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File: golden balls.jpg (108 KB, 695x572)
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>>
50%
>>
>>14713874
it's been proven to be 2/3
>>
>>14713866
Excellent, unambiguous text in terms of picking a box, then picking a ball. Only the diagram is still false.
>final box contains gold/silver
This is false, the middle box contains gold/silver.
>>
>>14713874
Correct, either it is or it aint
>>
>>14713954
But there are two of the three gold balls left
>>
>>14713954
>>14714110
https://youtu.be/ytfCdqWhmdg
>>
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66% according to python
i don't know shit about prob but the algorithm is sound and follows the instructions given by the problem
>>
You have identical twins, and can distinguish them only by their clothing.
But one day, while bathing them, you realise that you aren't certain that you've kept track of which is which. You guess, dress them, and try not to think about it.
Unfortunately, a week later, this happens again in exactly the same way.

What is the probability that the twins' identities have been confused?
>>
>>14714250
>you aren't certain that you've kept track of which is which
does this really happen to parents of twins?
>>
>>14713992
ignoring the third box:

box 1 : which you have a 2/3 chance of drawing from initially, as 2/3 possible gold balls are in this box.
[100% chances of gold-gold]

box 2 : which you have a 1/3 chance of drawing ball 1 from, as 1/3 gold balls are in this box
[0% chance of gold-gold]

therefore you have a 66.66666... % chance of drawing a gold ball next, when your first randomly drawn ball is a gold one. as 2/3 times you draw gold first, it will be in the box where you have a 100% chance of drawing a gold ball next. and the remaining 1/3 times you will never draw gold next.
I hope this helps.
>>
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>>14714250
Well, I got 50%
>>
>>14714325
There is a secondary proof to this; regardless of color, you have a 2/3 chance to draw another ball of the same color given a silver or gold ball on the first ball drawn randomly from the chosen box.
[g,G] 100%
[G,g] 100%
[g,S] 0%
[s,G] 0%
[s,S] 100%
[S,s] 100%
as OP post says, after picking one of the above boxes, the probability of the next ball you draw being the same as the first one is : 4/6 == 2/3.
>>
>>14714250
You confused their identities in the end iff you mistake them exactly once. So 1/2.
>>
g1 --> g2
g2 --> g1
g3 --> s
2/3
>>
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>>14713866
Chance of getting a gold ball from the mixed box after randomly picking(1/3) this box is half(1/2). If it only had gold balls then the answer of the problem would be 50%. Chances relative to a moment don't change after the moment passes. If you end up with a gold ball in your hand you got to consider how lucky you are if it's from a mixed box, and how certain it is from a gold ball only box. Quantify that and most of the time you will be getting the gold ball from the pure gold source.
>>
>>14713866
Statistics and probabillity are for faggots
>>
>>14714587
Quite the contrary. Faggots usually ignore risks and don't account for hiv, monkeypox, responsibility, when indulging in degenerate pleasure, similar to you.
>>
>>14714250
>you realise that you aren't certain that you've kept track of which is which.
we have to assume this one event(being uncertain) means 50/50 they were switched. And i don't think we can do that.
>>
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what about this, smartasses?
>>
>>14714684
i don't believe an analytical solution can be found
>>
>>14714684
I don't know it appears to be infinite but i may be just wrong.
>>
>>14714711
it is infinite
>>
>>14714684
Speed in the 1st mile = 15 mph
Let the speed in the 2nd mile = x mph
Average speed on entire 2 mile = (15 + x)/2
30 = (15 + x)/2
60 = 15 + x
x = 60 - 15
x = 45
Therefore, speed in the next mile should be 45 mph
>>
>>14714755
incorrect
>>
>>14714755
correct
>>
>>14714755
it would only be true if the driving time of the second mile was the same as the first which is impossible and you're a moron
>>
>>14714755
What if it's talking about the average time spent, not the average speed over the miles. You would be correct if we're talking about the average speed per unit distance, but not per unit time. This is because if you went 45 mph, you would cover the second mile in 1/3 the time; thus you spent 3/4 of the overall time going 15 mph and only 1/4 of the time going 45 mph. The time-based average comes out as 22.5 mph.

The problem here is the faster you go, the less time you spend on that part, so there is an upper limit to your maximum possible average velocity over the course of only 2 miles, given the first mile was fixed at 15. It's possible for the maximum to be less than 30. That's what you can tell right away.

If you want to maximize the average speed, you can try to find the place where the time-derivative of the function for the average speed is zero, but you'll notice there is no such solution for positive t. If we constrain our second half to a constant velocity (encompassing any other travel speed during the second half that averages to this) then the overall average velocity will be (4 minutes * 15 mph + t minutes * 60/t mph)/(4 + t) = 120/(4+t).

If t = 0, then we get an average of 120 / 4, which is 30, but that would imply the car is traveling at 60 / 0 mph, which is a zero denominator. Basically, it is possible to get infinitesimally close to an average of 30 mph by travelling at speeds that would cause your travel time for the second mile to approach zero, but never to reach 30 mph, because this would require travelling the second half in no more than 0 units of time.

However, a workaround to this word problem is the fact that it never required the driver to travel the second mile along the road. He could go offroad and take a longer route and go faster and there would be a whole family of solutions there.
>>
>>14714684
It takes 4 minutes to drive 1 mile at 15 mph.
It takes 4 minutes to drive 2 miles at 30 mph.
So he has to stop after 1 mile, because there is no time left and he will never arrive.
>>
>>14714878
>will never arrive
or have to accelerate infinite or to teleport to the destination
>>
>>14714834
uhmmm sweetie, i think you forgot to account for the effects of general relativity
>>
>>14713866
>It's a gold ball.
This means the 3rd box is out of question and there's either a gold or silver ball left. So the chance in 50/50.
>>
>>14713874
>>14713992
>>14714193
>>14714340
>>14714935
thank you anons, the answer is 1 in 2
>>
>>14714942
you didn't even read >>14714340's post you stupid fucking mouthbreather.
re-read the OP post as wel.
>>14714325
>>14714384
>>14714556
all put fourth extremely easy to grasp proofs of this being 2/3.
So stop trying to necro-bump your own troll thread.
>>
>>14713874>>14714205
It's more likely that the box you pick, is the first one, since that's the one with more gold balls. Therefore, probability should be greater than 1/2.
>>
>>14713866
Its 50% chance. The question is not asking the probability of pulling two consecutive gold balls out. Its asking once you take one gold ball out what is the probability it is gold. Since it can only either be gold or silver its 50%.
>>
>>14714684
Can the car change velocity instantaneously?
>>
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>>14714684
>>
>>14716117
>Its asking once you take one gold ball out what is the probability it is gold.
That would be 1, and is clearly not what the question asks.
>>
>>14713866
What's the point of the box with 2 silver balls?
>>
>>14714935
If the third box is out, that means there are more gold balls in total leftover.
>>
>>14716117
there are three scenarios. Two of them give you a win. Its 2/3.
>>
There are three boxes to pick from

but since you already have pulled a gold ball out of the box you can be sure that it isn't the box of two silver balls

thus the box you choose can be one of two boxes and your next choice is a 50/50
>>
>>14718366
>two boxes
aren't equal
>>
>>14713866
Shouldn't the answer be simply 1/3 ? There is 1/3 chance that you choose the box with 2 gold balls
>>
>>14718526
>>14714535
>>
>>14718366
The two boxes aren't equally likely when it comes to drawing another gold ball.
>>
>>14718528
The two gold balls are in a single box and the chance that you chose that box = 1/3
>>
I just love watching /sci/ always have a mental breakdown when trying to solve a Children's puzzle. It helps remind me of the caliber of people that visit this board.
>>
>>14718531
>the chance that you chose that box = 1/3
nah
>>
>>14718539
How? Probability of choosing one box randomly out of 3 equally likely boxes = 1/3
>>
>>14718546
500 times you pick from SS
500 times you pick from GS
500 times you pick from GG

SS: 0 favourable
GS: 250 favourable
GG: 500 favourable
--------------------------------------------
250+500= 750 favourable (from prompt: "It's a gold ball")

500 of those 750 times you have locked into the GG box.
500/750 = 2/3
>>
>>14718547
> GS : 250 favourable
How? It says that you picked a gold ball, this means that the probability that the next ball is G = 0, and if you picked silver ball first, then that is an invalid case anyway, there are no favourable outcomes in case of GS box.
>>
>>14718555
>How?
250 times you manage to get a gold ball
250 times you get a silver ball
>>
>>14718557
Dude the question says you have already drawn a gold ball, the question is about drwing *another* gold ball.
>>
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>>14713866
>https://en.wikipedia.org/wiki/Monty_Hall_problem
This is a variant of the Monty Hall problem.
For the door thing, imagine there are one million doors and one goat, and 998 empty doors were revealed. Do you REALLY think you just happen to pick the right door on first try? Is is best to opt to pick the second door.
>>14714325
Likewise, imagine two boxes, one with a million gold balls, and the other with one gold ball and 999,999 silver balls. You take a gold ball. Do you REALLY think you took the gold ball from the 999,999 silver balls box on first try? It is more likely you took it from the 1M gold balls box.
>>
*999,998 empty doors but yeah
>>
>>14718564
you're the one who is insisting with including the silver balls with the 1/3 number.
all i'm saying is that then you've got to give them a full blown treatment.

personally i prefer the 'got a gold'-treatment, like is done in >>14714535
>>
>>14718582
Just realised my mistake, the case where you draw the gold ball from GS box isn't a favourable but has to be counted in total number of cases.

Thus 500/750 = 2/3 = correct.
>>
>>14718580
Monty hall problem is quite a different problem, both are veridical paradoxes though.
>>
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>>14718573
>>14718582
>>14718591
>double edit

>>14718597
>https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
Apparently it's called Bertrand's box paradox.
Sure it's different, but similar in that the same actually-two-third-vs-one-half probability derives from the same reasoning that you're more/less likely to have picked something to begin with.
>>
>>14718609
>double edit
2/3
so meta
>>
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>>14718615
(73+91)/(73+82+91) = 2/3
>>
>>14713866
0% because I'm not reaching blindly into a box to grab your balls.
>>
>>14713866
if there are 3 golds you can label them gold1, gold2, and gold3
if you pick a gold at random then your options are
gold1 - gold2
gold2 - gold1
gold 3- silver
there are only 3 possibilities and 2 of them are gold gold
>>
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>>14719986
This implies you were already tricked into grabbing his bawls once and now kicking up a storm about how you ain't gonna touch no goddamn second golden bawls.
>>
>>14720122
>if you pick a gold at random

Except you DIDN'T pick a gold at random, you picked a box at random. Since you picked up a gold ball, it couldn't have been the box with two silver balls, therefore it was either the one with two gold balls or the one with one gold ball and one silver ball. If you picked the one with two gold balls then the next ball will definitely be gold and if you picked the one with one gold ball and one silver ball then the next ball will definitely be silver. Therefore, you have a 50% chance of picking a gold ball next and a 50% chance of picking a silver ball next.
>>
>>14721967
> Except you DIDN'T pick a gold at random
How else did you pick a gold then?
>>
>>14721991
You picked the ball after you already picked the box, therefore you only had two balls to choose from, no more.
>>
>>14714587
probability is very important
statistics are in some respects propaganda, but statistics that significantly differ from probability are worth at least apprehension of significance and at most their own mid-minor model in considering the main models of the system you are trying to garner value and inefficiencies from.
back to the 'answer' to the first question, it depends on the wording/semantics of the problem and that interpretation. It's a fun provocative post, so thanks for who posted it.
>>
>>14721967
>>14722008
2/10 shitposting
>>
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>>14713866
well /sci/?
>>
>>14713889
my unironic first guess was 1/3 but now that you say it you're 100% correct.
>>
>>14713866
There is a 1/3 chance you selected the first box, and thus drew a gold ball
There is a 1/3 chance you selected the second box, and a 1/2 chance that you selected the gold ball first

So, 2/3 of the times that you draw a gold ball first, you picked the first box.
>>
anyone that legitimately believes it's 2/3 is a pseud
>>
>>14722703
0.25 in both cases.
>>
>>14724980
Wait, no shit, for b it would be higher. Hang on,I need a pen and paper
>>
>>14724983
Is the answer to b 0.5?
>>
>>14713866
I don't exactly understand the wording of the question
Doesn't it imply that the drawing of the first ball had already happened and it's probability shouldn't be taken into consideration?
It asks you about the probability of drawing a gold ball when you pick at random a box out of two where one contains a silver ball and the other one contains a good ball
>>
P(Second ball gold | First ball gold) = P(Both balls gold)/P(First ball gold)

P(Both balls gold) = P(You pick the gold/gold box) = 1/3
P(First ball gold) = P(First ball gold| You pick the gold/gold box) P(You pick the gold/gold box) + P(First ball gold| You pick the gold/silver box) P(You pick the gold/silver box) = 1 * 1/3 + 1/2 * 1/3 = 1/2

So
P(Second ball gold | First ball gold) = (1/3)/(1/2) = 2/3
>>
The question that's asked is; what's the next ball going to be? So since you've already chosen a box and have already taken out a ball, only two balls remain a gold one and a silver one, since you're going to take another ball from the same box.
It becomes a 1 in 2. 50/50 if you will.
>>
>>14725335
It says you pick from the same box. If they shuffled the boxes and you didn't know which box was the same, then yes this would describe that pick
> you pick at random a box out of two where one contains a silver ball and the other one contains a good ball
>>
>>14725440
Oh I see
Thanks
>>
Why does no one ever try to explain this with the same sort of million doors example they use for Monty Hall?
You have two boxes. One contains a million gold balls. One contains 999,999 silver balls and one gold ball. You pick a ball at random from one of the boxes - it's gold. What's the probability that a second ball picked at random from the same box will be gold?
>>
>>14725641
50%
>>
>>14725641
>Why does no one ever try to explain this with the same sort of million doors example they use for Monty Hall?
Because it's a dumb explanation and the people who spout it usually demonstrate a glaring lack of understanding.
>>
>>14725641
This analogy doesn't work because the second box contains the same quantity of silver balls as gold balls, not more silver balls than gold balls.

You have two boxes. One contains a million gold balls. One contains 500,000 silver balls and 500,000 gold balls. You pick a ball at random from one of the boxes - it's gold. What's the probability that a second ball picked at random from the same box will be gold?
>>
>>14725641
50%
>>
>>14725677
Approximately 5/6. How is that better? It's still a different answer to the original. The point of the example is to make it clear that if you pulled out a good ball your more likely to be pulling from the box with more gold balls. That point is better made if one box has only gold ball out of a million.
>>
>>14725702
>If anything, the analogy would be for one box to contain a million gold balls while the others remain the same,
That's not better. It encourages brainlets to get distracted by thoughts like
>but I'd be able to tell that this box has more balls in it
>>
>>14725641
>One contains 999,999 silver balls and one gold ball.
It makes no difference from the perspective of someone who doesn't understand the original problem. As far as 1/2ers are concerned, if they pick one box and they get a gold ball, they're guaranteed another gold ball, and if they pick the other and they get a gold ball, they're guaranteed another silver ball. How many silver balls there are, doesn't figure into it at all with that line of reasoning.
>>
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it's 66%
>>
>>14725301
Yes, now do this.
>>
>>14714319
About 20 years ago I was friends with twins and one of them was born without an arm.
It wasn't a problem for them, that's for sure.
>>
>>14727807
I'm the same anon who has absolutely no idea about probabilities, but the problem sounded too fun to pass
I did it in python and got this
a. it's always 66%
b. as p goes from 0.5 to 1, chances of getting the car by always switching go from 66% to 50%
c. as p goes from 0.5 to 1, chances of getting the car by always switching go from 66% to 100%

I probably fucked something up, but at least I tried kek
>>
>>14714684
30 mph and it must drive an additional mile. Then we'll launch a propaganda campaign discrediting the existence of the first mile and making anyone who promotes belief in the first mile an outcast.
>>
>>14725677
No, you have to stipulate that the balls are divided into bins that hold 500,000 balls each, and those bins always contain balls of the same color, and then you choose a bin. And that doesn't help the reader intuit the original problem, so why would you even bother?
>>
>>14728929
That’s correct; very good, anon!
>>
>>14713866
Holy fucking shit, there are so many illiterate nigger-faggots on this board. Posting the full 2\3-solution derivation.
>>
>>14730701
>using formulas to solve kindergarten problems
YWNB intelligent.
>>
>>14730701
Also, the P(A|B) notation reads 'the probability of event A occurring assuming event B occurred)
>>
>>14730704
I only used the definition of conditioned probability, you brainlet
>>
>>14730712
You don't need to calculate anything, know any formulas, or have any idea of what "conditional probability" is to figure this out out, brainlet.
>>
>>14730713
You sound like a fucking dropout. OP's problem is solved in under 50 seconds with mathematical rigour. Cope. Seethe. Dilate.
>>
>>14730720
You sound like a middleschooler plugging numbers into formulas you barely understand. OP's problem is solved in 10 seconds with common sense: you'll have twice as many cases of picking a gold ball from the leftmost box as from the middle box because you get a gold ball every time with one, but only half the time with the other. i.e. 2:1 cases where your second ball is gold.
>>
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>>14730730
Not sure how this translates to 66% probability, but regardless your reasoning is fallacious. Remember you a priori know you have already picked one gold ball, so if you want to arrive at the solution through quantitative analysis (not sure what to call your method) you need to list all options using the a priori knowledge which you do only half-way. If you want to know more about this google 'Bayesian Probability'.
>>
>>14730750
>Not sure how this translates to 66% probability, but regardless your reasoning is fallacious
Hahahahahahahaha. Thanks for demonstrating my point so perfectly.
>>
>>14730751
You need to be 18 to post here.
>>
>>14730753
You need to be non-retarded to post here. Please come back when you can grasp >>14730730. It should be fairly easy to grasp.
>>
Is there an actual name for the fallacy that causes people to think the answer is 2/3?

I know that it’s caused by muddling the probabilities of the first pick with the second one - like thinking that the previous flip of a coin affects the next flip - but I don’t know the name for it.
>>
>>14730760
Is there a name for the mental illness that causes 90 IQ brains to softlock when faced with kindergarten probability problems?
>>
>>14730316
(a)
What's the distribution of D? I will just assume that it's uniform under all the 2^100 subsets of C.
[eqn] P(A = D) \\
= \sum_{k=0}^{100} P(A = D | |A| = k) P(|A| = k) \\
= \sum_{k=0}^{100} \frac{1}{2^{100} } \frac{{100 \choose k}}{2^{100}} \\
= \sum_{k=0}^{100} \frac{{100 \choose k}}{2^{200}} \\
= \frac{1}{2^{100}}
[/eqn]
(b)
[eqn] P(A \subseteq B) \\
= \sum_{n=0}^{100} \sum_{k=0}^n P(A \subseteq B | |A| = k, |B|=n) P(|A| = k) P(|B| = n) \\
= \sum_{n=0}^{100} \sum_{k=0}^n \frac{n \choose k}{100 \choose k} \frac{{100 \choose k}}{2^{100}} \frac{{100 \choose n}}{2^{100}} \\
= \sum_{n=0}^{100} \frac{2^n {100 \choose n}}{2^{200}} \\
= \frac{3^{100}}{2^{200}}
[/eqn]
(c)
[eqn]
P(A \cup B = C) \\
= P\left( \bigwedge_{x \in C} \left(x \in A\cup B \right) \right) \\
= \prod_{x \in C} P(x \in A\cup B) \\
= \prod_{x \in C} \frac{3}{4} \\
= \frac{3^{100}}{4^{100}}
[/eqn]
>>
>>14730712
>>14730713
>>14730720
>>14730730
>>14730750
I agree. Using a formula for this is peak brainlet. There are two golds out of three which satisfy the condition, therefore the probability if 2/3. You don't need to use fucking Bayes' Theorem or whatever.
>>
>>14730780
>There are two golds out of three which satisfy the condition, therefore the probability if 2/3
Brainlets like you should do like him and stick to the formulas because this is worng.
>>
>>14730770
I think it's asking for any given D. It's the same answer in any case.
>>
>>14730761
Check your medical records, if you don’t find the diagnosis there, it probably doesn’t exist.


You confuse the probability of picking a box which contains two of the same color balls (which is 2/3) with the new probability of your second pick matching the first, but the second pick is under a new set of circumstances which technically create a 1/2 chance.

It’s 2/3 to me, but I like trolling the narrow-minded and do not think 1/2 is wrong either;
It’s just an ambiguously written problem that allows for interpretation that can lead to two right answers, like 6\2(1+2).
>>
>>14730858
Is there a name for the mental illness that causes 90 IQ brains to softlock when faced with kindergarten probability problems?
>>
>>14730755
>You need to be non-retarded to post here.
Unfortunately this is not the case.
>>
>>14730862
Fair enough, but even if you're retarded you should be able to comprehend >>14730730.
>>
>>14730760
I rate this bait a solid 8.
Unless you were being serious, in which case I just weep.
>>
>>14730783
>the answer is right, but this is wrong...because...because...it just is okay.
>>
>>14730879
The answer is right but the reasoning is wrong. You could have a million balls which "satisfy the condition" and the answer would still be 2/3.
>>
>>14730780
>Using mathematics to solve a mathematical problem
>Peak brainlet
>>
>>14730896
Yes and the problem with your mindless plug-n-chug approach is immediately demonstrated by your failure to understand >>14730730
>>
>>14730783
>>14730880
Nope, consider the case of: (GG),(GG),(GS).
Since you also turned out to be a brainlet who needs to use formula:

[eqn]
\mathrm P(B_1 \cup B_2 \mid G)
=
\frac
{ \mathrm P(G \mid B_1 \cup B_2) \quad \mathrm P(B_1 \cup B_2)}
{ \mathrm P(G)}
\\
\mathrm P(B_1 \cup B_2 \mid G)
= \frac
{1 \times 2/3}{1 \times 2/3 + 1/2 \times 1/3}
= 4/5
[/eqn]

Since, four balls out of the five satisfy the condition. Sigh anon, I thought you were different.
>>
>>14731068
>consider the case of: (GG),(GG),(GS).
I don't need to. Consider the case where the leftmost box has a million gold balls. How many balls are there that "satisfy the condition"? The answer is still 2/3.
>>
>>14731077
If you had so many balls in only box, I think person would figure out which is the required box when they put their hands in it. The fact that you even had to consider probability for this one, shows that you were the true brainlet all along.
>>
>>14731097
>I think person would figure out which is the required box when they put their hands in it.
There's nothing to figure out. The box is chosen at random. You're a legit retard and this is boring.
>>
>>14730868
It was 50% bait, 67% real.
The 2/3rds guys are fun to mess with because they don’t understand that it’s a riddle intentionally ambiguous enough to leave two valid answers because of personal interpretation:

1. Practically speaking, you have a 2/3 chance to pick a box with two of the same colors; if you picked a gold ball, there’s a 2/3 chance that you picked it from the box with 2 gold balls, hence a 2/3 chance that remaining ball is also gold.

That’s the obvious answer, the fun answer is the one that makes simpletons angry because they lack perspective:

If you grab a gold ball, you eliminate 1 box and it’s 50/50 which of the remaining two boxes the gold ball was taken from, so there is a 1/2 chance that it’s a silver ball and a 1/2 chance that’s it’s a gold ball.

Yeah one answer is less valid, but still not entirely untrue; it’s just a different interpretation of where to start the math- like 6/2(1+2).

Is it 1 or 9?

And why does it make you so angry that it’s 1?
>>
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>it’s a riddle intentionally ambiguous enough to leave two valid answers
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>>14732128
That’s literally the entire reason it was posted; its not an earnest math problem someone really needs the answer to, it’s just intended to seem that way to frustrate and confuse the normies who can’t hold two competing theories at once.

I would be sorry to spoil this surprise for you, but there’s a 2/3 chance that you will just think I’m lying or wrong or whatever you have to pretend to maintain your ego.

It’s ok to be wrong.

It’s being unwilling to admit you’re wrong that makes you ignorant.
>>
>>14732172
>That’s literally the entire reason it was posted
The reason it was posted is to bait people into arguing with clinical imbeciles like you.
>>
>>14732125
>Practically speaking, you have a 2/3 chance to pick a box with two of the same colors
Non sequitur.

>if you picked a gold ball, there’s a 2/3 chance that you picked it from the box with 2 gold balls, hence a 2/3 chance that remaining ball is also gold.
Correct, but so reasoning led you to this. I award you no points.

>If you grab a gold ball, you eliminate 1 box and it’s 50/50 which of the remaining two boxes the gold ball was taken from
Wrong. You only had a half chance of choosing the gold ball from the mixed box, so it's not 50/50, it's 1 vs 1/2, i.e. 2/3 vs 1/3.
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>>14714205
If you use ball = box.balls.pop(choice);next_ball = box.balls[0], no one needs to question what a negative index could possibly mean.
>>14714340
thanks
>>14714556
I didn't read your post, but the pic is what I needed to grasp what was so hard about the OP
>>14714684
Why must the driver have a sex?
>>
>>14732187
>there are only two boxes containing at least one gold ball and one box containing two and therefore a 1/2 chance the other ball is gold.
Doesn't follow. The chance your gold ball is from the box with only one gold ball is half that of the box with two. You're just ignoring that by assuming it's equally likely to come from one box or the other. This is not interpretation, it's about indcorrect assumptions.
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>>14713866

There are actually two valid answers depending on how you read and interpret the problem. Because it was vague. But I assume the asker intended scenario 2.

Interpretation one, imagine three real boxes, and I stick my hand in one and find a gold ball, there are only two boxes containing at least one gold ball and one box containing two and therefore a 1/2 chance the other ball is gold. Because this scenario implies that the silver box was still an option supposedly, initially, and therefore the information of a gold ball does not matter as there are only two scenarios:

Gold > Gold (Box 1)
and
Gold > Silver (Box 2)

If I am picking between three boxes my odds are 50%.

Scenario two, instead of picking from 1 of three boxes I pick from two boxes only, or... from 3/6 balls. Therefore the odds are:

Gold > Gold
Gold > Gold
Gold > Silver
Silver > Silver
Silver > Silver
Silver > Silver

Therefore in this interpretation there is a 1/3 chance the next ball is gold. Note you must pick up a gold ball and look at the other ball in the same box.

Therefore the odds are actually 2/3 because:

Gold > Gold
Gold > Gold
Gold > Silver

/thread
>>
>>14732187
>Therefore in this interpretation there is a 1/3 chance the next ball is gold.
No, 2 out of the 3 times you chose a gold ball, the next ball is gold. Fail.
>>
>>14714755
>>14714684
Average speed is total distance over total time.

Total distance = 2 miles.
Total time = time for first mile + time for second mile = 1/15 + T.
Average speed = 2/(1/15+T) = 30.
This gives T = 0.
To cross the second mile in 0 time requires infinite speed.

Moron.
>>
>>14732204
>there are only two boxes containing at least one gold ball and one box containing two and therefore a 1/2 chance the other ball is gold.
Wrong. See >>14732198
>>
>>14732198
>>14732205

Relax scienceheads.

Think about it this way.

I am not choosing between 4 balls I am choosing between 3 boxes, therefore each choice has a 1/3 chance of being the silver/gold box, now with the knowledge that the first ball is gold yes there is a 2/3 chance, but without that knowledge there is a 1/2 chance.
>>
Here is the smartest way to interpret it.

May I ONLY pick a golden ball? Then it is 33.333...%

May I pick a golden ball by coincidence? It is 50% because it implies that if I were to extend this experiment I could also pick the silver/silver box too or I could pick the silver ball in the silver/gold box.

Thereby up to interpretation on the wording.

Did OP imply I MUST pick a golden ball in this particular case? Or can I roll again and get a silver ball? Overall odds are 1/2 in such case.
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>>14732186
> Non sequitur.
This is the simplified concept - what are the odds that if I pick x color ball, the remaining ball will also be x color? Well, you have one of three boxes that contains different colored balls, so 2/3 chance that first pick color = second pick color.

> Correct, but so reasoning led you to this. I award you no points.

I don’t know how to uncomplicated such simple concepts; there are 3 gold balls, 2/3 of them are in the same box, so when you pick a gold ball, there’s a 2/3 chance you picked it from the box with 2 gold balls, so a 2/3 chance you will pick the gold ball next.


>Wrong. You only had a half chance of choosing the gold ball from the mixed box, so it's not 50/50, it's 1 vs 1/2, i.e. 2/3 vs 1/3.

You’ll never understand because you are still approaching the problem from the same starting point as before. You had a 1/3 chance to pick the box with 2 gold balls and a 1/3 chance to pick the box with 1 silver 1 gold and 1/3 chance to pick the 2 silvers box. You are just as likely to have picked the 1 gold 1 silver box as the 2 gold box.
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>>14732217
>I am not choosing between 4 balls I am choosing between 3 boxes
You're choosing between boxes abs then balls within those boxes.

>but without that knowledge there is a 1/2 chance.
Without that knowledge the question is meaningless.
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>>14732223
>May I ONLY pick a golden ball?
No, it says you pick randomly.

>It is 50% because it implies that if I were to extend this experiment I could also pick the silver/silver box too or I could pick the silver ball in the silver/gold box.
Doesn't follow. Go back to school.
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>>14732322
> Without that knowledge the question is meaningless.

Which is exactly why the original question is proposed as it is;

Including the NOTE: you can’t see into any of these boxes.

The problem intends to validate both the arguments for the sake of argument.
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>>14732289
>Well, you have one of three boxes that contains different colored balls, so 2/3 chance that first pick color = second pick color.
Again, non sequitur. Picking a box with two of the same colors is not the same as picking a box with two gold balls. The question direct say the first ball can be gold or silver, just gold.

>there are 3 gold balls, 2/3 of them are in the same box, so when you pick a gold ball, there’s a 2/3 chance you picked it from the box with 2 gold balls
Incorrect reasoning. If there were 100 balls in the box that was all gold, then according to your reasoning, the answer would be 100/101. But it's still 2/3. The number of balls is not relevant. Only the ratios of gold to silver, i.e. the probability of getting gold in each box.

>You’ll never understand because you are still approaching the problem from the same starting point as before.
I perfectly understand your mistakes.

>You had a 1/3 chance to pick the box with 2 gold balls and a 1/3 chance to pick the box with 1 silver 1 gold and 1/3 chance to pick the 2 silvers box.
Yes, now continue. You had a 1/3 chance of choosing box 1 and getting a gold ball. You had a 1/6 chance of choosing box 2 and getting a good ball. So it's twice as likely you're in box 1 as it is your in box 2. You just stop your reasoning halfway and don't reach the answer.
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>>14732346
>Which is exactly why the original question is proposed as it is
Yes, the original question is proposed exactly as it is to make your interpretation wrong. Ignore information in the problem and you get the wrong answer. It's that simple.

>Including the NOTE: you can’t see into any of these boxes.
Which just means you can't tell whether you're in box 1 or 2.

>The problem intends to validate both the arguments for the sake of argument.
No it doesnt. It's unambiguous as I've proven. Your interpretations all hinge on a fallacy or ignoring information, not ambiguity.
>>
>>14713866
50%

Proof. First suppose T is surjective. Thus W , which equals range T is finite-dimensional (by
Proposition 3.22). Let w1, ..., wm be a basis of W . Since T is surjective, for each j there exists
vj ∈ V such that T vj = wj . By Proposition 3.5, there exists a unique linear map S : W V such
that
Swj = vj
.
Hence for any w ∈ W , w = a1w1 + ... + amwm for some a1, ..., am ∈ F . We have
T S(w) = T (S(a1w1 + ... + amwm))
= T (a1Sw1 + ... + amSwm)
= T (a1v1 + ... + amvm)
= a1T (v1) + ... + amT (vm)
= a1w1 + ... + amwm
= w
Hence we have T S is an identity map on W , as desired.
To prove the implication in the other direction, assume that there is some S ∈ L(W, V ) such that
T S is an identity map on W. Then for any w ∈ W , we have that w = (T S)w = T (Sw) ∈ range T .
So w is in the range of T , so T is surjective
>>
>>14732349
> Again, non sequitur. Picking a box with two of the same colors is not the same as picking a box with two gold balls. The question direct say the first ball can be gold or silver, just gold.

Again, that’s why it is an ambiguous riddle: you (or most non-NPCs, at least) can interpret the question as asking what are the odds of picking the same color twice.

>Incorrect reasoning. If there were 100 balls in the box that was all gold, then according to your reasoning, the answer would be 100/101. But it's still 2/3. The number of balls is not relevant. Only the ratios of gold to silver, i.e. the probability of getting gold in each box.

Again, there is no incorrect reasoning, smooth-brain, it’s interpretation. If I wanted to say that 2 of the 3 remaining possible balls that can be picked are gold, so there is a 2/3 chance that my next ball is gold, that’s also equally valid.

> I perfectly understand your mistakes.
My mistakes are that there is a 1/2 chance that I’m arguing with a literal retard or 1/2 chance that I’m being trolled by someone whose pretending to be.

> Yes, now continue. You had a 1/3 chance of choosing box 1 and getting a gold ball. You had a 1/6 chance of choosing box 2 and getting a good ball. So it's twice as likely you're in box 1 as it is your in box 2. You just stop your reasoning halfway and don't reach the answer.

The chances of having picked a box with 2 golds and 1 golds were equal, and remain equal after eliminating one possibility, now the odds are 1/2 instead of 1/3.
>>
While there are three boxes that you could pick at random, we're told that we're in a situation where we've already picked either one of the first two of the three boxes, and to state the probability of an event occurring after the fact. Meaning the problem is asking us to consider the situation after picking a box that you know is one of the first two boxes listed in the problem.
Next, to ask "What is the probability that the next ball you take from the same box will also be gold" is the same as asking: "After knowing the box you've picked is one of the first two boxes (ones with two gold balls and one with a gold and silver ball), what is the probability of picking the box with two gold balls?" Since you're given two choices at this point where one is gold and the other is silver, the probability of picking up a gold ball is therefore 50%
>>
>>14732374
>Again, that’s why it is an ambiguous riddle
Your failure to read doesn't make the question ambiguous. The question says nothing about the first ball being either gold or silver, just gold. So the probability of choosing either the all gold or the all silver box is a non sequitur. We already know we're not in the all silver box.

>can interpret the question as asking what are the odds of picking the same color twice.
Technically you can interpret anything in any way. You can interpret yes as no and gold as silver. That doesn't mean your interpretation is valid and the question is ambiguous. It's not.

>Again, there is no incorrect reasoning, smooth-brain, it’s interpretation.
It's not even interpretation, you're trying to explain how you calculated the answer. Your explanation is wrong. The number of balls is irrelevant. Only the probability of choosing each box and the probability of getting a gold ball in each box matters. Are you actuality going to respond to this?

>If I wanted to say that 2 of the 3 remaining possible balls that can be picked are gold, so there is a 2/3 chance that my next ball is gold, that’s also equally valid.
It's not valid. The same reasoning would give you the wrong answer if there were different amounts of balls in the boxes. Try it yourself if you don't believe me. Calculate the probability of choosing a box and then choosing a ball from that box. Compare them. There is no "interpretation" here, it's just how reality works. You can even do it with real balls if you want and you'll see I'm right.
>>
>>14732374
>The chances of having picked a box with 2 golds and 1 golds were equal
CORRECT.

>and remain equal after eliminating one possibility
WRONG. After you eliminate the possibility that you chose a silver ball from the all silver box, it's now impossible that you chose it (1/3 changes to 0). After you eliminate the possibility that you chose the silver ball from the gold box, it's now half as likely you chose from that box as it is from the all gold box. The probability for that box remains the same because the elimination of the all silver box increased the chance you chose from the mixed box but the elimination of the silver ball from that box decreases it back to 1/3. The all gold box changes from 1/3 to 2/3.

Your argument is self-defeating. You realize elimination changes the probability you choose the all silver box to 0 but you ignore elimination of the silver ball in the mixed box. Again, you stop your reasoning half way and get the wrong answer.
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>>14732388
>Since you're given two choices at this point where one is gold and the other is silver, the probability of picking up a gold ball is therefore 50%
Doesn't follow. It's twice as likely you're in the all gold box because you had a 1/3 chance of choosing it and a gold ball from it but only a 1/6 chance of choosing the mixed box and a gold ball from it.
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>>14732443
>You had a 1/3 chance of choosing the all gold box
The question isn't talking about the probability of choosing one box out of the three initially. It's talking about the probability of being in a state of picking up 2 gold balls AFTER you know the third box isn't possible.
>You take a gold ball at random
This eliminates the third box
Now that you've picked up a gold ball, you're in a state which leaves you with a binary blind choice of either picking up a gold ball vs picking up a silver ball. You KNOW that after picking up a gold ball, that the box you've picked out of now either has a gold ball or a silver ball. it's 50%.
>>
Probability is like a mega brainlet filter. It's like they turn into the angry NPC meme when they read an explanation of it being 1/3
I believe this problem is like a sub-90 IQ cutoff test
>>
>>14732459
>The question isn't talking about the probability of choosing one box out of the three initially.
I didn't say it is. I said you HAD a 1/3 chance of choosing the all gold box.

>It's talking about the probability of being in a state of picking up 2 gold balls AFTER you know the third box isn't possible.
And you know picking the silver ball from the mixed box is also impossible. That's why it's only half as likely you chose from that box. You're failing because you ignore that.

>This eliminates the third box
And half of the mixed box.

>Now that you've picked up a gold ball, you're in a state which leaves you with a binary blind choice of either picking up a gold ball vs picking up a silver ball
Yes, but those are not equally likely. It's twice as likely you chose from the all gold box. 2/3.
>>
>>14732128
How would you make this "riddle" less ambiguous?
>>
>>14732372
>wall-of-text schizo
...and wrong
lol
>>
>>14718547
/thread
>>
>>14718547
>500 times you pick from SS
You pick from SS ZERO times. The question assumes that you picked either the first box or the second, therefore the third box is irrelevant.
>>
>>14734615
>The question assumes that you picked either the first box or the second
It doesn't assume, it says that that's what occurred. Of course it didn't have to occur.
>>
>>14734615
it is a full analysis, first task is to figure out how many " it's a gold ball " happens to each box
clearly GG : GS has a 2:1 ratio
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>>14732437
Don’t hurt yourself on some long response or putting things in cups;
every formula or long response can be reduced to the fact that 2/3 boxes contain two of the same colored balls, so whatever color (silver or gold) you pick first will always have a 2/3 chance to be the same color as the second.

It’s literally that simple.

The troll answer is that you had a 1/3 chance to pick every box, then you eliminate 1 possibility
> 1/(3-1) = 1/2
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>>14736667
>every formula or long response can be reduced to the fact that 2/3 boxes contain two of the same colored balls, so whatever color (silver or gold) you pick first will always have a 2/3 chance to be the same color as the second.
Doesn't follow. If there was no all silver box, then by that logic the answer would be 1/2. But removing the silver box changes nothing since we know we didn't choose it. The answer would still be 2/3 since it was twice as likely to choose the all gold box (1/2 chance) as it was to choose the mixed box and then the gold ball from that box (1/4 chance). You're an idiot who thinks that two things are connected just because they happen to have the same probability in one case. It's literally that simple.

>The troll answer is that you had a 1/3 chance to pick every box, then you eliminate 1 possibility
> 1/(3-1) = 1/2
Pure gibberish. There are 6 possible balls,
all equally likely to be chosen. You eliminate 3, and 2 of the remaining 3 have a gold ball in the same box.

I guess you're never going to respond to the proof you're wrong and you're just going to post more nonsense over and over. That's fine, it's trivial to show how you're wrong.
>>
>>14736667
>The troll answer is that you had a 1/3 chance to pick every box, then you eliminate 1 possibility
>> 1/(3-1) = 1/2
No, you eliminate the all silver box AND half of the mixed box.

>1/(3-1.5) = 2/3

lmao
>>
>>14713866
2/3, there's a 50/50 chance you picked the box with 2 golds. But since there's now 2 gold and 1 silver. It makes the answer 2/3
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>>14713866
I like how this trips people up. The reason why 2/3 is wrong is because the question is "What is the probability that the next ball you take from the same box will also be gold?", not "What is the probability that you will pick two gold balls in a row?"
The reason why it's 50% is because you've already eliminated one of the three boxes by picking up a gold ball, the options have been reduced to only two boxes, and the one you picked the gold ball from is the same you must take the next ball from.
>>
>>14737480
>why 2/3 is wrong
lol retard
>>
>>14713866
0%
Because I'm not putting my hands in any fucking box that I can't verify the contents of.
>>
>>14737416
>there's a 50/50 chance you picked the box with 2 golds.
Nope, there was a 1/3 chance initially and then a 2/3 chance after it was revealed you chose a gold coin.
>>
>>14737480
>"What is the probability that the next ball you take from the same box will also be gold?", not "What is the probability that you will pick two gold balls in a row?"
Same question.

>The reason why it's 50% is because you've already eliminated one of the three boxes by picking up a gold ball, the options have been reduced to only two boxes
And you've also eliminated half of the mixed box. You stopped halfway through solving the problem.
>>
>>14713866
Another way to think about it is simply getting rid of the 2 silver ball box since you know that it's out of the question and just treat the rest of the balls as being in one container. You know you drew one gold ball, and there's 2 gold balls left out of 3 potential balls. Therefore the probability is 2/3.
>>
>>14738014
Now try the same problem except in the all gold box there are 100 gold balls instead of 2.
>>
>>14738014
Now try the same problem with all silver balls removed.
>>
>>14713866
>You choose one ball, it's gold
>Two of the three gold balls are in the leftmost box
>The probability that you chose the leftmost box is 2/3
Remember this the next time you try to pick on CSfags you physics undergrad niggers. This is basic conditional probability, you're more retarded than you think you are.
>>
>>14738031
Let's say you have A gold balls in one box, B gold balls and C silver balls in another box. By your logic, the probability you chose the first box given you chose a gold ball is A/(A+B). It isn't. Can you tell me what it actually is?
>>
>>14738045
Gee, it's almost as if generalizing the problem makes it more complicated. Try Bayes' Rule if you really want to know.
>>
>>14737680
No literally, you are the retard. you're a grown ass nigger and you don't know conditional probability, stupid uneducated fuck. Go back to highschool.
>>
>>14713908
your methodology is incorrect.
When you have three boxes you always count and describe them left, right, middle.
>>
>>14713908
The order of the boxes or order in which they are described has zero impact in solving this problem. You'd have to be a serious autist to even give a shit about that.
>>
>>14738204
Is not more complicated. If you actually know the reason why your argument works for that example then it's trivial to solve the general problem. But you apparently don't. You don't even need Baye's theorem. Give up yet? The answer is (B+C)/(2B+C)
>>
>>14738262
It's you that doesn't know it. See >>14737951
>>
>>14732189
>no one needs to question what a negative index could possibly mean.
thats one of the best features of python: a negative index reads from the back, so -1 is the last element in the list, -2 is the second to last, etc.
>>
>>14738284
Wow, you've really amazed us with your towering intellect in solving a slightly more general problem than the one that could just as easily have been solved with Bayes' by now. Are you done mentally masturbating yet you stupid prick?
>>
>>14713866
Do you return the ball after taking it out?
>>
>>14738325
Why didn't your logic work?
>>
>>14738386
The problem's already been solved faggot. Maybe try another hobby, this one reeks of desperation.
>>
>>14738429
Not a response. So you admit your logic is faulty. Thanks.

Hint: the number of gold balls doesn't matter, only the probability of getting a gold ball in each box matters. Each box had an equal probably of getting chosen, so it doesn't matter how many balls are in the all gold box, the chance of getting a gold ball from that box is always 1 regardless.
>>
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>>14738466
I admitted no such thing and you are indeed a massive faggot, I will now proceed to call you a nigger and invite you to lick my balls.
>>
>>14738488
>I admitted no such thing
Then why doesn't your logic lead to the right answer in the general case? And why can't you explain why it works in this specific case? You post nothing of substance, just childish insults.

Thanks for again admitting I'm right.
>>
>>14737951
How in the flying fuck could it be the same question?
The question already states that you've picked a box. You yourself did not get to pick it, and due to the question already establishing that the ball you picked up was gold, that already eliminates one of three boxes for this question.

It's the same as saying: "You've taken a ball from one of seventy boxes, you must take the next ball from the same box", since the question limits you into a single box, the rest might as well not exist.

>And you've also eliminated half of the mixed box.
You haven't, because you don't know if it's the mixed or the gold box. You only know the contents of the boxes beforehand, but you do not know which box contains those contents. You have eliminated the third box by simply already picking a gold ball beforehand.
>>
>>14738737
>How in the flying fuck could it be the same question?
The next ball you pick will be from the same box. So two gold balls in a row. Why did you think anyone was answering a different question?

>The question already states that you've picked a box. You yourself did not get to pick it, and due to the question already establishing that the ball you picked up was gold, that already eliminates one of three boxes for this question.
It also eliminates half of the mixed box. You keep ignoring that, so you get the wrong answer.

>You haven't, because you don't know if it's the mixed or the gold box.
Doesn't follow. You don't know which box you're in, because you could have chosen from the all gold box or half of the mixed box. You have eliminated half of the mixed box by simply already picking a gold ball beforehand.

It's pretty simple if you just calculate the probabilities instead of using vague heuristics. The initial chance of choosing the all gold box 1/3. The initial chance of choosing the mixed box and then the gold ball from that box is (1/3)(1/2) = 1/6. So it's twice as likely you're in the all gold box vs. the mixed box after choosing a gold ball and those are the only possibilities. 2/3.
>>
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>>14738756
The initial chance of picking a gold ball from all boxes is 50%, 3/6. Once you have picked a ball, and as the question says, it is gold, the probability of the next ball being gold is also 50%, since you know you couldn't have picked the all silver box, you know that you have either the all gold or all mixed box.
The next ball you pick has to be either gold or silver, you don't know which it can be and the chances of you picking a gold one are exactly 50%, since you could have the gold box or the mixed box.
>>
>>14738800
>The initial chance of picking a gold ball from all boxes is 50%, 3/6.
Correct, but irrelevant.

>Once you have picked a ball, and as the question says, it is gold, the probability of the next ball being gold is also 50%, since you know you couldn't have picked the all silver box, you know that you have either the all gold or all mixed box.
Doesn't follow. It's twice as likely you chose a gold ball from the all gold box as it is you chose from the mixed box. You always choose gold from the all gold box but only had a 1/2 chance of choosing gold from the mixed box. They are not equally likely, you just assume they are with zero thought.
>>
>>14738800
The guy you're arguing with isn't wrong but he's being a complete fucking fag about it. It's more promising that you draw from the first box since you can count on it to be gold every time, it's less likely that you drew from the other box since only half the time it's not silver. You get into a car and it starts, it's more likely that you got into the car that starts every time than the one that starts only half the time. Someone robs a store and it's more likely to be a nigger, it's really common sense when you think about it.



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