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/sci/ - Science & Math

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2516802 No.2516802 [Reply] [Original]

There are 40 chips in a bag.
Of these, 5 have two blue sides, 10 have a blue side and a red side, and the rest have two red sides.
You pick one of them at random, and look at one of the sides. It is red.
What are the odds of the other side also being red?

Let's see how long it takes until someone delivers the right answer...

>> No.2516808

I could get it right, because I know how to.

This doesn't mean I can be arsed to do it. I'd have to grab a pen to write, and it's at least 30 cm away from my keyboard.

>> No.2516820

>>2516808
Then don't do it.

>> No.2516821

I would put mayo on them in a paper cone

>> No.2516824

5/8

>> No.2516829

Intuitively I'd say 0,714, or 25/35, but given how the question is posed, I assume it's not that simple.

>> No.2516827

>>2516824
show your work faggot

>> No.2516828

2/3

But that's probably wrong.

>> No.2516830

5/6

>> No.2516831

Bring red paint when you pull out the chips and you'll always be right.

>> No.2516833

40-5-10 = 25 = red/red

One side is red therefore it cannot possibly be a blue/blue.

It is either a red/blue or red/red.

25+10 = 35
25/35 = probability other side is red.

>> No.2516848
File: 132 KB, 686x686, Thisiscorrect.jpg [View same] [iqdb] [saucenao] [google]
2516848

>>2516830

>> No.2516852

>>2516833
this is incorrect because the fact that you got a red side when you randomly sampled one of the sides of the chip adds evidence to the hypothesis that the chip has 2 red sides.

the correct answer is 5/6

>> No.2516876

>>2516852

I can't really see why, but then again I shouted at a guy for not understanding the Monty Hall problem today, and in this case I'm just going by intuition, so... *Shrug*

>> No.2516891

>chips
>blue and red

what, are you eating paint chips or something?

>> No.2516893
File: 3 KB, 126x107, zx7.jpg [View same] [iqdb] [saucenao] [google]
2516893

>>2516876
it is because you are not picking a chip, you are picking a side.

you have to count the double sides reds (25 of them) as seperate. so you have 50 chances to pick a red side that also has a red side.

there are only 10 other red sides, and these have blue on the other side

so thats 50 pure reds, to 10 mixed, and you can ignroe the 5 pure blues because you know you can't have picked one.

50/(50+10) = 50/60 = 5/6

kudo's to these guys >>2516830
>>2516852

>> No.2516900

so this is like the car show problem i guess?
seems like it should be 5/7.
#hits /#possibilities = 25/35.

the prob that the other side is blue is 10/35, the prob that the other side is red is 25/35
where does 5/6 come from?

>> No.2516903

>>2516891
chips, as in poker chips, or playing counters.

not edible chips. (fries)

>> No.2516905

someone explain please?

>> No.2516898 [DELETED] 

Chance of the first random chip's random side being red: (10/2 + 25)/40 = 75%.

The odds that the other side is red also? Well, we already know that we haven't picked up a blue chip, so it's one of the other 35, each of which were equally likely. Of those 35, only 25 will show the other side also being red.

So the odds of flipping the chip over and also seeing red are 25/35 = 0.714285714.

>> No.2516917

>>2516900
>>2516905
see: >>2516893

>> No.2516918 [DELETED] 

>>2516833
This is the correct answer.

Let me explain

There are 40 chips with two sides each, so, 80 sides.
5 * 2 + 10 = 20 blue sides.
25 * 2 + 10 = 60 red sides.
So, the odds of getting a red side are of 6/8.

The chance of getting a red/red chip is 5/8.
Knowing you already have a red side, divide by that.
(5/8)/(6/8) = 5/6, which is the correct answer

What most people do is: There are 35 chips with a red side, and 25 with two red sides. 25/35 = 5/7. The error here is assuming that you always see the red side when picking a blue/red chip.

>> No.2516928

>>2516905
What we know from the first observation is that we've picked up a chip with at least one red side. Given that we know that, how many equally-likely scenarios produce that result? There's 25 double-red chips, and either side looked at first, so 50 arrangements there. There's the 10 red/blue chips, and only the arrangements with the red side up first (10). So there's 60 equally likely chip-arrangements that give us the observation. Of those, 50 will also show a red side when flipped over. That's 50/60, or 5/6.

Basically, you just make sure you eliminate all the things that are ruled out, and make sure you're looking at a set of equally likely things that remain before continuing.

>> No.2516947

>>2516918
you referenced the wrong post.
>says 'this is correct' referencing a 5/7 post, then goes on to explain why 5/6 is correct...

>> No.2516949
File: 33 KB, 640x480, Snapshot_20101204.jpg [View same] [iqdb] [saucenao] [google]
2516949

>>2516893
Alright, thanks.

>> No.2516974

>>2516947
Deleting in a sec...

>>2516830
This is the correct answer.

Let me explain

There are 40 chips with two sides each, so, 80 sides.
5 * 2 + 10 = 20 blue sides.
25 * 2 + 10 = 60 red sides.
So, the odds of getting a red side are of 6/8.

The chance of getting a red/red chip is 5/8.
Knowing you already have a red side, divide by that.
(5/8)/(6/8) = 5/6, which is the correct answer

What most people do is: There are 35 chips with a red side, and 25 with two red sides. 25/35 = 5/7. The error here is assuming that you always see the red side when picking a blue/red chip.

>> No.2516971

thanks allot, think I get it

>> No.2516977

5/6
PICKING UP A RED SIDE MEANS YOU PICKED UP ONE OF 35 CHIPS
THIS IS 70 SIDES, MINUS THE 10 BLUE SIDES ON THE RED/BLUE CHIPS, LEAVING 60 SIDES
OF THOSE 60 SIDES, ONLY 50 ARE MATCHED TO ANOTHER RED SIDE
50/60 HAVE ANOTHER RED SIDE
ODDS ARE 5:6

/ FUCKING THREAD

>> No.2516982

okay, there are two different rolls: one for the first color and one for the second.

p(blue)=20/80=.25
p(red)=60/80-.75

p(b/b)=5/40=.125
p(b/r)=10/40=.25
p(r/r)=25/40=0.625

So, probability for a red/red, would be p(r)*p(r/r)=.75*.625=.46875

>> No.2517013

I asked the Internet:

You're now chatting with a random stranger. Say hi!
Official messages from Omegle will not be sent with the label 'Stranger:'. Strangers claiming to represent Omegle are lying.
You: i have a challenge
Stranger: Hey!!!
You: are you smart?
Stranger: Yeah super smart
You: ok
You: lets see then
Stranger: OK
You: there are 40 poker chips in a bag
Stranger: ok
You: 5 have two blue sides
You: 10 have a blue side and a red side
You: and the rest have two red sides
You: you pick one of them at random
You: and look at one of the sides
You: it is red
Stranger: ok
You: what are the odds of the other side also being red?
Stranger: very good because poker chip have the same side of color
You: these have different sides
Stranger: whoo hoo!!!!!!!!!!!
Stranger: bye
Your conversational partner has disconnected.

>> No.2517029

>>2516982
Thanks, but it's not quite the question asked.

AFTER picking a chip, looking at one side, and seeing it is red, what is the chance that the other side is also red?

>> No.2517040

>>2516982
Uh, the chance of red/red is just 25/40 (or 5/8), before you pick up a chip. After you pick up a chip and see a red side, it's a 5/6 that the other side is also red, as explained earlier.

>> No.2517041

5/6
Prove me incorrect.

>> No.2517049

>>2517041
I am unable to do so, good sir.

>> No.2517059

>>2517013
You're now chatting with a random stranger. Say hi!
Official messages from Omegle will not be sent with the label 'Stranger:'. Strangers claiming to represent Omegle are lying.
You: There are 40 chips in a bag.
Of these, 5 have two blue sides, 10 have a blue side and a red side, and the rest have two red sides.
You pick one of them at random, and look at one of the sides. It is red.
What are the odds of the other side also being red?
Stranger: Guess what? I DONT GIVE A FUCK LOL!
Your conversational partner has disconnected.

>> No.2517075

my fingers have a purple residue after eating said chips. No solution.

>> No.2517078
File: 632 KB, 847x538, near 23456432.png [View same] [iqdb] [saucenao] [google]
2517078

Probability of one side you pick up being red: 75%

Probability of other side being red: 50/60 = 83.33%

Now get out.

inb4 trolls whining about "hurrdurr probability taken from all the chips not just red ones"

The fact that the other side is read has to depend on the instances in which the first side is red so shut up.

>> No.2517110

5/7

>> No.2517117
File: 269 KB, 466x320, near dart asdfa.png [View same] [iqdb] [saucenao] [google]
2517117

>>2517075

If you pick up a chip that's red on both sides that wouldn't be true :3

boy /sci/ sure is foolish today

>> No.2517123

You're now chatting with a random stranger. Say hi!
Official messages from Omegle will not be sent with the label 'Stranger:'. Strangers claiming to represent Omegle are lying.
You: There are 40 chips in a bag.
Of these, 5 have two blue sides, 10 have a blue side and a red side, and the rest have two red sides.
You pick one of them at random, and look at one of the sides. It is red.
What are the odds of the other side also being red?
Stranger: idfk dude this is not math class
You: try at least
Stranger: hmm lets see
Stranger: 35
Stranger: wait 25
Stranger: bye
You: it's not that hard
Stranger: its not 25?
You: 25 what?
You: a probability ranges bewteen 0 and 1
You: =/
Your conversational partner has disconnected.

>> No.2517128
File: 77 KB, 500x500, happy love rxn girl sketch.png [View same] [iqdb] [saucenao] [google]
2517128

>>2517123
>>2517123

aww, humans are so adorable :3

I don't have contempt for non-intellectual types, only compassion.

>> No.2517131
File: 46 KB, 415x311, casino-chips.jpg [View same] [iqdb] [saucenao] [google]
2517131

>>2517075
You would eat these? Because that's what OP is talking about.

>> No.2517143
File: 104 KB, 716x800, near head set.jpg [View same] [iqdb] [saucenao] [google]
2517143

>>2517131

Interesting..intuitively I thought the same thing as that other anon. I guess the fact they were in a 'bag' brought me into that train of thought. Either way, the math works.

>> No.2517159

>>2517128
This one at least tried, compare with >>2517013 or >>2517059.

>> No.2517163

0.8

>> No.2517170

.375

>> No.2517188

why out of 60 and not 59 if you already know for sure one of the red sides is chosen?

>> No.2517199

5/8

the 5 and 10 chips have no chance of the other side being red
the remaining 25 have a 100% chance of other side being red
25/40 = 5/8

>> No.2517207

wait, I was wrong.
the 5 chips can't be included in the sample.
25/35 = 5/7

>> No.2517206

1/1

because i'm a magician.

>> No.2517225

I was one of the 5/7 guys until I wrote this: http://pastebin.com/S6hM2Psh

:(

>> No.2517229

It doesn't matter how many sides are red, because you are picking a chip, not sides.
/omega = {red/blue, red/blue, red/blue, red/blue, red/blue, red/blue, red/blue, red/blue, red/blue, red/blue, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red, red/red}
P(red/red) = # of ways red/red can occur (25) / # of ways red/X can occur (35)

25/35 = 5/7

>> No.2517258

>>2517225
$ gcc chips.c -o chips
chips.c: In function ‘main’:
chips.c:9:4: error: ‘chipEnum’ undeclared (first use in this function)
chips.c:9:4: note: each undeclared identifier is reported only once for each function it appears in
chips.c:9:13: error: expected ‘;’ before ‘chip’
chips.c:12:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
chips.c:12:4: note: use option -std=c99 or -std=gnu99 to compile your code
chips.c:15:5: error: ‘chip’ undeclared (first use in this function)

wat

>> No.2517264

>>2517225

How does it feel to know that your ability to think for yourself has been outsourced to a machine? Mathfag here; did it by myself. Critical thinking ftw

>> No.2517289

>>2517264
There is nothing wrong with using a program to check if you did it right.

>> No.2517298

>>2517289

Sure, except for the fact that you got it wrong by yourself..

>> No.2517301

>>2517258
Sorry compiled it with c++
just change "chipEnum chip;" to "enum chipEnum chip;" and move the int i declaration out of the for loop.

>> No.2517308

tl;dr

>> No.2517321

it's 2/3 you morons

>> No.2517323
File: 11 KB, 247x204, full retard.jpg [View same] [iqdb] [saucenao] [google]
2517323

>>2517301
>c++
>stdio.h
>printf()

>> No.2517337

>>2516802

The answer is ~70%

>> No.2517341

>>2517323

Standard I/o dot H

FUCK YEA

>> No.2517353

>>2517225

The script uses 70 chips, not 40. With 40 you'll get 5/7, which is the correct answer.

>> No.2517356

>>2517323
Started c++ project in eclipse, wrote c code, didn't feel like changing project type.

>> No.2517360
File: 84 KB, 623x358, 1277477518669.jpg [View same] [iqdb] [saucenao] [google]
2517360

>>2517337
>>2517321
>>2517229
>>2517225
>>2517207
>>2517199
>>2517188
>>2517170
>>2517163
>>2517110
>>2516982
>>2516900
>>2516833
>>2516829
>>2516828
>>2516824

>> No.2517361

>>2517353
>>2517337

I win

>> No.2517368

Its 5/7

You picked up a red side. Thats 35 possible chips.
Of those 35, 25 are red on the other side
25/35 = 5/7

You cannot include the 5 blue/blue chips, since you already know you have a red side

>> No.2517376

The entire last half of this thread is trolls.

>> No.2517377

The answer is probably ITT, but not looking first. It seems possibly tricky due to the prior probability of getting a red side, so I'll use Bayes.
P(A|B) = P(A)*P(B|A)/P(B)
A = selected chip is all red
B = random side of selected chip is red
P(A) = 25/40
P(B) = (25*2+10)/80 = 60/80 =3/4
P(B|A) = 1
P(A|B) = 25/40 * 1 / (3/4)
= 25/40 * 4/3 = 100/120
= 5/6

The naive way would be to say out of 60 possible red sides, 50 of those sides have opposites that are red so 5/6. I would think that was wrong, because it ignores the prior probability of getting one with 2 red sides as opposed to 1 red side, but it's obviously the same answer.

>> No.2517380

>>2517353
Oh shit I read the problem wrong, well I guess I was right all along.

Fixed code: http://pastebin.com/YaUVqAj8

>> No.2517389

>>2517377
Actually, now I see that by saying "out of 60 possible red sides" instead of saying "out of 35 possible chips that have a red side" like this >>2517368 ...fellow who at least tried, my "naive" way does account for the prior probability of picking a 2-sided verses 1-sided chip.

>> No.2517390

>>2516802


there are two chips in a bag, red or black.

I decided how to arrange their proportions.

What are the chances of picking a red chip?


>hint: the answer is 50% you dumbfuck.

>> No.2517400

>>2517360

You mad?

>> No.2517409

100% because i ain't never seen no potato chip that's red and blue.

>> No.2517431

>>2517390

Congratulations you just broke the MAxwell-Boltzmann theory of entropy

>he thinks the change in arrangement of the possible molecular states has nothing to do with the temperature change of the water

>> No.2517445

>>2517380
>http://pastebin.com/YaUVqAj8
You're doing it wrong. You're assuming that every time you get a RB chip that you happen to look at the red side.

where you have RBNum++ you need to flip the coin to see whether you see the red or blue side, and if blue, i-- instead of RBNum++.

>> No.2517452

>>2517377
/thread

>> No.2517461

>>2517380
Your code is wrong in so many ways...

>> No.2517480

>>2517445
Or even better, randomly select from 80 sides instead of 40 chips. 10 sides are blue with blue opposites, 10 are blue with red opposites, 10 are red with blue opposites, and 50 are red with red opposites.

Although, if you frame the problem that way, you don't really need a program, do you? It's just 50/60.

>> No.2517496

>>2517461
>mfw he's manipulating his counter variable.
didn't think it worth mentioning though

>> No.2517527

>>2517496
>implying a B/B chip counts towards the count when it is given that it has a red size.

>> No.2517561

5/6

>> No.2517577

>>2516900
it comes from idiots with no statistical background whatsoever

>> No.2517627

>>2517377
the first good explanation of why its 5/6

>> No.2517632

>>2517527
By decrementing the counter when he gets a B/B chip, he's not counting them in ROLLS. However, by not decrementing it half the time he gets a R/B chip, he's over-counting R/B chips.

>> No.2517717

>>2517445
Thanks, that was the problem I was having when I did it the first time, before the program.

C code (hopefully correct) : http://pastebin.com/fVneKXeh

>> No.2517751

>>2517717

I think this is a more elegant solution, in PROPER C++: http://pastebin.com/ZfSuqfut

>> No.2517788

>>2517751
Rectifying: the "// is at least one of the sides red?" at line 31 is imprecise, it should be "//is a random side red?".

>> No.2517801

>>2517717
shit I am making to many stupid mistakes today (forgot to account for BB chips when I changed to while loop)
http://pastebin.com/5E7qME0H

>> No.2517829
File: 49 KB, 393x318, 1297467322137.jpg [View same] [iqdb] [saucenao] [google]
2517829

>couldn't prove a shit to save his life
>write a program instead
>it doesn't work.

it's like i'm on /sci/ or something

>> No.2517834

>>2517801
Still wrong somewhere, and >>2517751 looks a helluva lot better than your code.

>> No.2517842

ITT: People falling victim to gamblers dilemma.

If I flip heads 5 times in a row, the sixth time I flip the coin, there's still a 50% chance of heads. Derp derp

Answer is 5/7s.

Out of bag of 40 chips, 35 have a red side.

I picked one of those 35.

Out of those 35, 25 of them have 2 red sides.

25/35 chances other side is red.

If the answer was 5/6, then that would mean there is a 1/6 chance of the other side being blue, which doesn't add up.

5/7 red, 2/7 blue.

>> No.2517844

It's a trick question when you see all these blue and red chips you throw the bag in the bin cause chips aren't supposed to be blue and red something is wrong with them

>> No.2517864

>>2517842
There's a 1/ 8 chance of picking a red/blue chip and seeing the blue side first.
Read >>2516974

>> No.2517879

>>2517844
Casino chips, mind you.

>> No.2517894

25/35 = 5/7

>> No.2518132

How 5/7 people read the question:
"You pick chips at random until you see a red side."

How 5/6 people read the question: "You pick one chip, look at one side, and see that it is red."

As the question was stated, the 5/6 people are correct.