The dice-rolling board knows how to think about odds, right?
(ignore twitter idiots, couldn't be bothered to crop the image)
 It's All Fucked Shirt $22.14 |
 DMT Has Friends For Me Shirt $21.68 |
It's All Fucked Shirt $22.14 |
Probability and stats questions
You math nerds and your min max shit trying to solve games ruin everything
FPBP. You can predict the odds but in practice they'll never line up with what you predicted. You WILL roll the minimum on your dice 5 times in a row despite it being astronomically impossible, you WILL NOT succeed, and you WILL NEVER roll the best you can.
Only nogames homosexuals pretend like probabilities and statistics mean shit in a TTRPG.
the worst I've ever had that I remember was a series of rolls which were 0.01% likely (3 NATTY 20s in a row which sent the table into spasms of laughter [it was pretty funny in the moment admittedly])
You'll find that 99% of viral math questions like this boil down to problems with English, not math. Those answers could be interpreted in 2 ways:
>More likely to be green than the previous draw
>More likely to be green than red
the answer is the same, so not really a problem in this case
yes, there's 100 balls that are completely randomly allocated, with an equal chance of each ball being green or red
I'm not entirely moronic then, phew
still get the distinct feeling they are intentionally wording the question in a more confusing "fancy" manner to make themselves come off as smarter. fairly obnoxious
>the answer is the same, so not really a problem in this case
It's not.
>More likely to be green than the previous draw
The previous draw is already established as red. The probability of it being green is 0. That would be a completely idiotic interpretation of the question.
You are not good at reading. Or logic. The result of the previous draw was red, the previous draw was a probability of unknown proportion which was different from the post-result, post-discard draw now must be.
>I'm not dyscalculic, I'm illiterate!
If you say so.
I'll try to rephrase unambiguously
>He's randomly (and secretly) chosen a number "n" in 0-100 (so a 1/101 chance of each). You don't know n
> n red balls and 100-n green balls in the urn, for 100 total
> He picks a ball at random from the urn. It's red.
> He throws that ball away, there are 99 balls left
> He's going to pick another one, again at random.
The question is "what's more likely, that the next ball is red, or that the next ball is green?"
i don't even fully understand the question tbh
are they saying there are 0-100 green balls chosen randomly and the rest of them are red, equalling 100 total?
There are 100 balls in an urn. Some may be red and some may be green. The balance is chosen randomly, so it could be 41-49, 27-73, or even 0-100 or 100-0. You pick up one ball out of the urn without looking inside and then you see that it's red. The ball is discarded and you grab another, again without looking. Is it more likely to be red or green?
The question does involve some more thinking than it might seem at first, but is a fairly low-level statistics problem. The tweeter just phrased it with light math jargon ("100-n" and "chosen uniformly at random in [0, 100]" rather than a plain English sentence with conversational grammar) because either:
1. He's a mega-autist who does not grasp that this phrasing is unintuitive to anyone not already initiated in statistical math
Or, more likely given that he's put this fairly entry-level statistics problem (i.e. it's not something that will be worth asking at all to people who know the jargon already) out there on a website used mostly by members of the general public with low attention spans,
2. It's phrased intentionally oddly to trip people up and his interest is more in making people feel dumb so that he can feel smarter for knowing basic statistics.
It's phrased in that way to remove ambiguity, and the people who get it wrong understand it and then proceed to incorrectly solve the problem. You can see like ten different people in this thread say, "well, must be about 50-50 since the average chance over all possible urns is about 50-50."
The intuitive answer is the right one here. For all purposes, you should assume the balls are 50% green, since that would be the average distribution if we did this experiment infinite times.
So removing a red ball means there are more green balls left. I can't be arsed to do the math, but it'd be something like 50.5% chance of green.
>since that would be the average distribution if we did this experiment infinite times
Not so. Any number (of n) from 0 to 100 is equally likely. The selection model does not imply that the numbers will tend toward middle over time. It's equally likely that you get both high and low rolls with infinite repeats. You should not assume average distribution given time because that's a fool's errand. If, for example, n ended up being in the upper 50% every time even given infinite rolls, then your numbers will still be skewed. Infinity does not imply that there will be equal distribution or that you will see every result given enough time. It is possible, no matter how unlikely, that you never see some numbers. Assuming that you will is a liberty you have taken for the sake of simplicity.
Removing a red ball does not mean there are necessarily more green balls left. If n was something like 83 out of the 100, then there are still significantly more red than green remaining. It may actually still be more likely that you draw red than green. You have no way of knowing.
Mathematician here. It would actually be more likely that it's red with unknown n. But it's definitely not something you have a great confidence interval with.
If OP meant to joke about those voting things other than green, OP is a moron. The answer is more complicated than his small brain can understand same as all idiots here who think they're good at science because they were good at it in high school.
Eugh. Okay well on the extreme that i got the only red ball there's a 100% certainty ill draw green, and if its the inverse there can be no green balls. It follows that the odds of each trend linearly towards n=50.5, where my next draw is 50/50
Assuming each value of n is equally likely then each outcome is equally likely then that leaves 50 green favoured sets to 51 red favoured sets, so the next draw is more likely to be red by like 0.9%. Unless you mean its more likely to be green than the previous draw, in which case obviously
Ah whoops ive miscounted. Green wins at n=50. Theres 51 outcomes between 50 and 100. That leaves n=0 to n=49 as red favoured, which is only 50 outcomes. Green is more likely by a very small percentage
I think your original reasoning was correct. The n=0 case can't occur as we always have at least 1 red. I tried considering the general case of m balls in total and somewhat loosely came to your original conclusion.
So n red balls and m-n green.
Ah oops. I WAS right but i dont deserve credit for it, i thought n was the green balls this whole time
What? No thats... can he do that?
Shouldn't you calculate per red ball, not per case?
If the ball you pull out is red, it is more likely that you pulled out of a heavily red case.
Yep, I ended up making the same mistake
corrected. So my analysis there is bunk.
>100% - N where N is zero, chance of moronation.
With the shit interval of confidence equally likely cannot be marked as wrong.
Well, obviously it can't be 0 after the test
I think at lot of you are getting this wrong: he's not *choosing* a red ball to show you first, he's picking a ball at random - and it turns out to be red.
That means you're much more likely to be in one of the cases where there's lots of red balls than the case where there's only one.
Nah senpai when I get an apple from a random box it's equally as likely to be an orange next dw
>you're much more likely to be in one of the cases where there's lots of red balls
N is chosen UNIFORMLY so taking out a random red ball means nothing and 50/50 is still the correct answer if there's no certainty on the interval of confidence.
You're not taking into account conditional probabilities. You're 100x more likely to draw a red ball in the "it's all red balls" case than in the "it's all green balls with one red ball" case.
There's no reason to shift to either side of the normal distribution if only 1 ball was taken before the question.
Yes there is. Also it's not a normal distribution, it's a uniform distribution.
https://www.online-python.com/I7fCzl28rd
Here's a simple python script to simulate all 999,900 combinations of n/first draw/second draws. Result is ~67% chance that the second ball is red, given the first ball is red.
Oops, made a tiny error. Corrected version: https://www.online-python.com/sx6VNUijoe
Comes out to exactly 2/3rds in this case.
I'm not sure your script is working properly. Range is nonexclusive so i in [0,100). Furthermore for the 2 balls in total case i get the same 2/3 for red. I would tentatively expect that case to be equally likely for red and green.
>Range is nonexclusive so i in [0,100).
Uch. You're right. I don't know why they made Python that way, it's just confusing. But it doesn't actually matter for the results, you get the same result no matter how big/small the jar is.
>Furthermore for the 2 balls in total case i get the same 2/3 for red. I would tentatively expect that case to be equally likely for red and green.
2 balls means 3 x 2 x 1 cases total, so actually pretty easy to hand calculate.
0 reds: Ignore these cases, obviously.
1 red: 1 case where first ball is red and second ball is green. Ignore the case where the first ball is green (because we know the first ball is red).
2 reds: 2 cases where first ball is red and second ball is red.
1 case where the first ball is red and second is green, and 2 cases where first ball is red and second is red, so 2/3.
huh.
I had to go look this up. Really surprising.
Welp, guess I'm just moronic.
That's why ppl with no education should stop talking about shit they don't and can't understand.
And that's why no one likes talking to math nerds and you always eat your lunch alone.
Oh I'm not a math nerd, I'm a mathematician. I hate doing maths and would never actually fricking get on Python to prove idiots on /tg/ they're wrong.
Right you should be using R
I'd probably use MATLAB cuz I'm a numerical analyst actually, but I was referring to that anon who used Python
R fricking sucks
>Yes there is. Also it's not a normal distribution, it's a uniform distribution.
Sum of series n, n from 1 to 100 = 5050. (Total number of cases where you draw a red first out of all 10,000 possible cases.)
(100-n)/100 = the probability in each case n, that the next ball is green.
Sum of series n*(100-n)/100, n from 1 to 100 = 1666.5.
1666.5/5000 = 1/3
So p(green|first ball red) = 1/3, p(red|first ball red) = 2/3.
>from 1 to 100
But zero is a possibility
It is not, since I drew a red ball, so the probability there are only green balls = 0. In any case it would just add 0 to the denominator and numerator, and (x+0)/(y+0) = x/y
It's not after you've already tested that it's at least 1.
Pretty sure this is right (but it should be "/99" on the second line).
2/3 doesn't have any clear relation to 100, I bet the number cancels out.
(and what a surprise, Twitter is full of morons)
>2/3 doesn't have any clear relation to 100, I bet the number cancels out.
> I would tentatively expect that case to be equally likely for red and green.
Frick it, this is the easy case, let's do it explicitly
Three ways to start: RR, RG, GG, equal 1/3 chance of each.
Chance of pulling a red ball first time: (1/3)*1 + (1/3)*(1/2) = 1/2
Guy pulls a red ball first so, conditional probabilities, using P(A given B) P(B) = P(B given A) P(A)
Chance we're in case GG given R: 0
Chance we're in case RG given R: (1/2)*(1/3)/(1/2) = 1/3
Chance we're in case RR given R: 1*(1/3)(/1/2) = 2/3
there's only one ball left, so we're done 2:1 odds it's red, just like when there's a hundred.
Wait, I think I can do the general case with almost no math: The number that was 100 in the original question, I'll call "k"
P(n red balls given the first ball was red) = n * (some shit that doesn't depend on n)
P(second ball is red given we're in case n) = (n-1)/(k-1)
P(second ball is green given we're in case n) = (k-n)/(k-1)
So
P(second ball red) = sum n*(n-1) *[some shit]
P(second ball green) = sum n*(k-n) *[the same shit]
Where the sum's over n from 0 to k (inclusive)
The two things we're summing are n^2 -n and kn - n^2.
Like any nerd, I know the sum of integers from one to N is N(N+1)/2
Can't remember the result for squares, look it up: apparently N(N+1)(2N+1)/6
So the ratio of those two probabilities is
P(next ball red) / P(next ball green)
=[k(k+1)(2k+1) /6 - k(k+1)/2] / [k*k(k+1)/2 -k(k+1)(2k+1) /6 ]
There's "k(k+1)/2" everywhere, cancel it
[(2k+1)/3 -1]/[ k -(2k+1)/3]
= [2k -2]/ [k -1]
= 2/1
The answer is more likely to be green.
In this hypothetical, the randomization occurs at the start when there are 100 balls.
The randomization effect is basically a coin flip for either color.
The odds that the coin flip would land on either color is the same, so the most likely scenario is 50 green balls and 50 red balls.
If the randomization effect would occur again after each ball is picked, the answer would be equally likely.
Why an urn. Why not just a jar or a basket or box? Why an urn?
>the urn is for the ashes of the braindead twit on Xtweet.
https://en.wikipedia.org/wiki/Urn_problem
Right, that's it. Specifically referencing this:
https://en.wikipedia.org/wiki/Ellsberg_paradox
wherein people prefer clear risk probabilities to ambiguous/unknown probabilities. This is probably what Daniel Litt was trying to demonstrate.
When n is in [0,50], it's always more likely to be green. When n is in [51,100], it's more likely to be red.
Since [0,50] is of size 51 while [51,100] is of size 50, the probability that n is a case where it's better to be green is higher than the probability that n is a case where it's better to be red.
However, as other anons have said, you drew a red ball already, which means that it's actually in [0,99] after the first red ball is removed. Then, if n is in [0,49], then the next ball is more likely to be green. If n is in [50,99], the next ball is more likely to be green.
Since both intervals are of size 50 on a uniform distribution, the probabilities should be equal.
That's only true if they show you the red ball at the start on purpose.
If the first red ball was picked randomly (like the problem said), then you have to consider how it affects the combined probability. Since randomly picking a red ball means those n with a higher red ball count are also more probable.
You can simulate the problem with any programming language, and you'll see that when you only consider the cases where you pick a red ball at random first, the chances of the second ball being red is about 2/3.
>That's only true if they show you the red ball at the start on purpose.
The way I interpreted the question is that, by conditioning, red has already been selected. At this point, you're just drawing a ball where the number of red balls is a random variable uniformly distributed on [0,99] and the number of green balls is 99 minus the number of red balls.
brainlet here
its a nonsense question with no true answer, we don't live in a magically combined multiverse where you are reaching into all 100 variations of the urn at once. there is one correct answer that is randomly decided and you can't know it unless you know the ball ratio
>brainlet here
The only true part of the post.
Repeat the experiment many times and the mean of the real value will converge to the theoretical value. Even if we ignore the fact that this is in fact a question about probability and not a phyisical ball and urn, the answer to the theoretical problem still says something about the real world.
While true, that deterministic nature is what allows us to make predictions when the true state is uncertain. All the statistics is doing is weighing deterministic events so we can predict which one is happening more reliably from an incomplete understanding of the situation
So true.
Incidentally, we should play cards. For money.
I flip a biased coin. It comes up heads. Since there's no way to know the distribution, you should take 50-50 odds on it being tails, right?
Okay, I've got a game for you:
I'm selling tickets for a strange lottery, each person only gets to play once.
The rules are simple:
> 1. the prize pot starts at $1
> 2. I'll flip a fair coin, if it's heads, you get the prize pot and the game ends
> 3. If it's tails the prize doubles and we go back to step 2
How much would you be willing to pay for a ticket for my game?
The prize rises inversely proportional to my odds of winning it, so i only really need to consider the first flip. Its only worth 50c but id pay a dollar for the sake of curiosity. Regardless of the rules i would never wish to play more than once.
>Either I get the money
>Or the price doubles and we flip again
Nice try, genie, but you aren't going to trap me in an eternity of coin flipping with your """fair""" coin.
Here, take an infinite amount of my money, kind gypsy
Rolled 1, 2, 1, 2, 1, 1, 2, 1, 1, 1 = 13 (10d2)
1's are tails, 2's are heads. Let's see what you've won
I'm going to pay you my infinite debt in 1 cent coins.
I hope you like it
You've won
>four dollars
!
with your initial outlay of
> infinity dollars
this puts your balance at
> Minus infinity dollars!
Do you feel you've made the right choice?
Well, I already had an infinite amount of money, or else the carny would have not let me play.
I'll build my second infinite fortune, no problem. I'm a hard worker.
That would be a sensible thing to assume. But then, "you take a random ball out the urn" would not make any sense.
That's why these problems usually have a guy showing you a red ball, instead of picking it at random.
I report you to my state's gambling commission and get a $500 reward
The unwashed mass is right, green is more likely
Roll 1d101-1 = Red Balls (0 to 100 is a d101-1, averaging 50)
100-Red balls = green balls, since average of red = 50, average of green = 100-50=50 is also 50
-1 red ball (the ball pulled in scenario)
on average will leave you with more green balls in the remaining 99
We're playing a game, and there are two closed boxes in front of you.
One box has two hundred dollar bills, and one box has a hundred dollar bill and a one dollar bill.
You reach into one box and - without being able to look - pull out one bill. It is $100. You get to pick to reach into either box one more time, and I offer you $1 to switch boxes and pull out a bill from the other box instead.
You are statistically illiterate so you take me up on it and lose money.
>[0, 100]
Haha did this pompous shit really write an off-by-one error in his dumb bullshit?
No, since [0,100] means it includes the cases where there are 0 and 100 red balls.
well if n is 50 there's 50/50 red/green. If you take out a red ball that means there's now 49 red balls and 50 green balls. so there's more green balls. am I missing something? Is this meant to be a Monty Hall scenario? it's draws WITHOUT replacement, if it was with replacement then it would be even odds since it's basically flipping a coin at that point, but here it's more like counting cards as they're played out of a deck, just with only 2 options
>am I missing something? Is this meant to be a Monty Hall scenario?
If there are lots of red balls, you are more likely to draw a red ball first. If there are lots of green balls, you are more likely to draw a green ball first. Since you drew a red ball, it is likely to be a red-ball-dominated urn, and unlikely to be a green-ball-dominated urn.
>Since you drew a red ball, it is likely to be a red-ball-dominated urn,
But since it's a normal distribution you're just as likely to have a red dominated urn as a green dominated urn, meaning you should treat it as if there's an average number of each given we're told it's a normal distribution, no? In which case you're back to treating it like 50/50, since it's just as likely to be red dominated as green dominated
The distribution tells you about the ball distribution in 10000 urns but not in one urn in particular.
>But since it's a normal distribution you're just as likely to have a red dominated urn as a green dominated urn, meaning you should treat it as if there's an average number of each given we're told it's a normal distribution, no?
No.
If your wife gets pregnant in the first month of trying, she's probably on the more fertile end of women. If a used car has a major issue within the first month of you buying it, it's probably going to have more. If you find and kill a wienerroach in your house, there are probably other wienerroaches in your house.
And if you pull out a red ball from the urn, it's more likely to be a red ball dominated urn.
And the average roll of a d6 is 3.5 yet I never stop rolling snake eyes. Go back to your pit, fiend; math is sociology for numbers.
Too bad the James Randi paranormal challenge was cancelled or you could make a cool million on your superpower.
Born too late to participate in Randi's challenge.
Born too soon to trick the devil in a game of dice.
Born just in time to never penetrate armour.
Word games like this is why we have common core. Grug flips board.
It's not a word game, every single person who gets it wrong and explaining their thought process is clearly getting it wrong because they aren't taking into account conditional probability, a concept that comes up in real life constantly.
You are just mathematically illiterate, which is in fact why we have common core - a desire to try to teach math well enough that even bozos like you can get it.
America was doing just fine until we decided to copy the Euros. Word games. Grug flips board.
>t. moron who doesn't understand basic mathematics and cries when people try to slowly insert the knowledge into his thuggish, stupid brain.
Education is mostly signaling and your lack of education is signaling that you are stupid.
Less than 1% more likely to get a green ball.
I'm taking a statistics class right now and it blows ass. Statics is way more fun and interesting frick you prob dickheads and frick you again for making me take this class for an unrelated degree.
It's not unrelated, you're just dumb.
you are correct but I'm still mad I had to sacrifice all my hobbies and free time to pass a course I don't enjoy.
Is it saying n can be zero through one hundred or zero or one hundred?
You fell for the trick.
Picking red is a bigger part of the question than you might guess.
Think about the fact that for cases in which the urn is mostly green balls, drawing red at random is very unlikely.
f(x) = (x - 1) / 99 is the probability that the 2nd draw is red.
The expected value of the function when x ~ uniform(1, 100) is sum ( f(x) / 100 ) for x in [1, 100], so E[f(x)] = 0.5.
x=0 is impossible because if there are no red balls then the first draw cannot be red
so what happen here if n=0, supposedly zero red in urn, yet I did took out red one?
Found the QA tester