video game mechanic.

[Anonymous No.15760798]

>>15760468
It is possible if you get two chests on the first roll with a bar in each