>>720122476 What algorithms do you think artificial intelligence use to learn?
They use probabilistic models, more often than not Bayesian models. This question is a basic Bayesian probability question. If you don't understand this you cannot build any probabilistic machine (control system, AI, recommendation system like Netflix, ranking system like Google.)
Enjoy your "man's" job in the woods like a savage while I continue running your shit.
>>720122406 Even if Box 1 has a million gold balls, if you choose a box with one gold ball, the only two options are either the next ball is or is not gold, since there are only two boxes, one with another gold ball to pick, and one without.
>>720122875 Lol, protect and provide. As though we live in a feudal society. Protect and provide means having money. Tech makes more money than your ancestral occupation. Enjoy your alcoholism and white trash.
>>720122761 It's actually 75%. It's impossible for you to be on the 3rd box. If you're on the first box (which you have a 50% chance of being in), then you have a 100% chance of pulling a gold ball. If you're in the 2nd box (again, 50% chance), you have a 50% chance of pulling a gold ball. So it's 75%.
It's impossible to answer because it was not specified if the ball was replaced or not.
For any half educated mathematician, this is the equivalent of setting a question where you forgot to put a pen in your hand before you wrote it, and also somehow accidentally getting your dick caught in the desk fan.
>>720123083 Simpleton white trash women enjoy only good dick. Even those enjoy financial security more than "good dick". Despite me doubting you are above average in bed given this puzzle is too demanding for you I am sure you will never achieve more than working a dead end blue collar job for the rest of your life.
>>720123033 >Learning how to read But you have to draw a gold ball for it to work, so you must eliminate the silver ball box, making it only 2 possible boxes you could have, one with silver, and one with gold
>>720123195 And to add to this, the people counting the 3rd box in their equations are the actual morons, even though you may know math, you don't have any critical thinking. You already know you must be in box 1 or 2, even though there is a 3rd box, you can safely ignore it, because there are no gold balls in it. There could be 500 boxes, but if there is 1 box with 1 gold and 1 silver, another with 2 golds, and 498 with 2 silvers each, you can safely ignore them all because you will NEVER be in any of them (you already pulled a golden ball, for god's sakes).
>>720122476 If you manage to achieve some moderate success in this life despite you're shit attitude, you can look forward to sucking actuary dick for the rest of your life because you don't actually know how anything fundamentally works.
>>720123494 You are right, then it's 50%. Either you're in the 0% probability box (the 2nd) or you're in the 100% probability box (the 1st). I was actually the moron, thanks for pointing it out (but so was everyone else).
>>720123522 It's incorrect. Since we know you pulled a gold ball, there's a higher chance you picked box 1 as opposed to box 2. Keep restarting the experiment where you pick a random box and then a random ball. In the cases where you picked a gold ball, 2/3 of the time you will have chosen box 1.
>>720123658 If you drew the gold ball out of the box containing silver and gold, how would you then draw another gold without being in box 1? Thus, you cannot pull a second gold ball out of the second box. If you drew one initially, that would be the only one there (the remaining ball would be silver).
So I've taken a box, in that box is a gold ball. Knowing this we can discard the silver box entirely. We now have 2 boxes left that can be the one I have chosen. In this box that I have picked, there remains either 1 gold ball or 1 silver ball as we have already taken 1 gold out. The odds are 1/2 (50%).
>>720123473 >>720123473 >That's the wrong idea. You got the correct answer by fluke. But, the way you justified it is incorrect. No, you're incorrect.
It's really quite simple. You draw a gold ball first. There are 3 gold balls, and there's an equal chance you drew any one of them. So the easiest way to visualize the solution is to look at the 3 gold balls independently. And for each of them ask, what color is the other ball in the box?
>>720123915 So there is a possibility that the second ball drawn will be silver. However, it is more likely that it will be a gold ball, since it is more likely that the first gold ball came from the box with two gold balls inside.
>>720124198 Except there's a 2 in 3 chance the gold ball you just drew came from the first box. So the first box counts twice, the second box counts once. (If you're counting boxes, which is a very misleading approaching.)
>>720124239 Yes, but my statement was exclusively referring to the 2nd box, and the fact that you can't draw two gold balls out of a box containing silver and gold. I never referenced the probability that the second one will be a gold ball. Incorrect assumption.
The following reasoning appears to give a probability of 1/2:
Originally, all three boxes were equally likely to be chosen. The chosen box cannot be box SS. So it must be box GG or GS. The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 1/2.
The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:
The probability that GG would produce a gold coin is 1. The probability that SS would produce a gold coin is 0. The probability that GS would produce a gold coin is 1/2.
>>720124654 That's the wrong intuition but the correct answer.
>>720124764 Yup and wiki specifically says in what you posted that, the reasoning leading to 1/2 is flawed. It then continues to state that the correct answer is 2/3.
The correct answer of 2 / 3 can also be obtained as follows:
Originally, all six coins were equally likely to be chosen. The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS. So it must come from the G drawer of box GS, or either drawer of box GG. The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is 2 / 3 . Alternatively, one can simply note that the chosen box has two coins of the same type 2 / 3 of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 2 / 3 of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".
>>720124845 It's not the same problem. Very similar but the way the conditional statement affects the problem space is different. https://en.wikipedia.org/wiki/Bertrand's_box_paradox
>>720122251 50% The probability is no longer between 3 boxes, but between a box that has 2 gold and a box that has 1 gold, one silver. As we know that we have drawn a gold ball, the box has equal chance of having either the gold or the silver ball left in the box for us to take next. The box of 2 silver balls doesn't play into this chance.
>>720125058 I misunderstood the when you said 2/3 coming from the first. A mistake often made with this problem is people do not consider it as 3 separate actions. Rather they just assume that if you ignore the third box. It is 2/3 because that's the number of gold balls left.
>>720125404 do you realise how stupid you sound? when the Monty Hall problem was first proposed, all the mathmaticians (people a lot smarter than you) disagreed with it. you're here spouting your obnoxious vitriol like you invented the fucking solution. you are nothing, you simply learned a problem to a solution. get over yourself.
>>720122251 The answer is 50% or 1/2. Since the ball you took was gold, you can discard the box with silver balls as it is impossible that is the box you chose. So now the two possible boxes are the gold and silver box, and the gold and gold box. Remove 1 gold from each box (because you took a gold out) and now u have either silver or gold. 50%.
>>720124483 I cant understand WHY G2 and G1 have to be unrelated instances. The question is what's the probability of picking another gold ball if you picked a gold one in the first place, so it wouldn't matter which one of the golden balls in the box with 2 of them you took.
>>720126046 nothing you say can change the fact that you are an abrasive niggard. you are here to prove to yourself that you are better than everybody, when in fact you've proven yourself to be lesser.
>>720122251 This shitstorm again. Since amerifags in particular but pretty much everyone is illitterate in maths and won't "believe" a correct proof, i'll put it in a way so that those who say 50% realise they're wrong. Say you have a box with 1 million gold balls, another one with 1 million silver balls, and a third one with 999999 silver balls and 1 gold ball. You extract a gold ball from one random box. Even a retarded dog can understand that the possibility is waaaay over 50% to have picked it from the all gold box (with these numbers it's about 99.999%). Now take one ball from every box, you'll still admit it doesn't change the fact that it's not 50%. Keep doing it. Eventually you'll have 3 balls per box, GGG, GSS, SSS. It is easy to see that the chance is way higher for the G ball to come from the first box. Take out one last ball, the result won't suddenly shift to 50%. This is not a rigorous proof, but it gives the idea of why you're wrong. 2/3 is the right answer. If this still doesn't convince you, try it yourself. Take 4 coins, mark 3 with an X. Put two of those with the X together in a glass and the other two (one marked, one not marked) in another. Mix them up and then pick a random coin from one. If it has the X, check the other one in the same glass. Do it 20-30ish times and you'll see that in about 2/3 of the attempts the X coin will come from a glass with he other marked coin in it. The reason is simply that you'll pick each glass 50% of the times, but every time you'll pick the unmarked coin, the extraction won't count an you'll just put it back in, because we only care about extraction of the marked coin.
>>720126443 People usually get mad when their idiocy is exposed. It's okay just go back to living in your bubble of ignorance.
>>720126599 If it doesn't say it then don't assume it. The question also doesn't state whether or not a car comes in and kills you while simultaneously Moses grants you 30 gold balls that fall directly on your face. Fucking idiot. Both of your answers are wrong regardless.
>>720126284 Maybe this will help: The initial conditions assume you drew a gold. You're drawing blindly, so you could end up with any of the three gold balls. There are three gold balls, so there's a 1 in 3 chance of drawing any one of them.
That's what matters. Because in the second step, you're just looking at the ball you already drew (1 of the 3 golds). It doesn't really matter which box it came from, just what the OTHER ball in the box is. And if you iterate through all the possibilities (all 3 gold balls), 2 out of 3 times the other ball is gold.
>>720126798 i never actually put an answer forward, you are falsely assuming i got the answer wrong. you claiming that the people who answered wrong are idiots makes you look like the king of idiots. you can dish out criticism, but you can't take any. you are a coward. as i said in an earlier post, the entire mathematical community disagreed with the monty hall problem when it was put forward. think about that for a second. now think how stupid you sound calling anonymous people on the internet "mongoloids" and "retards" for coming to the same conclusion as them. get over yourself, you learned some trivia, you are not smart for knowing this.
>>720127199 The boxes make this orders of magnitude harder. I was confused because I was thinking:
Since you already took a gold ball that means box 3 is eliminated. You now have 2 eligible boxes. Since one box has one gold ball and one silver and the other one has two gold balls I was calculating the probability of choosing the all-gold box instead.
I always count the two separate inverted instances as one (GB1 and GB2 in the all gold box should be differentiated and counted as different scenarios)
>>720128273 i've likely been here longer than you, kiddo. and in this case, context is important. when someone pretends to be better than someone on /b/, they need to be brought down to earth. you claim i was defending the people he was berating, when in fact i was dishing out criticism to someone who thinks he's king dick. check yourself before you claim im the cancer killing /b/ you one-dimentional faggot. i was being meta, you were being cancer.
>>720122979 Fuck off you faggot cunt, the first given fact is that you've grabbed one golden ball. There is no probability in this as of now. This does mean, the third box, the two silver ones, is excluded.
Now, probability plays a role. There's two possible boxes.
>>720128861 >i've likely been here longer than you, kiddo. stopped reading right there
look, idiot, you can go on screeching about people calling other people names as long as you want if it makes you feel better. in this case at least the people being called names are actually factually wrong
You believe there are three options: gold, gold, or silver.
That is not true, though. There are actually only two options, either your hand is in the box with two golds (and hence, the other ball in the box is gold), or your hand is in the silvergold box and the other ball is silver.
>>720129899 there aren't 5 balls after a gold ball pull because you know you aren't picking from the box with 2 silvers in it, thus you must be pulling from either the double gold or the silver/gold box - 3 balls remaining (2 of which are gold).
When you pick a box and remove a ball, you're committed to that box and that ball. So the ball you pick first completely determines which ball you'll pick second. There's no way to draw ball three then get ball two; there's no way to draw ball three then get anything but ball four.
So we know you drew ball 1, 2, or 3 (because the ball you drew was gold). This means you absolutely have to get ball 2, 1, or 4. Balls 2,1 and 4 are G, G, and S, which means you have a 2/3 chance of having picked a ball that guarantees you will get a second gold ball, and a 1/3 chance of having picked a ball that guarantees you get the silver one.
The second ball you get isn't even a probability thing: you already rolled for it when you picked the first ball.
>>720129962 Well that's Bertrand's Paradox (that it seems to be 1/2) but it doesn't actually work out like that mathematically or in practice.
From the wiki: >Since we know his probability is 2 / 3 , not 1 / 2 , we have an apparent paradox. It can be resolved only by recognizing how the combination of "observing a gold coin" with each possible box can only affect the probability that the box was GS or SS, but not GG.
>>720130103 Uh but that's not how the question is constructed. If you have to pick from the same box again you *know* it isn't the double silver box thus those 2 balls aren't being counted in the odds.
Anyway the right answer has been posted with both a math proof (which prolly isn't helpful to anyone who doesn't already know it but anyway) and in plain english with wiki sources so at this point we're just slinging shit and ima go back to fapping
>>720129583 It's not 50%, because the second ball isn't a roll.
If you pick one ball from a box, you are 100% guaranteed to get the other ball from the box. You will never get any ball other than the other ball in the box. You're not picking boxes at all; that's a decoy to lead you to the wrong answer. You're actually picking one of the six balls to pull out first, and then the ball you get second 100% depends on which ball you got first.
Therefore: If you pick the silver ball in the middle box or either gold ball in the first box, you're guaranteed to get a gold ball second. If you pick the gold ball in the middle box or either silver ball in the last box, you're guaranteed to get a silver ball second.
Because the ball you picked was gold, we know it's not the silver ball from the middle box or either ball from the last box. Two of the balls you could have picked will guarantee you a gold ball, and one of the balls you could have picked will guarantee you a silver ball.
This is simple you have a 50% chance of grabbing a good or silver ball at this point you are guaranteed to have either the two fold out the gold and silver because you've already grabbed one gold ball so you have a 50/50 chance that it's going to be the gold ball or the silver ball
In the problem of the coins, this makes sense. In this problem, so far it doesn't seem to make sense to me. Because I didn't understand the problem of the coins at the beginning, I'm cautious this time.
Anyhow: the question here is the opposite. It doesn't ask "what's the chance you draw a gold ball, and then another gold ball?". You KNOW you have drawn a gold ball already. Therefore, you could be in either box. So you either draw a gold ball, or not.
second pulling the first ball as gold eliminates the silver silver box so you have one of 2 boxes, a gold gold or a gold silver, thus you have a 1 in 2 chance (50% 50/50, however you faggots wanna put it)
Easiest way to understand these problems is to try them yourself. Just like the 3 door problem.
GG -- GS -- SS
If you reach into GS, there's a 50% chance of getting the gold. But, there's a 100% chance of getting a gold ball from GG. But since we know that we got a gold ball, that means it's more likely that you reached into GG than GS.
The options are as follows:
1. Reach into GG, get G. (Win) 2. Reach into GG, get other G. (Win) 3. Reach into GS, get G. (Lose)
3 options, each with 1/3 chance of happening based on the setup of the problem, right? So 2/3 times, you end up with a gold ball.
>>720129670 this would be true if the odds were based on balls (heh) rather than the box but they aren't (if they were all 3 would be in one box but you would also still have the silvers thus it would be 2/5 chances) the odds however are based on the boxes themselves and after pulling the gold and eliminating the double silver leaves you with, as explained, 2 chances of you have the gold/gold and the gold/silver box
>>720123473 Techincally, it's not that far off. The probability that you'd draw from the 1G1S box is ½. The probability you'd draw gold from the 2G box is 1. The probability of the 2S box is 0. That's literally what it says in the wiki.
>>720131081 Man I have to get a two-headed and a two-tailed coin. I'd make bank.
You guys are all getting confused because of the boxes. The boxes are mathematically identical to the coins: if you pick one ball (face) in one box (coin), you absolutely have to then get the other ball that was in the box (other side of the coin).
The problem can be rephrased as "You have three coins: one has two heads, one has two tails, and one is a normal coin. You pick one without looking, toss it without looking, and when you look down, you see the coin is heads-up. What are the chances that, when you turn it over, the other side will be a head?" and would be exactly the same problem.
>>720131464 No, it's 2/3 because there are six balls, and 2/6 of them are in the same box as another gold ball (this is where the "2/" comes from), and three of them are silver so you can't have picked them.
so 2/(6-3) = 2/3.
If you added another box with two silver balls, there'd be 8 balls total, 2 gold balls sharing with another gold ball, and 5 silver balls, so it would be 2/(8-5) = still 2/3
If you added another box with one ball each, there'd be 8 balls total, 2 gold ball sharing with another gold ball, and 4 silver balls, so 2/(8-4) = 2/4 = 1/2.
That's the mistake. The question, just like the coin question, has unnecessary information. The third box is pointless. It doesn't matter. I have a gold ball already. I know I'm in either box 1, or box 2. Now, if I have a gold ball in my hand, the remaining ball is either a gold ball (Box 1) or a silver ball (Box 2).
IF there were three balls in each box, three golds in one, two golds and one silver in another, the answer would be different.
>>720132389 1 2 3 GG GS SS pick 1 of 3 take G from 1 or 2 IF G was taken from 1, 1 now has XG IF G was taken from 2, 2 now has XS take second ball from 1 or 2 (whichever you were in before) it can either be G or S 1/2
>>720131507 You've already drawn a gold ball from one of the boxes, that means it can be two of the three boxes. Two choices = 50/50 This is how I was taught math. Am I missing something? Am I bad at english and not understanding?
>>720132170 you are retarded. There are only 4 balls. Just because you can still see the other two doesn't mean they are not exactly like a number that you scratched off an algebra question and then forgot about because your teacher told you to simplify the question.
>>720132700 The first criteria gets rid of box 3 entirely, so of course it is the same, still your 'logic' is wrong.
| GOLD GOLD | GOLD SILVER |
Now you take one out of these boxes, that one is a golden one, there now are two possible scenarios: 1: | GOLD EMPTY | GOLD SILVER | 2: | GOLD GOLD | EMTPY SILVER | Now, depending on which box you first picked the golden ball from, you either get a golden ball (Scenario 1) or you get a silver ball (Scenario 2).
>>720132487 For the first draw, it doesn't matter whether you drew from the box with the golds or the box with a gold or a silver. It just matters that you drew a gold. And for the second draw, the odds are 2:1 that the gold you already have came from the gold/gold box, so there's a 2/3 chance the other ball in the box is gold.
>>720132788 You're not picking anything in the second round, you just get the ball that comes with the ball you picked in the first round.
So however many balls are "left" for you to "pick from" in the second round doesn't matter, because there is no second round and you always, 100% of the time, unavoidably, get the roomie of the ball you picked in the first round.
>>720133023 But it said in OPs picture that you had picked up a golden ball. There are two boxes that contain at least one gold ball. SO it has to be one of them. Since you've taken out a gold ball. It means there is either a silver or a gold. That's 50% chance.
>>720132185 It's not a mistake, it is exactly the point. You either chose ball A from the gold/gold box, ball B from the gold/gold box, or ball A from the gold/silver box. That gives you a 2 out of 3 chance that you're in gold/gold.
>>720133758 Good idea, but you don't even have to actually write the program to see it comes out to 2/3. Your code would look something like this:
>pick a random number from 1-3 >if it's 3 (silver/silver), do nothing and pick again >if it's 1 (gold/gold) add one to the total goldBallPicked, add one to the total goldGoldPicked >if it's 2 (gold/silver), pick a random number from 1-2 >if it's 1 (gold), add a total to goldBallPicked, add a total to goldSilverPicked >if it's 2 (silver) do nothing and pick again
See how you add to goldGoldPicked 100% of the time you are in it, and to goldSilverPicked only 50% of the time you are in it? That right there is 2/3.
Right. I thought so too as well. But you have to remember that Gold ball got to you somehow. And that somehow is "you had 2 out of 3 possibilities to land in the box with 2 gold balls".
In other words, it is more likely you are in the gold box than not. The answer is still "the next ball is either gold or silver", which is true. But the real question is "what are the ODDS the next ball is gold?". And the odds are stacked in Gold's favor.
>>720138104 General dictionaries aren't very good with technical terms. Computers process in 0s and 1s (or offs and ons, or any other representation of a switch). Hex is a way to present those 0s and 1s in a more compact, more human-friendly way. But it's still a layer on top.
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