I made a puzzle for you guys, maybe you can solve it, but I doubt it, you fucking autists
>>27657615
I know this is clearly designed to promote trolling and everything. And that's all fair and well, but the answer is 50%. You don't need autism for figure that out. You don't even need to be good at maths.
either 0% or 100%, on average 50%
>>27657615
50%, I'm pretty sure most anyone on this board would be able to figure that out. I guess I fell for the bait though.
>>27657721
>>27657714
How is it 50%? Isn't it more likely you picked the first box if the first lightbulb you pull out is gold?
>>27657615
(1/2)/(11/20) = 20/22
10/11 chance it's another gold
>>27657684
>>27657714
>>27657721
https://en.wikipedia.org/wiki/Bertrand's_box_paradox
please stop being retarded thanks
>>27657796
this seems correct tbqh
Try my problem, it's an infamous brain teaser that has confused brilliant mathematicians.
It's much better than OPs'.
>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
>>27657877
It would be better than OP's if everyone hadn't already seen it hundreds of times.
75%
Originallity desu senpai squad
>The answer is 50%
>faggots from /sci/ come in and bring trick questions to bait with
>they're surprised when people not indoctrinated into their group don't give the expected answer
>declare a board is unintelligent because they don't know the answer to a trick question
every fucking time you faggots do this shit
>>27657877
the answer is to switch doors
anime taught me about this trick question
>>27658027
>He didn't get the right answer.
Let me make this even easier for you retards.
There's two boxes. One has 999,999 silver balls and one gold ball. The other has 1,000,000 gold balls. You pick a box and randomly select a ball from it. It's gold. If you randomly select another ball from that box, what is the probability that it's gold?
Are you really so stupid as to think that there's an equal chance of drawing the first gold ball from each box? There's virtually no chance you selected the box with silver balls based on your random drawing of the gold ball. Thus, even though there's two possibilities (selecting the box with all gold vs. box with one gold) you must weight each probability based on how likely it was to occur.
>>27658061
it's as stupid as the faggots on /g/ who ask you to make a fizzbuzz or reverse a linked list
it's a trick question that's insanely easy if you've been taught the trick, but anyone who hasn't been taught the trick won't get it no matter how smart they are
just a bunch of pseudo-intellectuals jerking off about how smart but lazy they are
>>27657615
(1/2)*(1/10)+(1/2)=0.55
"(1/2)*(1/10)" is the probability to draw a gold ball from the second box if the ball drawn during the first draw was from the first box
"(1/2)" is the probability of drawing a gold ball from the first box if the ball drawn during the first draw was from box 2
>>27658102
Except this is literally logic, the smarter you are the easier it is for you to get it.
>>27658134
55% if you autists can't convert.
>>27658179
solving it logically doesn't get you the answer the /sci/ fags want
logically it's 50% becasue you're locked into a box and there's a 50% chance you picked the right box
but in the academic circlejerk the answer is >>27657800 and the problem behaves counter to logic because it has to follow the rules of probability
>>27658134
>calls other people autists
>didn't even read the question correctly
>>27658234
No need to see in the box
>>27658027
OP here, I'm not mad people get it wrong, I'm mad people tell the people who get it right that they're wrong. Fucking retards I swear, also I've never been to /sci/
>>27658261
>"(1/2)*(1/10)" is the probability to draw a gold ball from the second box if the ball drawn during the first draw was from the first box
The question doesn't ask about the drawing balls from boxes other than the one you initially picked from.
Hypergeometric distribution and bayes theorem
>>27657615
>puzzle
that's not a puzzle, that's a math probability problem
fuck you
Suppose you play this game 20 times. You randomly pull a ball from box A 10 times, and you randomly pull a ball from box B 10 times. If these pulls are truly random, approximately 9 of the pulls dealing with Box B will initially yield a silver ball. We can ignore those.
That leaves the 11 times where we pull a gold ball. 10 of those correspond to Box A, and 1 of those corresponds to Box B. Clearly, we see that we are much more likely to pull a gold ball from Box A the first time around. In fact, if we pull a gold ball from one of the boxes randomly, we know with probability 10/11 that the box we chose has entirely gold balls, and suspect with probability 1/11 that the box we chose had only 1 gold ball.
The probability of drawing another gold ball if we chose box A (which occurs 10/11 of the time) is 1. The probability of drawing another gold ball if we chose box B (which occurs 1/11 of the time) is 0.
1 * (10/11) + 0 * (1/11) = 10/11.
let's say a 10/11
If you draw a golden ball it's one of the ten golden of box 1 or the only golden of box 2
If it's from box 1 the next one will be golden, so we have and if it's from box 2 the next one will be silver so we have 10 cases where we pick golden and 1 case where we don't pick golden.
>>27657615
The only way you can get a second gold ball is if you got the first box.
So 50%.
I don't get it. Where's the catch?
>>27658426
>what is the probability that the next ball you take from the SAME box will be gold
It's a 50% chance assuming this is drawing without replacement. Either you picked the first box and it's a 100% chance to draw a gold ball or you picked the second box and it's a 100% chance to draw a silver ball.
The outcome of what ball you grab after the initial gold ball is entirely a factor of which of two boxes you choose - hence, a 50% chance.
>>27658514
That's what he's talking about, he's right.
>>27658563
>>27658479
lmao, just read the thread, goddamn
>>27658572
>assuming this is drawing without replacement.
There's a reason I said this. Drawing with replacement yields different odds.
>>27658225
How fucking dense can you be?
If you pick from the second box, then half of the time it'll be a silver ball, which means that half of the times you pick the second box, it won't count. Do you understand this? if you pick from the first box 10 times and from the second box 10 times, then half of the times you picked from the second box won't count, so only 5 of them will. and 10/(10+5) is fucking what again?
Please tell me if you still don't understand this as I really want to fathom how fucking stupid you are.
>>27658614
Not him but I don't get it... You're not a good teacher...
>>27658606
But the problem doesn't say anything about replacement. If we count replacement then we have an exact 91%
>>27657615
P(2nd ball is gold | 1st ball is gold)
= P(2nd ball is gold | 1st box is chosen & 1st ball is gold) * P(1st box is chosen | 1st ball is gold) + P(2nd ball is gold | 2nd box is chosen & 1st ball is gold) * P(2nd box is chosen | 1st ball is gold) [Law of Total Probability]
= P(2nd ball is gold | 1st box is chosen & 1st ball is gold) * P(1st box is chosen | 1st ball is gold) [2nd box has 1 gold ball]
P(2nd ball is gold | 1st box is chosen & 1st ball is gold)
= P(2nd ball is gold & 1st ball is gold | 1st box is chosen) / P(1st ball is gold | 1st box is chosen) [Definition of Conditional Probability]
= 1 / 1 [1st box has only hold balls]
= 1
P(1st box is chosen | 1st ball is gold)
= P(1st ball is gold | 1st box is chosen) * P(1st box is chosen) / P(1st ball is gold) [Bayes' Theorem]
= P(1st box is chosen) / P(1st ball is gold) [1st box has only gold balls]
= (1/2) / P(1st ball is gold) [boxes are chosen at random]
= (1/2) / (P(1st ball is gold | 1st box is chosen) * P(1st box is chosen) + P(1st ball is gold | 2nd box is chosen) * P(2nd box is chosen)) [Law of Total Probability]
= (1/2) / (P(1st ball is gold | 1st box is chosen) * (1/2) + P(1st ball is gold | 2nd box is chosen) * (1/2)) [boxes are chosen at random]
= 1 / (P(1st ball is gold | 1st box is chosen) + P(1st ball is gold | 2nd box is chosen))
= 1 / (1 + 1/10) [balls are chosen at random from their box]
= 10/11
Thus
P(2nd ball is gold | 1st ball is gold)
= 1 * 10/11
= 10/11
TL;DR: The answer is 10/11
>>27658681
not that guy, but look at it like this:
you got a golden ball, right? this gives you information about what happened and where you are, and you have to use it.
to make this clearer, lets look at it from a different perspective.
what are the odds that you will get a golden ball? there are 20 balls in total, and 11 of them are golden. therefore, you have a 11/20 probability of getting a golden ball. but is this still the case if you draw a silver ball first and then draw again from the same box?
no - the probability changes, as you hopefully figured out, because getting a silver ball first gives you crucial information about which box you drew from - with a 100% probability, you are in the 2nd box. in this box, there is only one golden ball, and there are still 9 balls left in total, so your odds of getting a golden ball from the same box is only 1/9.
see what you did there? you didnt just say "its equally likely that I draw from any of the boxes, so I'll give both of the scenarios a 50% chance", but instead you used the information you have - you calculated the conditional probabilities (0% box 1, 100% box 2) and multiplied those with the probability of getting a golden ball if you are in box X (1 for box 1, 1/9 for box 2).
so, the probability of getting a golden ball from the same box after getting a silver ball is:
p = 0 * 1 + 1 * 1/9 = 1/9
>>27658879
Guess I'm not really That good at math... I don't get why everyone is using formulas here. Mind explaining the original question now?
>tfw brain sicks now
>>27658681
>>27658879
if you understood that, then lets look at the problem at hand.
we got a golden ball, and now we want to find out the probability of getting another golden ball.
first: what info do we get from getting gold?
there are 11 golden balls. 10 of those are in box 1, 1 of them is in box 2. getting any of those is equally likely. you'll understand intuitively that the probability that you're in box 1 is exactly 10/11. if not, look at it this way:
give each of the golden balls a number. the balls with number 1 to 10 are in box 1, number 11 is in box 2. you either got 1, 2, 3, ..., or 11. each of these scenarios is equally likely (likelihood 1/11), and only one of those scenarios has you end up in box 2. therefore, the probability of ending up in box 2 must be 1/11, and the probability of ending up in box 1 must be 10/11.
now that we figured that out, what are the chances that we get a golden ball?
if we are in box 1, we are guaranteed to get another golden ball (probability of 1)
if we are in box 2, it is impossible to get another golden ball (probability of 0)
let's calculate it now:
q = 10/11 * 1 + 1/11 * 0
>>27658681
Just to make it clear, I'm talking about the 2 balls per box example as it's easier to imagine.
Imagine that you do the exercise 30 times, right? You pick a box at random 30 times. 10 times the first box, 10 times the second box and 10 times the third box. Since none of the balls in the third box are gold, these wouldn't matter. So we're gonna discard these 10 attempts. 5 of the times you pick out of the second box would also be silver, and as such these wouldn't matter as well. So you're left with 15 out of 30 times where you pick a gold ball the first time. And so we're only gonna look at these 15 times. 10 of these are from the first box and 5 from the second box. So out of 15 times you got the golden ball, 10 of them were from the first box. People get too hung up on the likelihood of picking a box, but if you pick the second box, it's not 100% chance you'll pick a golden ball, and so it won't matter the times you pick up the silver ball, and you will pick the silver ball, 50 % of the time. Which means that if you picked a golden ball, the likelihood you picked from the second box is half of the chance you picked from the first box. 1/3 and 2/3s.
>>27658874
backed up via simulation
>>27658181
Probability, i'd be shitting myself if a retard couldn't convert that, but incase they are retarded
>Fraction 1/100
>Decimical 0.01 - 1.00
>Percentage 1% - 100%
>>27658874
>only hold balls
*only gold balls
It's also clear that
P(2nd ball is gold | 1st box is chosen & 1st ball is gold) = 1
Just by observing that the 1st box only having gold balls means the 2nd ball can only be gold. The conditional probability reduction was unneeded.
Regardless the answer is still 10/11.
>>27659019
thanks senpai
>>27658971
What's the q formula btw? I don't remember that from algebra class. Also mind also explaining the other thread on this? The one with 3 boxes and 2 balls in each?
You can simulate this for yourself: Set up two piles of playing cards, one with only red cards and the other with just one red card. Flip a coin to select between the piles
>>27658062
thanks brah I hadn't thought of that
so what's the real answer
>>27658967
>>27658971
if that does not help, think about what assumption you make when you say that the probability of 50%.
first of all, you agree that the following is true:
"If I am in box 1, I will get a golden ball with a chance of 100%."
"If I am in box 2, I will get a golden ball with a chance of 0%."
so you say that the probability GIVEN that you are in box 1 is 100%, and GIVEN that you are in box 2 is 0%.
notice how you don't have one number but two numbers. but you want to have only one number because you cannot actually see what box you are in, right?
therefore, you have to combine these numbers in one way or another. you give both possibilities a certain "likelihood number" or "weight" and just calculate the probability from that. the sum of all weights must be 1 (because one of those scenarios has to happen, no matter what).
so, if you believe that you will be in box 1 every single time no matter what, then the weight for that will be 1, and the weight for box 2 will be 0. if you believed that, then the probability of getting a golden ball is 100%, and you can calculate it like this: p = 1 * 100% + 0 * 0% = 100%
if you believe that you will be in box 2, you'll do the opposite. p would then be: p = 0 * 100% + 1 * 0% = 0%.
most people who dont know about this type of problem intuitively say "hey, there are two scenarios, I know the chance is either 100% or 0%, lets just take the middle of that!", because our brains are somehow trained to give every possibility the same probability. but is that really true? in this case, its not.
the CRUCIAL part of this type of problem is:
you DONT know the probability of getting the golden ball after your first golden ball. the ONLY thing you can immediately tell is the probability of getting a golden ball GIVEN that you are in box 1 or 2 (but not the "average" probability).
to calculate a number from that, you need to weigh those two scenarios correctly. to do this, you must calculate the prob. of the two!
>>27659084
You'll be inclined to look at it intuitively and say "oh, you can only have chosen one of two boxes if the first ball is gold, thus the probability of the second ball being gold is 50%, since one of the two boxes has another gold ball."
The problem is the same as the one here; each box containing at least one gold ball is not equally likely to be chosen. If you choose a box at random and see that one of the balls is gold, you're more likely to have chosen the box with two gold balls than the one with a single gold ball.
Consider the six possibilities that can play out after you've randomly selected the box and looked at a single random ball:
1. Pick box 1, see the first (gold) ball
2. Pick box 1, see the second (gold) ball
3. Pick box 2, see the first (gold) ball
4. Pick box 2, see the second (silver) ball
5. Pick box 3, see the first (silver) ball
6. Pick box 3, see the second (silver) ball
We only care about situations where the first ball is gold, so we eliminate possibilities 4, 5, and 6. Now we see that there are two possibilities where we selected the first box and saw a gold ball, and only one possibility where we selected the second box and saw a gold ball. Thus, since we are more likely to have selected the box with two gold balls, the probability that the second ball is gold is greater than 1/2. It's actually 2/3.
>>27659227
>>27659237
Ok, thanks. I thought I was good at math, but I'm probably just good at following instructions the teacher gave me. From what I can gather is that probability isn't equal in all situations correct? Now Where would you use this knowledge? Like what should I look for in a question to apply this kind of thinking?
>>27659084
it's not really a formula. there is no "q formula". I just made up on the fly that q is "the probability of getting a golden ball, given that you first got a golden ball", and I hoped you'd understand (I study this shit and all my peers do to, I totally forgot that people normally wouldnt understand lel)
I wrote about the key to the "formula" in this post: >>27659227
basically, you can easily calculate the odds of getting another golden ball GIVEN that you are in either of the boxes. this is called a conditional probability.
to get to the unconditional probability, you have to find out how likely it is that any of the scenarios happen, and weigh the conditional probabilities with the probabilities of the scenario taking place.
by the way, this is called Bayes' rule.
-------------------------
the solution to the problem is the following:
probability of the second ball being gold if we are:
in box 1: probability is 100%
in box 2: probability is 0% (since we already took out the one golden ball from the box for us to end up in box 2, so the remaining ball is the silver one)
in box 3: probability is 0% (this is a technicality, we cannot even end up in box 3 but I want to make the general path to solving these problems clearer. there are no golden balls here, so the probability must be 0)
what are the probabilities of ending up in:
box 1: 2 out of a total 3 golden balls are in here => probability is 2/3
box 2: 1 out of a total 3 golden balls is in here => probability is 1/3
box 3: no golden balls here => probability is 0
lets compute this:
probability p = 2/3 * 100% + 1/3 * 0% + 0 * 0% = 66,67%
>>27659417
Don't sweat it, these problems are designed to go against intuition.
This board is fucking great. nothing like a bunch of early twenties people that dropped out of high school trying to argue over mathematics that they don't understand.
50%.
Both boxes contain a gold ball from first retrieval.
As one gold ball is removed, the possibility of getting gold or silver for just one more turn is 50/50. On retrieval of the second you can determine what colour each ball will be.
>>27659417
to be honest, this concept is very useful if you really make it your own. this must become intuitive for you, you have to understand it and also intuitively think the way the people who explained the problems did.
for that, really just look at some of these problems and solve them. it becomes really easy at a point, after a while you just "get" it and from that point on you start naturally thinking that way.
from that point on this will come in so fucking handy (imo) in your life. people get tricked so often by stupid as fuck fallacies, like "there are two possibilities - the chances must be 50%-50%" and the like.
seriously, before you learn and understand Bayes' rule, you wont be able to make educated and well-informed financial decisions. any resonsible adult should know this, but hardly anyone even gets that they might be wrong in the way that they think.
like, imagine that someone is offering you a certain service, like a insurance or the like. how the fuck are you even supposed to calculate whether or not the insurance is actually good for you? you have to think about the money you'll lose IF shit happens, and think about the likelihood THAT shit happens, and calculate the "expected" losses from that, and compare that with the loss you get by paying for the insurance.
often times you'll have to make guesses for the numbers, but it will still help to find out if an insurance policy might be sensible or just a scam.
so many people just dont know about this shit and they either get scammed or just dont get an insurance at all because their parents also didnt have one or they just cant handle the numbers.
>>27659084
66%?
>>27659534
why would you think anyone dropped out
>>27659534
what the fuck did you just say, you little bitch?
I'll have you know in doing my PhD in economics and I had some basic maths and statistics courses in my undergrad. just because you're too retarded to even stop watching cartoons for children does not mean that the entirety of r9k are dropouts or retards. some people in this thread might be uneducated in a specific field that they dont work/study in, which is completely normal, but anyone can understand this if they try.
you're literally a caricature of a typical robot - you're so fucking pathetic that you're not just a loser but also convinced yourself that getting educated is pointless because you couldnt deal with just how pathetic you are otherwise.
kys
>>27659612
correct
>>27657615
The intuition isn't too hard.
First, notice the second ball will be gold if and only if the first box was picked. That is, if the first box IS picked, the second ball will definitely be gold, and if the first box IS NOT picked, the second ball will definitely NOT be gold.
This means you're really asking for the probability that the first box was picked, given that the first ball was gold.
Now realize that before a box is chosen and a ball is drawn, each ball has an equal probability of being selected (1/20), since the boxes and balls are chosen at random.
With 11 gold balls in total, and 10 of those belonging to the first box, the odds are 10:1 the gold ball came from the first box, i.e. the probability is 10/11.
>>27659600
You're wrong. The fact that you drew a gold ball implies you have a better chance of drawing another gold ball.
not OP but I like the attention that this thread got, so I've got something else for you.
the numbers in the brackets are the so called "payoffs" of each player. the higher the number, the better. both players try to maneuver themselves into a scenario where they get the best payoff.
let the number be the time the prisoners have to spend in jail in years. the first number is the payoff of the first prisoner, the second number is the payoff of the second prisoner.
if you dont get how to read this:
if both players confess, they end up at payoffs of (-8, -8). if both lie, they get off with a conviction for some minor shit and only get (-1, -1).
what will happen in this game, if both players are trying to get the best payoff for themselves and dont care about the other guy?
>>27659488
>>27659603
Thanks guys, and all the other smart robots in this thread. I may not completely understand it but I'll run through this thread a few times and try to make it intuitive. You guys got any more puzzles like this if I feel up to the challenge?
>>27657615
50/50
too easy, all you retards are over thinking because of the presentation
who the fuck can solve this. this is some shit right here how do you even start. idk if im right or now but i thnk the answer is like 99%
>>27659839
is this assuming that they basically never talk to each other and both assume the other person has a 50% chance of either confessing or lying?
>>27659839
wait they should just both end up lying since you get either 1 year or 0 years in jail if you do that, so they'd both get 1 year each
is this meant to be a trick question
>>27658614
You fucking idiot it clearly says that the first one you pick IS gold so it doesn't fucking matter if there's silver balls in the box or not. The probability is 50 because it just depends on which box it was and that is 1/2
>>27658062
Gambler's fallacy detected
>>27657800
Isn't this question slightly different phrased? Because in this case you have already choosen a box and a ball. While in the wiki the question includes choosing a box and coin that will be a golden.
For an example, let's say you go to a casino and waits for the roulet to be black 3 times in a row. How big is the chance next roll will be red? (ignoring 0,00 etc)
>>27660072
they can't talk to each other. what would they have to assume in this scenario? would they really assume that what the other guy does is totally random?
>>27660116
if you yourself lie and the other guy confesses, you get 10 years in jail while the other is freed. your payoffs are either -10 or -1, not 0 or -1. check the table again.
SOLUTION
look at this problem from the perspective of any of the two prisoners.
imagine you are prisoner 2.
you can either confess or you can lie. which is better for you? well, lets analyze the situation: under which circumstances is it better to confess or to lie?
for now, assume that prisoner 1 confesses.
if you yourself confess as well, you get a payoff of -8. if the other guy confesses but you lie, you get -10. -10 is strictly worse than -8, so: IF THE OTHER GUY CONFESSES, YOU WANT TO CONFESS.
now, assume that prisoner 1 lies.
if you yourself confess, you get a payoff of 0. if you also lie, you get -1. 0 is strictly better than -1. this means: IF THE OTHER GUY LIES, YOU WANT TO CONFESS
NO MATTER WHAT THE FUCK THE OTHER GUY DOES, YOU CONFESS.
the tricky part of these types of problems is: your optimal action depends on what the other guy does. but in this case, confessing is ALWAYS optimal. in game theory, confessing would be called "dominant", and lying would be called "dominated". lying is a strictly dominated strategy, and no player will ever play a strictly dominated strategy. therefore, the only possible outcome is that both players confess.
>>27660245
jesus christ, stop posting about things you dont understand. do you even know what gambler's fallacy is? gambler's fallacy means that you think that tails becomes increasingly likely the longer a streak of heads becomes, for example, which is wrong - the probability of tails is always 1/2, the coin has no memory.
totally unrelated to this shit.
>>27660301
why'd you give the answer already baka
I just misread the payoffs
cool question though thanks
completely random question...
say that i am having a multiple choice test. one question has 4 possible answers, thus the chance of randomly getting the right answer is 25%
so what if i choose to disregard an answer? say that i randomly disregard an answer. my chance of not disregarding the correct one is 75%. then suddenly my chance of getting the right answer is 33%. i suspect that statistically there is no difference?
50%
Either it is gold or it is not
>>27659218
10/11ths, some other anons post rigorous formulas and shit if you want to see the work.
>>27660245
lmao
>>27660301
to add to this, this also depicts a social dilemma.
it would be better for the two to lie together and only get a minor sentence, but both of the players know that they'll be exploited by the other if they agree to doing this. if you'd agree to lying, the other guy would just sell you out to the cops and get back his freedom.
this is the case for all socially preferrable outcomes whenever at least one of the players can get a profit from deviating.
the best scenario would be to have no nuclear weapons on the planet because using them fucks up the planet. but if all countries agreed to remove all nuclear weapons, every country would have huge incentives to just not get rid of their nuclear weapons and be the only nuclear powerhouse left on earth that can now dictate politics or fuck shit up. one of the countries is bound to exploit the situation, and since every country knows about this, there will never be an agreement about removing all nuclear weapons.
>>27660364
sorry, I thought I made it clear that the post would have the soltuion when I wrote SOLUTION
>>27660444
yeah, the chance of getting the right answer is 33%, IF you disregarded a wrong answer.
the likelihood for that is
0,75 * 0,33 + 0,25 * 0 = 0,25
with a 75% chance, you have a 33% chance.
with a 25% chance, you have a 0% chance.
the probability is 25%.
>>27660477
oh whoops
I just kept reading automatically
seems easier than the OP question desu
