probability and statistics

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probability and statistics Anonymous Sat Dec 10 22:45:10 2022 No.15043983 [Reply] [Original]

why do basic probability questions trip up the rest of 4chan so easily, especially compared to questions from other mathematical fields? picrel triggered at least 10 different wrong answers

>>	Anonymous Sat Dec 10 22:46:13 2022 No.15043985 50%. You either find gold or you don't.

>>	Anonymous Sat Dec 10 22:47:30 2022 No.15043988 >>15043985 fpbp!

>>	Anonymous Sat Dec 10 22:48:15 2022 No.15043990 >>15043983 Midwits love doing little problem puzzles. Imagine wasting time even reading this.

>>	Anonymous Sat Dec 10 22:48:32 2022 No.15043991 >>15043983 >Hey guys, I have extremely low self-esteem so I'm going to virtue signal my distain for a group you also distain. Now you like me and we're friends, right? Absolutely pathetic.

>>	Anonymous Sat Dec 10 22:49:56 2022 No.15043993 >>15043983 Somehow you managed to combine Zeno's Paradox with the box of balls question. Congrats.

>>	Anonymous Sat Dec 10 23:08:50 2022 No.15044015 [math]P_G=0.3+\lim_\limits{n\to\infty}0.3\sum^n_{i=0}0.2^i[/math] I think it's something along these lines. I might be terribly wrong, however.

>>	Anonymous Sat Dec 10 23:09:53 2022 No.15044016 Actually is [math] P(1 gold bar) = 0.3 \sum_{n=0}^{\infty} (0.4)^n = \frac{0.3}{1-0.4}=\frac{0.3}{0.6} = 0.5 [/math] 50% chance.

>>	Anonymous Sat Dec 10 23:11:41 2022 No.15044022 >>15044015 >>15044016 See >>15043990 and kys.

>>	Anonymous Sat Dec 10 23:12:55 2022 No.15044024 >>15043983 x = 0.5+0.2x^2 (1+sqrt(1-0.4))/0.4 ~ 56.4%

>>	Anonymous Sat Dec 10 23:13:02 2022 No.15044025 >>15044022 >t. seething brainlet who doesn't even attempt to solve the problem because he can't conceive of the answer.

>>	Anonymous Sat Dec 10 23:14:53 2022 No.15044029 >>15043983 x = 0.5+0.2x^2 x = (1+sqrt(1-0.4))/0.4 1-x ~ 43.6%

>>	Anonymous Sat Dec 10 23:20:06 2022 No.15044037 >>15044029 >>15043983 Exact answer is (sqrt(15)-3)/2 if you need it.

Anonymous Sat Dec 10 23:23:00 2022 No.15044040
File: 324 KB, 882x889, NJW.png [View same] [iqdb] [saucenao] [google]

This problem has severe logical issues since it requires the possibility that there are an infinite number of chests contained in an infinite number of chests, all of which are large enough to contain at least one gold bullion. That is completely absurd.

Anonymous Sat Dec 10 23:24:00 2022 No.15044044

>>15043990
>>15044022
Most of the world's field's medalists and elite researchers participated in competition math and placed at the IMO. This is some hot copium that reeks of insecurity and projection.

Also, this question takes about 15 seconds if you aren't a midwit

>>	Anonymous Sat Dec 10 23:35:18 2022 No.15044061 >>15044040 Based physicalist.

Anonymous Sat Dec 10 23:37:26 2022 No.15044067

>>15044040
>This problem has severe logical issues since it requires the possibility that there are an infinite number of chests contained in an infinite number of chests
No it doesn't. There is only a defined probability for each finite odd number of chests. And it says "gold bar" without specifying a size.

Anonymous Sat Dec 10 23:41:15 2022 No.15044074
File: 663 KB, 1421x957, NJWW.png [View same] [iqdb] [saucenao] [google]

>>15044067
Then potentially the smallest gold "bar" is exactly one gold atom. The smallest a chest can be therefore is just large enough to exactly fit one gold atom. That is assuming that a chest that size can even exist in the first place. Apply the previous criticism and this remains an ill-conceived illogical problem.

Anonymous Sun Dec 11 00:01:06 2022 No.15044112

>>15043983
so when the problem states "30% chance of the chest containing a gold bar", is it also including the gold bars possibly contained in the subsequent chests? Also does "30% chance of a gold bar" mean exactly one or at least one. I might be overthinking this puzzle

>>	Anonymous Sun Dec 11 00:06:24 2022 No.15044118 >>15044112 Wait you're onto something. The answer has to be 30%, plain and simple

Anonymous Sun Dec 11 02:52:51 2022 No.15044465

>>15043983
The chance of getting at least 1 gold bar in one chest is x
>x = 0.3 + 0.2((1-x)^2)
(1-x)^2 is from both chests that come out not having any gold.
>x = 0.3 + 0.2 - 0.4x + 0.2x^2
>0.2 x^2 -1.4x +0.5 = 0
Two solutions discard x being outside the range 0-1.
x is around 0.3775

>>	Anonymous Sun Dec 11 02:56:36 2022 No.15044473 >>15044465 Error in the setup. Should be >x = 0.3 + 0.2(1-(1-x)^2)

>>	Anonymous Sun Dec 11 03:04:08 2022 No.15044486 Then approximate it with that assumption Retard.

>>	Anonymous Sun Dec 11 03:05:00 2022 No.15044487 >>15044473 Ah thanks for that I was wondering what the distinction was between me and >>15044029. So i inserted the probability of not having any gold rather than not not having gold (aka at least 1).

>>	Anonymous Sun Dec 11 03:11:24 2022 No.15044495 >>15044465 >>15044473 set up is easier to use p(g>0) = 1 - p(g = 0). let x = p(g=0) x = 0.5 + 0.2x^2

>>	Anonymous Sun Dec 11 03:16:00 2022 No.15044502 >>15043983 72.2222(repeating)% chance.

>>	Anonymous Sun Dec 11 03:20:10 2022 No.15044510 >>15044502 Woops read the question wrong. Thought was 50% of gold bar. It's roughly 32%. 30% + 30.2.3 + 1.8.2.3 + etc... Not a formal formulae as wrote down the successive percentages hamfistedly, but it works for an estimation.

>>	Anonymous Sun Dec 11 04:26:21 2022 No.15044653 >>15044025 I have better things to occupy my mind with. >>15044044 >field's medalists and elite researchers participated in competition So midwits.

>>	Anonymous Sun Dec 11 04:51:01 2022 No.15044687 >>15043983 got 47,(6)%

>>	Anonymous Sun Dec 11 05:03:48 2022 No.15044710 3/8

>>	Anonymous Sun Dec 11 10:43:30 2022 No.15045215 I got 38.4% (0.3) + (0.70.2(0.3+0.3)) = 0.384

>>	Anonymous Sun Dec 11 10:47:59 2022 No.15045222 >>15044044 >appeal to authority total midwit post

>>	Anonymous Sun Dec 11 15:09:18 2022 No.15045497 >>15043983 37.5%

>>	Anonymous Sun Dec 11 20:01:48 2022 No.15046237 >>15043983 1/2.

>>	Anonymous Sun Dec 11 20:19:45 2022 No.15046288 File: 14 KB, 518x422, monte.png [View same] [iqdb] [saucenao] [google] >>15044029 Nice work. Didn't realize you could model it with a quadratic like that.

>>	Anonymous Mon Dec 12 14:55:10 2022 No.15048066 >>15044118 Yes, obviously. If something contains x but is itself contained in y x is also contained in y. The answer is 50%, since you have infinite attempts to get a gold bar, but only need one (not infinite).

>>	Anonymous Mon Dec 12 14:56:18 2022 No.15048068 >>15045215 the new chests can have chest themselves

>>	Anonymous Mon Dec 12 15:05:15 2022 No.15048082 >>15046288 why is 0.8 30%

>>	Anonymous Mon Dec 12 15:13:46 2022 No.15048093 >>15048082 nta but it's an else if, so it only runs when 1/2<p<4/5

Anonymous Mon Dec 12 15:13:55 2022 No.15048094

50%

I cannot trust anything /pol/ says since they lie incessantly and are always making fake screen caps and edited photos.

Thus, I need to go with maximal entropy, in this case a uniform distribution, the likelihood of gold or nothing, my two discreet outcomes, being equal.

After opening such chests several times I can continue updating and come to the correct conclusion.

>>	Anonymous Mon Dec 12 15:17:33 2022 No.15048100 People getting more than 30% should flip the problem around such that: P(gold) = 50% P(nothing) = 30% P(two chests) = 20%

>>	Anonymous Mon Dec 12 15:19:20 2022 No.15048104 >>15043990 military-grade cope

>>	Anonymous Mon Dec 12 15:23:25 2022 No.15048112 >>15048100 It's 50%, the idea is every chest is: .5 you lose .3 you win .2 play again and since you only have to win once, "play again" is just "you win"

>>	Anonymous Mon Dec 12 15:26:38 2022 No.15048116 >>15044653 >I have better things to occupy my mind with. Like thinking about the anons in this thread? cope harder and kys

>>	Anonymous Mon Dec 12 15:28:54 2022 No.15048123 >>15048112 Not every play again results in a win. .5 you lose .3 you win --‐------------- .1 you lose .06 you win .04 play again ----------‐----------- .02 you lose .012 you win 0.008 again

>>	Anonymous Mon Dec 12 15:37:27 2022 No.15048139 >>15048123 there are two chests in the try again option, the chance for loss would be 0.25

Anonymous Mon Dec 12 15:37:34 2022 No.15048140

Selection continues until you get one of two discrete outcomes.

Ignore the 0.2 resample. This is the same as saying, "when you go to take a marble from your Bernoulli Jar, you slip and fall and don't take a marble, so you try to take a random marble again."

Chance of actually sampling the distribution is 0.8. Chance of nothing = .5/.8, or 62.5%. Chance of gold = .3/.8 or 37.5%.

There is no need for taking limits, just think of it in terms of entropy in a message where "try again" is not an outcome.

Unless you don't open the chests when you find them, in which case it is .5/.3/.2 obviously.

>>15048112
Play again isn't "you win," play again is "select randomly from the same exact distribution again."

Selection stops if either gold or empty comes up.

>>	Anonymous Mon Dec 12 15:39:59 2022 No.15048146 >>15048139 Do you get to open both chests? I figured you would pick one.

>>	Anonymous Mon Dec 12 15:41:07 2022 No.15048147 >>15048146 I guess you would open both, but I like >>15048140 explanation the most, and here it doesnt even matter

Anonymous Mon Dec 12 16:42:52 2022 No.15048256

>>15048147
No if you open both it's 43.8%. To see why, consider that you have a 50% chance of empty and then if you get two more chests you get a 25% chance that both of those are empty. .2 x .25 = 5%.

The chance of not gold is at least 55%. >>15044029 has the most eloquent answer.

However, >>15048140 is correct if you have to pick one chest of the two.

>>	Anonymous Mon Dec 12 18:41:18 2022 No.15048518 File: 722 KB, 853x1000, 1669321368483158.png [View same] [iqdb] [saucenao] [google] >>15044040

Anonymous Mon Dec 12 19:15:41 2022 No.15048598

>>15044040
Nonsense. A gold atom is 3 × 10^-10 M in diameter so you just have to start with a 1.5 meter large chest (larger and it is no longer really a chest anymore) and realize each inner chest can at most each be half the size of the prior chest, and go from there. This allows you to go 31 levels down before your chest is too small for an atom of gold. EZ PZ.

>>	Anonymous Mon Dec 12 19:57:57 2022 No.15048709 >>15044022 .5 is clearly wrong. After a single iteration, the answer is bounded by 1 - .5 - .2.5.5 <.4375

>>	Anonymous Mon Dec 12 20:16:20 2022 No.15048736 >>15048139 You're right, I missed that.

>>	Anonymous Mon Dec 12 20:17:04 2022 No.15048738 >>15044044 winning a fields medal is like winning best furry suit at fur-con. Nobody cares. Everyone knows mathematicians only want the fields medal so they can eat the chocolate inside anyway, just like a fag eating the ass of a nigger in a fur suit

Anonymous Mon Dec 12 20:21:01 2022 No.15048740

>>15043983
No, the official pol IQ test is the cookie monster math quiz

>If it takes one hour to bake a batch of cookies and Cookie Monster has 15 ovens working 24 hours a day every day for 5 years, would Cookie Monster be able to bake 6 million cookies?

Anonymous Mon Dec 12 21:29:31 2022 No.15048894

>>15048709
It's not "+ .2 × .5 × .5", it's a branching process with a 20% chance of reproduction and a declining chance of extinction in any given iteration as the number of boxes increases, i.e. 50% at n=1, 25% at n = 2, etc.

To see why your explanation is wrong, recall that you don't not get gold if just one of the two new boxes is empty, you only don't get gold if both are empty (extinction). You're looking for the probability that, given iteration 1 has two boxes, that the two boxes are both empty. That's .25 × .2, or 55% likely not to get gold considering just the first three possible boxes. But obviously each of the first two boxes to have been reproduced has a 20% chance to spawn two new boxes, and so you have a 4% chance to spawn four new boxes and a 36% chance that at least one set of new boxes gets spawned given you have two boxes.If you end up with four boxes, you have a 59% chance of spawning at least two more boxes.

Probability of a loss is 50% at the first iteration, + a 5% chance that you got a box first and then had two empty boxes. So the odds must be worse than 45%.

There is obviously an infinite number of paths a decision tree on the problem could create, although even getting past 16 simultaneously existing boxes from the same parent is very unlikely.

>>	Anonymous Mon Dec 12 21:49:23 2022 No.15048937 >>15048894 Sorry my math was right, but the number at the end was wrong. we're in agreement .2 * .5 * .5 (chance all branches terminate =.05

Anonymous Tue Dec 13 01:34:39 2022 No.15049479

>>15048740
>If it takes one hour to bake a batch of cookies
It doesn't take that long to bake cookies. Also, you can put multiple batches of cookies in the same oven if you don't care about creating perfect cookies.

>would Cookie Monster be able to bake 6 million cookies?
Why would Cookie Monster need to bake 6 million cookies in only a few small ovens? Cookie Monster could build a large fire pit to bake his cookies in, amd there are surely other desserts he could make simultaneously to the cookies if he needed 6 million desserts.

>>	Satoshi Nakamoto Tue Dec 13 01:36:49 2022 No.15049486 >>15043983 1/2 either you get it or you dont.

>>	Anonymous Tue Dec 13 02:43:58 2022 No.15049628 >>15044016 basically the first post said in a fancy way>>15043985

Anonymous Tue Dec 13 03:45:22 2022 No.15049754

Chance of not getting gold = 50%
+ if box on roll 1 (.5*.5*.2) = 5%
+ if at least one double box is rolled in either of the second two boxes (.20 chance of box in round 1, in round 2 .36 * .20 = 7.2% chance of a third and fourth box). 1.8% chance both these are empty.

Chance not gold: 56.8% not even considering the potential to get four empties after rolling two boxes on stage 2.

Chance of winning.
30%
+51% chance of winning on second round if box was rolled * 20% = 10.2%
As stated before, the chance of at least one box at round two is 7.2%. in 7.2% of interactions you will reach these third and fourth boxes and at least one will have gold at 51% odds. Total chances here 7.62% * 51% = 3.67%

Total chances to win = 43.87%

Total % chance of outcomes 100.67%, and I didn't even get into the possibility of box/box in round two,

Anonymous Tue Dec 13 10:56:39 2022 No.15050175

>>15049479
I think anon should look into what cookie baking experts say about asinine methods like
>you can put multiple batches of cookies in the same oven if you don't care about creating perfect cookies
and
>Cookie Monster could build a large fire pit to bake his cookies in

>>	Anonymous Tue Dec 13 12:10:19 2022 No.15050243 >>15048140 It appeared to me that this was the right and astute answer, but upon reading question closely there is nothing about how many treasure chests may be opened at any level. Therefore answer is simply 100%.

Anonymous Tue Dec 13 12:42:08 2022 No.15050286

>>15043983
Suppose the probabilities are a, b, and c with a+b+c=1, and that c gives the likelihood of there being d treasure chests each with the same odds as the original one. The probability of use not getting a gold bar is the solution to x = b+c*x^d. Plugging in the values a=0.3, b=0.5, c=0.2, d=2, we obtain the answer of roughly 1-0.5635.

>>	Anonymous Tue Dec 13 13:17:24 2022 No.15050339 >>15050175 I have, that's how I know your question is based on false cookie premises.

Anonymous Tue Dec 13 13:25:52 2022 No.15050350

You can't just calculate frequency. The results aren't iid, and you can't just calculate the frequency with which gold or empty occurs. Getting gold first in any nth box is dependant on not having received gold earlier. Your trials stop on one gold.

It's a problem that will trick you if you go with basic frequentist instincts, which it seems everyone has.

>>15049754
Sees the problem if you look at both sides.

Conditional probability works fine for Box 1 and Boxes 2 and 3.

Your chance of gold is 30%, then if Boxes 2 and 3 spawn (20%) of the time you get Boxes 4 and 5 31% of the time, 7.4% total.

But whether or not boxes 4 and 5 can contain the first gold is dependant on whether at least one box was rolled on Boxes 2 and 3 AND if NO gold's were rolled on Boxes 2 and 3. If 2 Boxes gets rolled on 2 and 3, then any good to this point is logically exclude and there is no dependence on prior trials.

Taking frequency won't work and the conditional probabilities quickly reproduce to levels that are going to be unsolvable. Just consider every possible dependence and condition for all the cases in which 25 boxes are opened, which can happen via many different paths.

Anonymous Tue Dec 13 13:30:04 2022 No.15050355

>>15050350
To see the dependence vs merely conditionality, consider going several levels down from one box, rolling more boxes in a probability tree. You now have 10 boxes to open. If your chance of finding gold for the first time The chance of getting all boxes and empty on these 10 boxes?

No. If gold is found on any higher levels, then the chances for a first gold on any lower one drop to zero. It's not conditional probability of gold in any X stop alone but rather co dictionary probability dependant on past observations.

Anonymous Tue Dec 13 13:31:42 2022 No.15050359

>>15050339
>I have
Alright, this is how I know you're arguing in bad faith. First you're making stuff up, not even knowing the official claim, but the excuses you made don't even hold up under layman understanding, let alone under expert scrutiny. You don't even need to deep dive, you can ask any honest mortician experienced in cremation how it works and what's possible. Either way, I won't engage with you until you at least show some good faith.

>>	Anonymous Tue Dec 13 14:06:27 2022 No.15050397 >>15050350 It's a trivial problem since the chests are all the same. See >>15044029

Anonymous Tue Dec 13 14:29:46 2022 No.15050424

>>15050359
>Alright, this is how I know you're arguing in bad faith.
Ironically you're displaying bad faith by assuming without any reasoning I'm arguing in bad faith based on an innocuous claim.

>First you're making stuff up, not even knowing the official claim
This is some grade A projection. You didn't even know the official claim is not that 6 million cookies were baked in ovens.

>the excuses you made don't even hold up under layman understanding
I can't wait to hear your brilliant layman understanding.

>You don't even need to deep dive, you can ask any honest mortician experienced in cremation
I have no idea what you're talking about. We're talking about cookies. Baking cookies one batch at a time in a manner respectful to the cookies is not the same as baking cookies en masse. Why do you keep ignoring this?

Anonymous Tue Dec 13 14:38:18 2022 No.15050434

>>15044029
Why does this answer work?

It looks about right, but how does it deal with the fact that, for every box rolled, all possible boxes have to come up empty to terminate the game, while the probability of any box being rolled actually mattering has is conditional on no golds being rolled in any prior selections.

Anonymous Tue Dec 13 14:42:48 2022 No.15050442

>>15050434
>It looks about right, but how does it deal with the fact that, for every box rolled, all possible boxes have to come up empty to terminate the game
x is the chance of a box containing no gold bars, including in any sub-box. The key is realizing this is the same for every box, and that the complement of x is the answer to the question.

Anonymous Tue Dec 13 14:50:23 2022 No.15050453

>>15046288
So, in this case it doesn't seem like the simulation stops when any one chest = 1.

If you ran a simulation where the first box has boxes, then chest B1 has boxes and chest B2 has gold, it will still calculate the outcome of chests C1 and C2. Perhaps these are both gold (9% of such cases) or one is gold (30%) of cases.

These outcomes should not be added to your simulation. They are not instances of having found gold first.

This simulation seems like it would show you the frequency for gold versus empty, which I believe >>15044029 answers as well.

But the question: "what is the frequency in which iterations of this game terminate without a single gold being selected," is not the question "what will be the relative frequency of gold versus empty."

I would think this would require some sort of branching survival models. It's a variant on a Gallon Watson process where you are looking at the percentage chance to terminate without the condition G occuring.

You can reformulate it like this. You have one bacteria. It dies 50% of the time. 20% of the time it reproduces itself. 30% of the time it mutates into an immortal, indestructible bacteria. In what percentage of branching survival processes does G occur before extinction. The answer is not the frequency of G vs E as additional Gs in cases that include at least one G do not matter.

>>	Anonymous Tue Dec 13 15:01:05 2022 No.15050476 >>15050453 The question is what is the chance of finding at least one gold, which is the same as the chance of not finding zero gold.

>>	Anonymous Tue Dec 13 15:11:20 2022 No.15050492 >>15050424 >pilpul post nose

>>	Anonymous Tue Dec 13 15:31:05 2022 No.15050531 >>15050492 Not an argument. Thanks for admitting Cookie Monster could have baked his cookies.

>>	Aether king of sci Tue Dec 13 15:33:23 2022 No.15050533 63.725%

>>	Anonymous Tue Dec 13 15:33:34 2022 No.15050534 This is how /v/ tackled the problem >>>/v/621038980 Thoughts?

Anonymous Tue Dec 13 15:37:12 2022 No.15050543

This thread is fucking cracking me up!

You cannot recurse probabilities because each step of the recursion has to equal one for the probability to be valid.
Since the probability contains the same distribution inside it, either the the sum of original probability does not equal one, or its recursion does not equal one.

Trying to force the limit by making the gold smaller is hilarious, and shows you are both trying to mix continuous and discrete in the same distribution AND equivocating the limit of the recursion with the physical gold bar.

It cracks me up when people get lost in the math and forget the problem the math is only pointing to.

Thank you all for a truly schizo thread!

inb4 "Oh no its not!" Seriously, my sides can't take anymore.

>>	Anonymous Tue Dec 13 15:39:41 2022 No.15050554 >>15050534 Not a single right answer, shocking.

Anonymous Tue Dec 13 15:43:21 2022 No.15050562

>>15050476
Right, but in that code you might roll chest first.

Then you have two chests. Say one is also chest, spawning a fourth and fifth chest and one is gold.

That is still going to roll a random number for chests 4 and 5. If they are both gold, you will have 3 chests = 1 in your numerator.

It should go sequentially and terminate calculating chests whenever a single chest = 1 regardless of how many chests have spawned.

Anonymous Tue Dec 13 15:46:35 2022 No.15050570

>>15050543
This is pure gibberish. Of course the probabilities of each mutually distinct event sum to 1. 0.5+0.2+0.3 = 1. The probability the question is asking for is not mutually distinct for each chest, since the event of a chest containing gold includes the event of of having a sub- chest that contains gold. So there is no issue using the simple recursion x = 0.5+0.2x^2 where x is the chance of a chest containing no gold.

Chance of a chat containing gold anywhere = (sqrt(15)-3)/2

Anonymous Tue Dec 13 15:48:13 2022 No.15050575

Ok /v/ is gaslighting me I think
The way I see it you have a 50% chance to open nothing, then a 20% chance to open two new chests. Since these new chests may also come up empty, 25% of the time your run ends there.
So for one step in your chances are already less than 45% to get gold. Repeat to infinity to whittle it down to some number between 40% and 45%

>>	Aether king of sci Tue Dec 13 15:50:18 2022 No.15050582 >>15050575 No because herp herp herp herp herp herp herp herp herp herp herp

Anonymous Tue Dec 13 15:53:45 2022 No.15050587

>>15043983
>/pol/ IQ test
I could easily derive the infinite series for this and simplify it to get a number, but it's not worth my time:
A real chest cannot contain two chests exactly like it. A fucking two-year-old could figure that out. The problem is completely stupid.
What do I win?

Anonymous Tue Dec 13 15:58:13 2022 No.15050590

>>15050562
As far as I can tell, it is doing the same thing by only counting a run where gold is found anywhere in the range of chests as one result. Maybe I'm reading it wrong, but if not it shouldn't be getting the correct answer calculated here >>15044029. That is not calculating anything about the number of chests inside the first chest, it's just finding the probability of a chest containing gold anywhere in it.

>>	Anonymous Tue Dec 13 15:59:14 2022 No.15050595 >>15050587 >I could easily derive the infinite series for this and simplify it to get a number, but it's not worth my time: Then you would fail the IQ test. See >>15044029

>>	Aether king of sci Tue Dec 13 16:07:05 2022 No.15050609 File: 215 KB, 680x695, 1668894586953890.png [View same] [iqdb] [saucenao] [google] >>15050595 It's over 50% so that answer is wrong mcshithead

>>	Aether king of sci Tue Dec 13 16:08:24 2022 No.15050613 >>15050609 The 20% is a boost to a normal 20% 30% success Plus booster 20% 63.725

>>	Knight of sci Tue Dec 13 16:23:04 2022 No.15050642 >>15050613 Lol aether they thought it was 43%

>>	Aether king of sci Tue Dec 13 16:24:22 2022 No.15050645 File: 1.65 MB, 2860x3750, Hatsune.Miku.full.3844628.jpg [View same] [iqdb] [saucenao] [google] >>15050642 I know loyal knight of sci that's because most people who post here just pretend to know math

Anonymous Tue Dec 13 16:43:44 2022 No.15050674

>>15050570
Jesus Christ. You can always tell those that think they know the math from those that know the math.

The probability is NOT one.

30% + 50% + 20% is supposed to equal one.
But the 20% is equal to one itself by being equal to the 30% + the 50 % + the 20% where THAT 20% equals one, and so on and so on

The probability of the entire thing is infinite because it adds 100% in the place of the 20% all the way down. This equivocates the probabilities with itself. This is why probabilities, like space itself, cannot be iterated upon themselves without creating more space than is there when you start.

And you cannot resolve that infinity with limits because there is no one view that can watch it converge from all views.

This IS Zeno's paradox of Achilles and the Tortoise in that from the one view from Achilles equals one, vut is dependent upon the other view of the Tortoise that also equals one, but NO POSSIBLE VIEW can resolve to be both EXCEPT the view from that is is a paradox.
At least in the Tortoise and Achilles it is a race that we as the observers can see what is happening from our conservation.

This is an oscillating paradox plain and simple because there are two stories with their own narrators that cannot be reduced to one narrator because each narrative view is dependent upon a part from the other that cannot be seen from itself.

Anonymous Tue Dec 13 16:57:52 2022 No.15050700

>>15050674
(cont)
Look. Watch how the story changes in the middle of its making by attaching a narrator to each iteration of the story itself. The story makes itself as its narrator and can only see what it sees.

Now the Achilles story narrator can only see where the Tortoise is then closes his eyes and runs. The Tortoise narrator sees the ground and moves. When the Achilles narrator reaches where the tortoise was, the Tortoise has moved, resetting the Achilles narrator to that point, then closes his eyes. The Tortoise, meanwhile, moves again. This goes on for infinity, and Achilles never reaches the Tortoise.
Notice the iteration in the making.

This is what is going on here!. You stand in the unity of the first probability, close your eyes then go to the 20% that itself is a unity of the of the first probability distribution, except for its 20%, and so you go to that unity. And so one and so on. Unity is NEVER achieved for the whole problem. Therefore, there is NO probability distribution whatsoever.

Anonymous Tue Dec 13 16:58:57 2022 No.15050703

>>15050590
That seems correct yes. But the code seems wrong and comes to a very similar answer which makes me suspicious that the prior answer is also looking at the frequency of empty to gold. But the frequency of faliure is conditional on the actual number of boxes generated in a permutation.

Try graphing it with a conventional probability tree and you quickly realize you need a forest for all possible iterations.

I'm also suspicious because when very similar problems come up in science the solutions involve branching processes, unless it's the condition that any gold = victory that reduces the complexity of the problem. It seems to make it more difficult for graphing though.

Anonymous Tue Dec 13 17:08:34 2022 No.15050720

>>15050700
(cont)
In the case of Achilles and the Tortoise you have an outside story with its own narrator that see the race. That story and its narrator can contradict the story that Achilles never reaches the Tortoise. That forms another paradox in that the outside observer story narrator can see one story of Achilles passing the Tortoise while Achilles cannot.
In the case of this fake probability problem, you cannot form an outside story with the story of math because there is no math story that exists, except the math story that this is not a conserved or convergent iteration, or except for the story that this is a paradox and no solution exists.

But I can see that people aren't getting this because you are all still posting about nonsense, trying to solve a paradox with more paradox.

Good luck to you all. Make sure you repeat this nonsense of shrinking gold when I give you this as an employment test. It really cuts down on the incompetence in my company.

>>	Anonymous Tue Dec 13 17:33:56 2022 No.15050786 >>15050609 The chest has a 50% chance of containing nothing at all, so how can it be greater than 50%? lmao, is negative IQ possible?

Anonymous Tue Dec 13 17:38:37 2022 No.15050800

>>15050674
>The probability is NOT one.
I didn't say it is. Are you illiterate?

>30% + 50% + 20% is supposed to equal one.
That's what I just said. Fucking hell.

>But the 20% is equal to one itself by being equal to the 30% + the 50 % + the 20% where THAT 20% equals one, and so on and so on
Gibberish. Take your meds.

>>	Aether king of sci Tue Dec 13 17:39:17 2022 No.15050802 >>15050786 I read the question wrong, whoops stupid me. Still, you're retarded.

>>	Aether king of sci Tue Dec 13 17:40:34 2022 No.15050804 >>15050802 The answer is 37%

Anonymous Tue Dec 13 17:45:13 2022 No.15050808

>>15050703
The code is correct, you're misinterpreting it. It doesn't count chests, it only indexes them up to a large number to determine if there's gold in the entire chest. If there is, that counts as one result. The number of results divided by the number of attempts is the correct probability.

>which makes me suspicious that the prior answer is also looking at the frequency of empty to gold.
It's the frequency of initial chests containing gold anywhere within it to the total number of initial chests, the probability of a chest containing at least one gold.

>Try graphing it with a conventional probability tree and you quickly realize you need a forest for all possible iterations.
Right, that's why instead of looking at each possibility it's much easier to use a recursion.

>>	Anonymous Tue Dec 13 17:47:30 2022 No.15050814 >>15050802 No, you're the retard who can't even read a question correctly, let alone answer it correctly, let alone answer it in a way that shows mastery of recursive probabilities.

>>	Anonymous Tue Dec 13 17:48:30 2022 No.15050817 >>15050804 Wrong. It's (sqrt(15)-3)/2 ~ 43.6%

>>	Aether king of sci Tue Dec 13 17:48:44 2022 No.15050820 >>15050808 Look nerd, no-one reads your hazy posts. You don't even understand yourself. It's like reading pure confusion and self-peity leading to depression. Honestly noose up and kys nerdlet

>>	Aether king of sci Tue Dec 13 17:49:49 2022 No.15050824 >>15050817 I WAS TROLLING.

>>	Aether king of sci Tue Dec 13 17:51:13 2022 No.15050825 >>15050824 I don't practice doing math with my hands I do it in my head. I know what the answer is, I just haven't the trained hand eye coordination. Lick my balls handmath nerds

>>	Anonymous Tue Dec 13 17:58:50 2022 No.15050856 >>15043983 But in the two treasure chests there are four chests and so on and so on for ever. Creating a black hole which fucks your mother

Anonymous Tue Dec 13 17:59:06 2022 No.15050857

just find the probability of never finding a gold bar, lets call that P(x), and the probability of finding at least 1 is 1 - P(x).

P(x) is the sum of all the times you get nothing. So, P(x) = 0.5 + 0.5*0.2 + 0.5*0.2^2 + 0.5*0.2^3... etc. this is the sum of an infinite geometric series with r < 1. P(x) = 0.5/(1-0.2) = 0.625

1 - P(x) = 0.375

>>	Anonymous Tue Dec 13 18:01:23 2022 No.15050867 >>15050820 Amazing projection, coming from the schizo who couldn't even understand the problem.

>>	Anonymous Tue Dec 13 18:02:24 2022 No.15050872 >>15050824 You were schizoing

>>	Aether king of sci Tue Dec 13 18:03:06 2022 No.15050875 File: 17 KB, 512x512, d352fdd03aae7423cbbfeb5261be9c32.png [View same] [iqdb] [saucenao] [google] >>15050867

>>	Anonymous Tue Dec 13 18:03:11 2022 No.15050876 >>15043983 37.5% 1-Sum[Power[0.2,k],{k,0,inf}]*0.5

>>	Anonymous Tue Dec 13 18:03:25 2022 No.15050878 >>15050857 There are two chests 20% of the time. You're calculating as if there is only one chest.

>>	Anonymous Tue Dec 13 18:06:22 2022 No.15050885 >>15043983 I crunched for 1min, Around 39%.

>>	Anonymous Tue Dec 13 18:07:29 2022 No.15050891 >>15050876 Wrong.

Anonymous Tue Dec 13 18:36:19 2022 No.15050963

>>15050878
Oh shit I misread the question. If it has two chests then every scenario where you get nothing goes like this. Layer 1: open first box its nothing. Layer 2: open first box its 2 chests, open both chests it nothing. Etc.

This can be represented as Sum of 0.2^(2^n -1) * 0.5^(2^n) as n approaches infinity. This is 0.5505. 1- 0.5505 is 0.4495.

44.95% chance of getting a gold bar.

Anonymous Tue Dec 13 18:48:25 2022 No.15050995
File: 50 KB, 901x255, Not IID.jpg [View same] [iqdb] [saucenao] [google]

The variables are not independent and identically distributed. The existence of later samples is dependent on the values taken by earlier samples and so the frequency is not .20 - .30 .- .50.

Anonymous Tue Dec 13 18:48:45 2022 No.15050997

>>15050963
Just to add on because I don't think I explained this very clearly. Adding the probability of every scenario where you get nothing goes a little like this:

Scenario 1: Open box, its nothing 0.5 percent chance

Scenario 2: Open box, its two chests (0.2), open both chests and its nothing (0.5*0.5), overall 0.2*0.5*0.5

Scenario 3: open 1st chest, its two chests (0.2), open the 2 chests, both reveal two different chests, (0.2*0.2), there are 4 chests, each chest opens to nothing, (0.5*0.5*0.5*0.5) overall probability (0.2^3*0.5^4)

essentially for each layer of opened chests, you first open 1 then 2 then 4 then 8 and obtain double the chests. This is basically 1 + 2 + 4 + 8 ... + 2^n. This is 2^n - 1, starting with n = 0. Then we open the amount of chests present at that layer, that being, 1, 2, 4, 8, aka 2^n. so the probability of obtaining nothing at layer n is 0.2^(2^n - 1)*0.5^(2^n)

adding them up you get 0.5505
and 1 - 0.5505 is 0.4495

>>	Anonymous Tue Dec 13 18:57:29 2022 No.15051016 >>15050963 Still wrong. It's 43.65%

Anonymous Tue Dec 13 19:17:18 2022 No.15051068

>>15050963
>>15050997
You are excluding scenarios like
>open box, it's two chest, open both chest one has two chest (both which are empty) the other is empty.
Which requires 3, 0.5 chances on each empty chest, something that 0.5^(2^n) won't reproduce.
This is why your probability of no gold bars is smaller hence why your chance of at least 1 gold bar is larger than >>15051016

Aether king of sci Tue Dec 13 20:11:09 2022 No.15051209

The answer is 63.725

This is because of the boosting 20%

It was 50% of correct UNTIL the logic happened 3+ times in 1 sitting.

You can't tell me that all adds up to 50% nothing if it has that booster effect.

/Thread
/Checkmated
/Midwits debate me now

>>	Aether king of sci Tue Dec 13 20:12:59 2022 No.15051217 >>15051209 IT GIVES YOU EXACTLY 50% OF AN EXTRA CHANCE SO THERES LITERALLY 25% CHANCE OF NOTHING LICK MY CRUSTY HOLE MIDWITS

Anonymous Tue Dec 13 21:16:31 2022 No.15051387

I couldnt remember calculus and i dont code but i made this and get 44.3 to 44.4 percent chance of finding gold

import random;

trial = 0
success = 0
failure = 0

while trial < 1000000:
gold = 0
chests = 1
while chests > 0:
chests = chests - 1
roll = random.randint(0,100)
if roll <= 30:
gold = gold + 1
if 30 < roll <= 50:
chests = chests + 2
if gold > 0:
success = success + 1
if gold == 0:
failure = failure + 1
trial = trial + 1

print (success/1000000)
print (failure/1000000)

>>	Anonymous Tue Dec 13 21:21:03 2022 No.15051399 >>15043983 50%

Anonymous Tue Dec 13 21:32:47 2022 No.15051430

>>15051387
I don't know for loops in Python well enough to vet >>15046288, but it is very weird that you two would be that far off.

I can follow yours easier.

I did it with DAX variables but fucked it up so that it only reported success if the first chest had gold or if there was more than one gold in the chests total. That ended up at 35-6% over 150,000 runs, but I had to pack up for the day before I realized how I had fucked it up.

Anonymous Tue Dec 13 21:55:48 2022 No.15051472

>>15051387
Careful with the boundaries
roll is presumably in the range of 0-99 hence 100 entries.
>roll <= 30
Is slightly over 30% chance as it will accept 0-30 hence 31% chance of triggering.
>if 30 < roll <= 50:
Is fine for a double chest, but it makes the chance of no gold be the values 51-99 which is 49 values so 49% chance

it should be
>roll < 30
>if 30<= roll < 50

Otherwise, it looks fine (ignoring minor details on efficiency) assuming the loops end where i think they do
>IE While chest > 0 ends after chest = chest+2

>>	Anonymous Tue Dec 13 22:44:27 2022 No.15051552 >>15043983 It's an infinite series right? >>15044029 Is this correct?

>>	Anonymous Tue Dec 13 22:48:14 2022 No.15051560 >>15051552 No. See >>15048598 It's quite a finite problem. Don't be fooled by mathematical headcanon.

>>	Anonymous Tue Dec 13 23:13:59 2022 No.15051612 >>15043983 /pol/ is like eight smart kids who occasionally do stuff of no real import for laughs... and hundreds of thousands of rednecks, deadbeats, schizos, boomers, and larptards who want to feel special and high-iq while being completely useless.

>>	Anonymous Tue Dec 13 23:16:59 2022 No.15051624 >>15043983 0% because da joos already took it

Anonymous Tue Dec 13 23:21:04 2022 No.15051638

>>15051552
No. The probability of n+1 chests gets infinitely smaller as n increases.

Now if the chest contained a 25% chance of empty, 15% chance of gold, and 60% chance of more chests... then you would tend towards infinite population growth of chests.

>>	Anonymous Tue Dec 13 23:40:21 2022 No.15051694 >>15051612 kinda true tbqh when I browsed there it was really a constant reminder that everybody was either 16 or they were 30 and had the worst ability to act like a functioning adult Still some funny people though

>>	Anonymous Tue Dec 13 23:56:58 2022 No.15051745 >>15043983 [math]1 +2/\sqrt 3[/math] so more than 100%

>>	Anonymous Wed Dec 14 00:09:18 2022 No.15051784 File: 545 KB, 736x1550, wat.png [View same] [iqdb] [saucenao] [google] did some digging, what was his end game...?

>>	Anonymous Wed Dec 14 00:11:19 2022 No.15051791 >>15051784 Sliding threads he doesn't like off the page.

>>	Anonymous Wed Dec 14 00:17:53 2022 No.15051808 >>15051209 >>15051217 Schizo is off his meds

>>	Anonymous Wed Dec 14 00:27:58 2022 No.15051850 30+20x where x is less than 1 and greater than 0 such that 50+20x + 30+20x + c = 100

>>	Anonymous Wed Dec 14 00:43:12 2022 No.15051893 >>15051472 import random; x = 0 y = 0 while x < 100: y = 0.3 + 0.2*y x = x + 1 print (y) oh maybe it is 37.5

Anonymous Wed Dec 14 00:44:31 2022 No.15051900

>>15043983
The chance of me getting a gold bar is independent of the probabilities relating to the chest, because I don't even know where said chest is or how to open it. So I'm gonna go with 0%. Because I don't intend to get a gold bar in the first place.

Anonymous Wed Dec 14 01:06:50 2022 No.15051943

>>15051893
I'm assuming you are trying to reach the value via iteration?

There's only 1 y there when opening up two boxes would mean two instances.
If y is meant to be the chance of gold in a box, then the chance of gold in 2 boxes (IE what multiplies 0.2) is
>(1-(1-y)^2)

37.5 is the answer if each box can only produce one box not two under 20% chance.

>>	Anonymous Wed Dec 14 01:36:17 2022 No.15052015 # find total chance of nothing x = 0 y = 0 while x < 1000: x = x + 1 y = 0.5 + 0.2(yy) print (y) turns out to be 1- 0.564 43.6%

>>	Anonymous Wed Dec 14 01:38:38 2022 No.15052021 >>15051209 Just two more boosters.

>>	Anonymous Wed Dec 14 02:08:33 2022 No.15052085 >>15044029 This equation equals ~.56 https://www.wolframalpha.com/input?i=x+%3D+0.5+%2B+.2*x%5E2

>>	Anonymous Wed Dec 14 02:15:43 2022 No.15052098 >>15052085 Anon isn't working out x. They are working out 1-x x is the chance of no gold which is ~.56

>>	Anonymous Wed Dec 14 02:17:01 2022 No.15052100 >>15052098 Oh shit I'm retarded

>>	Anonymous Wed Dec 14 02:26:30 2022 No.15052113 >>15051943 >>15051472 Thank you

Anonymous Wed Dec 14 02:34:20 2022 No.15052127

>>15043983
This can be modeled as a discrete time markov chain.

There are 4 key states:
>1) p_S, Initial state, 1 box, this is purely transient
>2) p_G, Gold state, at least 1 gold bar found, this is absorbing
>3) p_E, Empty state, all unopened boxes end up empty, absorbing
>4) p_i, Multiple box state, state p_i represents the state with 2i boxes

Transitional probabilities:
[eqn]p_{ij} = \left\{\begin{matrix}
\begin{pmatrix}
2i \\j\end{pmatrix}*.3^j*.5^{2i-j}: j\le 2i \\
0:\text{otherwise}
\end{matrix} \right\}\\\\
p_{iG} = 1-.8^{2i}\\\\
p_{iE} = .5^i[/eqn]

Any state in the multiple box state can return to a state with less boxes than it but cannot go directly to a state with more than double it's current amount of boxes.
In theory this is solvable but I'm not sure how. It will take advanced algebra.

>>	Anonymous Wed Dec 14 04:32:36 2022 No.15052328 >>15052127 You're way overthinking it. Let x = the chance of a chest containing no gold anywhere in it. Then x = 0.5+0 2x^2 x = (5-sqrt(15))/2 The chance the chest contains at least one gold bar is 1-x = (sqrt(15)-3)/2 ~ 43.65%

>>	Anonymous Wed Dec 14 05:02:35 2022 No.15052362 File: 1.59 MB, 498x213, pete-holmes-batman.gif [View same] [iqdb] [saucenao] [google] >>15043985 How could anybody possibly determine that? We need to flip like, 50 coins. And then 50 more. And then analyze them, with SCIENCE.

>>	Anonymous Wed Dec 14 12:58:22 2022 No.15053015 >chest has 30% chance of having a gold bar >whats the chance of have a gold bar 30% The 20% chance for two chests is a red herring because it's already defined that each chest has 30% chance to contain a gold bar.

>>	Anonymous Wed Dec 14 13:28:59 2022 No.15053047 >>15043983 30%, any extension beyond the initial opening is gambler's fallacy.

>>	Anonymous Wed Dec 14 15:50:29 2022 No.15053302 >>15053015 >>15053047 You guys are fucking retards. The problem says "at least one gold bar". You have to add the probability of two gold bars, three goldbars, and so on. You can't even read; you're not gonna make it.

Anonymous Wed Dec 14 16:03:51 2022 No.15053323

>>15053047
The fallacy in gambler's fallacy refers to the human belief that one can double their money to recoup losses at the same rate they lose.

The gambler's fallacy is a simple discrete time Markov chain, with easily calculatable long run probabilities.

>>	Anonymous Wed Dec 14 16:21:32 2022 No.15053350 >>15044016 That's definitely wrong. U have a 50% base chance to not get gold. Then atleast a extra 5% chance for getting 2 more chests and both having nothing. That's already 45% chance to get gold at most

Anonymous Thu Dec 15 01:10:13 2022 No.15054553
File: 12 KB, 474x322, OIP (1).jpg [View same] [iqdb] [saucenao] [google]

>>15052127
Shifting probabilities would be very difficult. You would need something like a Galton-Watson Process.

But since this is static you can use the quadratic formula.

You can visualize this intuitively. Envision a 10 by 10 square with 50 squares shaded red that have E (Empty) in them and 30 squares shaded gold with G in them.

Then the perimeter of the 20 "Chest" squares in the bottom right has a darker boundary of its own 40x50 subsection of the grid. This subsection is bisected into two down the middle, which unfortunately for visualizing makes for fractional squares, but oh well.

Now shade 50% of the two smaller squares red and 30% gold. Then turn the other 20% into a new square and bisect it.

As you can see, each smaller square has identical shares of red and gold, but the size occupied by each gets smaller and smaller at each level. This is your sample space. It contains itself. This is exactly what the quadratic formula is for, to deal with copies within copies. With the golden mean, you can get infinite copies of the same rectangle with the same ratios. Neat stuff.

Anonymous Thu Dec 15 18:14:40 2022 No.15055973

>>15043983
There's a really cool book called "The Drunkard's Walk" which explains why the human brain is so bad at comprehending the fundamentals of probability and why the absolute laws of probability actually contradict the way our brains have evolutionarily designed to tackle problems.

>>	Anonymous Thu Dec 15 18:18:37 2022 No.15055987 >>15043990 This thread and all the wasted activity on a stupid image proves it.

Anonymous Thu Dec 15 20:24:37 2022 No.15056271

>>15051387
>>15046288
It's 49,9% in my simulation though.

#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int chest()
{
int barnobar = 0;
int pos = rand() % 9 + 1;
if(pos <=3) barnobar = 1;
else if(pos > 3 && pos <=5)
{
barnobar = chest();
if(barnobar == 0) barnobar = chest();
}
return barnobar;
}

int main() {
srand(time(NULL));
int found = 0, iter = 0;
for(int i = 0; i < 1000000; i++)
{
if(chest() == 1) found++;
iter++;
}
double answer = (double) (found*100)/iter;
printf("%f", answer);
return 0;
}

>>	Anonymous Thu Dec 15 23:32:44 2022 No.15056632 >>15056271 >int pos = rand() % 9 + 1 There are 9 values pos can take here: >1,2,3,4,5,6,7,8,9 Your chances are: >pos<=3 -> (1,2,3) is 3/9 >pos >3 and pos <=5 -> (4,5) is 2/9 >pos > 5 -> (6,7,8,9) is 4/9 So your probability of getting gold is inflated.

>>	Anonymous Fri Dec 16 00:41:26 2022 No.15056757 >>15051016 >>15052328 I did 10 million trials in matlab and these answers are correct. I have no idea why.

>>	Anonymous Fri Dec 16 00:42:31 2022 No.15056759 File: 158 KB, 902x856, Treasure.png [View same] [iqdb] [saucenao] [google] >>15056757

Anonymous Fri Dec 16 01:43:04 2022 No.15056831

>>15056757
Well the approach of >>15052328
Is thinking it's easier to find out the chance of no gold
Then subtract that away from the total probability for the chance of at least 1 gold.

So to that end, they let x be the chance a chest has no gold.
This is the chance the chest is empty (0.5) or that the two chests inside both have no gold (0.2*x*x).
Crucially "No gold" doesn't mean there aren't chests inside but whatever number of chests are inside have no gold.
We then have an equation for x, as the two probabilities are separate events in the probability space, and the only way the original chests has no gold.
>x = 0.5 + 0.2*x*x
This solved for x gives you the chance of no gold and hence the chance of at least 1 gold as
>1-x

>>	Anonymous Fri Dec 16 01:50:02 2022 No.15056839 >>15043983 32% (I rounded up)

>>	Anonymous Fri Dec 16 02:23:43 2022 No.15056891 >>15043983 What if you get 2 chest in one of those 2 chests?

>>	Anonymous Fri Dec 16 02:26:47 2022 No.15056895 >>15056891 Then you continue with the recursion. You can also get two more chests in each of the two chests simultaneously.

>>	Anonymous Fri Dec 16 02:33:45 2022 No.15056907 >>15056891 Then you keep going as you have a better of getting at least 1 gold.

>>	Anonymous Fri Dec 16 14:51:30 2022 No.15057832 >>15056632 Fuck. You are right. It's 43,65% in that case.

>>	Anonymous Fri Dec 16 15:02:21 2022 No.15057841 File: 832 KB, 900x900, 1588535696988.png [View same] [iqdb] [saucenao] [google] >>15043983 Ah, yes. The famous Goldbar conjecture.

>>	Anonymous Fri Dec 16 15:42:07 2022 No.15057906 >>15043983 How do I get my potential gold bar? Can I 3D print it?

>>	Anonymous Fri Dec 16 17:08:41 2022 No.15058056 >>15043983 43.6492% [math] 3/10 + (2/10) (1-(1-x)^2) == x [/math] [math] x = {\sqrt{15} - 3 \over 2} [/math]

>>	Anonymous Fri Dec 16 20:32:23 2022 No.15058382 >>15050492 If pilpul is "having correct arguments that follow correct reasoning" then pilpul is good

>>	Anonymous Sat Dec 17 05:34:16 2022 No.15059222 File: 67 KB, 1458x475, the_solution.jpg [View same] [iqdb] [saucenao] [google] you guys are morons. just do a monte carlo simulation. also it's lim x->inf 3/10 + 2/10 * (1 - (7/10) ^ x)) = 50%

>>	Anonymous Sat Dec 17 05:35:12 2022 No.15059224 >>15059222 Dumb and wrong faggot. People already posted this.

>>	Anonymous Sat Dec 17 05:37:16 2022 No.15059227 >>15059224 fuck you retard it’s the right answer you’re probably a fucking engineer

>>	Anonymous Sat Dec 17 05:38:41 2022 No.15059230 >>15059227 >probably an engineer You obviously don’t understand probability.

>>	Anonymous Sat Dec 17 05:40:41 2022 No.15059233 >>15059230 sorry. there is a 100% chance you're fucking gay.

>>	Anonymous Sat Dec 17 05:42:11 2022 No.15059234 >>15059233 No shit newfag, we’re all fucking gay here.

>>	Anonymous Sat Dec 17 05:43:51 2022 No.15059237 >>15059222 you play until you win or lose. there is a 50% chance of losing on every chest you open. what's the complement of 50%? that's your win probability. go to school faggots

>>	Anonymous Sat Dec 17 05:45:50 2022 No.15059239 >>15059234 > be me >new to 4chan > get answer right > call someone gay this is the life

>>	Anonymous Sat Dec 17 05:56:54 2022 No.15059253 >>15043983 Hey here's a better question: What are the chances of you continually opening chests and acquiring gold bars indefinitely? Do the chances remain a consistent 50%?

>>	Anonymous Sat Dec 17 06:02:50 2022 No.15059262 >>15059253 what? as if you couldn’t have made this any more retarded than it already was… the chances of opening chests and getting infinite gold bars is effectively 0

>>	Anonymous Sat Dec 17 06:04:44 2022 No.15059267 >>15059262 What about a quantifiable number like 10 in a row?

Anonymous Sat Dec 17 06:34:00 2022 No.15059299

>>15059267
what is the probability of getting 10 gold bars in a row? you’d have to start by opening a chest that contains 2 chests, one with gold AND one with 2 more chests, repeating this process until we have 10 gold bars. the probability that we get a chest with a gold bar and another chest is .3 * .2 * 2 = .12, as there are 2 possibilities where we get that outcome in this scenario. repeat 10 times: .12 ^ 10 = 0.000000000619174. not gonna happen

>>	Anonymous Sat Dec 17 07:41:22 2022 No.15059348 >>15043985 >>15044016 Horseshoe kek

>>	Anonymous Sat Dec 17 07:42:19 2022 No.15059352 >>15043990 collectively several hundred hours of smooth brains squandering their time in this thread.

>>	Anonymous Sat Dec 17 11:32:32 2022 No.15059600 >>15059222 >calls everyone morons >does a simulation instead of showing understanding of probability >falls to even do the simulation correctly lmao

>>	Anonymous Sat Dec 17 11:34:38 2022 No.15059602 >>15059237 There's a 50% chance of losing on each chest plus the chance of getting two more chests that also lose. So your chance of losing overall has to be greater than 50%. Dumby.

>>	Anonymous Sat Dec 17 14:42:53 2022 No.15059814 >>15043983 I gave it a try, 37,75%

>>	Anonymous Sat Dec 17 14:46:51 2022 No.15059825 >>15053350 It's always 50%, either one thing happens or another thing happens

Anonymous Sat Dec 17 16:48:17 2022 No.15060010

>>15059253
>What are the chances of you continually opening chests and acquiring gold bars indefinitely?
Not really something that will happen.
For this to be the case we need to be able to expect that on average each chest will give more chests than they use up.
But in the question for every 10 chests opened has around 4 chests.
This isn't stable at all and so we don't expect to be opening chests for long at any point.

For an opposite example consider a case every chest has
>0.9 chance of 2 chests, 0.1 chance otherwise
Let's consider all ways you can stop opening chests
Opening a chest either introduces 1 more chest to the available ones at 0.9 chance or removes 1 from the available ones at 0.1 chance.
As we start at 1 chest, if we introduce x more chests, we need to remove x+1 chests
So we are looking for probabilities of the form
> 0.9^(x) * 0.1^(x+1)
The difficult part is finding all the ways we can order these probabilities for any x.
If we consider a tree diagram denoting which chests contained what we can see a binomial distribution over x-1 for x >= 1 (x= 0 is just 0.1)
As it is a binary tree there already exist a means of getting the number of unique binary trees of a given size, which is the Catalan number.
Given we have (x+1) end points which are leaf nodes, x is also the number of internal nodes, which we need for the Catalan number.
The catalan number for a given x is then
>(2x choose x)/(x+1)
So the probability that a tree ends is given by summing all the ways a tree can end
>[math]0.1 + \sum_{x=1}^{\infty} \frac{(2x \choose x)0.9^{x}0.1^{x+1}}{x+1} = \frac{1}{9}[/math]
Hence there is a more than likely chance you will open chests forever

By contrast let's consider the original problem in this same manner, which is changing 0.9 to 0.2 and 0.1 to 0.8
>[math]0.8 + \sum_{x=1}^{\infty} \frac{(2x \choose x)0.2^{x}0.8^{x+1}}{x+1} = 1[/math]
So it's garaunteed to end at some point

Anonymous Sat Dec 17 17:08:54 2022 No.15060047

Everybody is wrong.

Answer this:
If the probability of the last 20% can't be known because its distribution has 20% that is dependent on the distribution that has 20% dependent on the distribution and so on and so forth, how do you know that the chance of the original distribution of no gold is 50% and the chance of gold is 30%?

The answer is you cannot.

This is a paradox. The distribution used to determine the distribution itself cannot be known until the distribution of the unknown part is known.

There is no math that can overcome this paradox. All of you who are trying to find a limit to the iteration have either become stupid and and have equivocated the gold with the probability space and are trying to reduce the mass of the gold, or have a priori set an arbitrary sum of 100% that cannot be determined.

Please, for the sake of the world, take the time to puzzle this out. We don't need more stupid smart people.

>>	Anonymous Sat Dec 17 17:18:37 2022 No.15060062 >>15059600 no, it’s right, stupid. run it yourself and follow the results

Anonymous Sat Dec 17 17:33:57 2022 No.15060093

>>15060047
So to clarify: The original distribution you used to define the missing part is itself dependent upon the missing part to be solved in order to give you the original distribution used as parameters to the iteration. So even if you can find a limit to the iteration, what you used to determine the iteration itself is dependent upon its own answer.

You have committed the fallacy of circularity in that the answer you can’t yet know that the problem is asking for is given in the assumption that the probabilities will add to 100% that lets you set the probability of nothing equal to 50% and the probability of gold equal to 30% even though you can’t know that until the last probability of 20% can be known, which it can’t until it is iterated. So it is not yet 20%, but what you are using as the parameters of the iteration are the very numbers of percentages that have not been determined yet. It only equals 20% in relation to the 50% and the 30% you assumed but have not yet figured out. It is only by putting this arbitrary limit on the total distribution by setting the 50% and the 30% that cannot yet be known, that you can the iterated unknown part can converge.

Anonymous Sat Dec 17 17:37:40 2022 No.15060101

>>15060093
(cont)
So all of you trying to find a convergence have circularly defined the parameters of that convergence by the answer of 50% and 30% that you use in the iteration.

Again, this is a paradox that you are trying to solve with the fallacy of circularity.

It has no solution.

>>	Anonymous Sat Dec 17 17:58:56 2022 No.15060124 >>15060047 There is no way that youre right bro. It shouldnt be that hard to at least approximate it very accurately. Its something like 40% chance just based on my high IQ intuition alone.

>>	Anonymous Sat Dec 17 18:16:31 2022 No.15060150 >>15060124 Hence the problem with intuition based on paradoxical thinking that cannot see its own fallacies.

Anonymous Sat Dec 17 18:18:26 2022 No.15060153

>>15060047
>If the probability of the last 20% can't be known
But it can?
It's specifically defined as two instances of the initial distribution.
All the simple solutions use this to their advantage to create a quadratic equation and solve for the initial distribution.

If you don't trust any answers run some code yourself and see the probability be approximately what is expected from the prior mentioned approach.

Anonymous Sat Dec 17 18:26:41 2022 No.15060172

>>15060153
>It's specifically defined as two instances of the initial distribution

The initial distribution cannot be known until the last part is known. Assuming it -- or specifically defining them -- is the circular fallacy that proves it is the original distribution by now allowing the iteration to converge.

This is why I said in the original post >>15060047
to PLEASE take the time to puzzle this out.

Coding a fallacy will still produce the fallacy. It has been awhile since I coded but at least in Fortran, the program is not going to give you an error or call you an idiot for making a circular fallacy.

If you want to code it and see the error, leave the percentages of no gold, gold, and the iteration as variables themselves, run the iteration, and watch the program crash into an infinite loop.

>>	Aether king of sci Sat Dec 17 18:27:57 2022 No.15060173 30% the double chest chance is gamblers fallacy

>>	Anonymous Sat Dec 17 18:28:32 2022 No.15060175 >>15060172 >Assuming it -- or specifically defining them -- is the circular fallacy The problem came with that definition in hand, there is no assumption being done outside of the problem specification.

>>	Anonymous Sat Dec 17 18:35:49 2022 No.15060196 File: 40 KB, 573x509, Pepe Chuckle.jpg [View same] [iqdb] [saucenao] [google] >>15043983 >"/pol/ IQ test" >posts it in /sci/ Fucking retard.

Anonymous Sat Dec 17 18:37:27 2022 No.15060200

>>15060175
There most certainly is hidden in the definition of a probability distribution. The probabilities have to add to unity, or there is no distribution.

This is why there are no variables in probability, only Random Variables that all have the property that their sum equals one or unity.

You can't know the percentage of gold, no gold or whether there is a chest, because you can't know they add to one until the contents of the chest are known.

Simply giving the probabilities of 50% and 30% is makes them NOT random variables because we can't know that that is the distribution of the final whole until the final whole exists. Forcing them forces the last iteration of the last 20% to converge, making the choice of 50% and 30% and 20% the conclusion that is proven in the end.

This is why this is a paradox. Solving it with a fallacy does not solve the paradox; it only commits a logical fallacy.

>>	Anonymous Sat Dec 17 18:47:07 2022 No.15060220 >>15060062 >run the flawed simulation so you can get the same flawed result Oh no... it's retarded. See >>15044029

Anonymous Sat Dec 17 18:56:21 2022 No.15060243

>>15060200
> because you can't know they add to one until the contents of the chest are known
The way the problem is constructed pretty much means they must add up to one.
Within each chest, the 3 probabilities add up to 1.
Then for every 2 chests within a chest, the probabilities are built up from multiplying the two instances but no matter how you distribute sums of 1 their products will still be one.

Anonymous Sat Dec 17 19:02:54 2022 No.15060258

>>15060093
>You have committed the fallacy of circularity in that the answer you can’t yet know that the problem is asking for is given in the assumption that the probabilities will add to 100% that lets you set the probability of nothing equal to 50% and the probability of gold equal to 30%
That's what the problem says. It's not a 30% chance of gold being anywhere in the chest, it's a 30% chance of gold being in the chest and no more chests.

Anonymous Sat Dec 17 19:11:10 2022 No.15060271

>>15060243
>The way the problem is constructed pretty much means they must add up to one

Only because the givens are chosen to a priori add up to one, which is the conclusion the problem is trying to find out. Hence circularity.

This is a puzzle. It is meant to be hard. This is actually a difficult problem in that it is a paradox AND the fallacy of circularity:

You can't know the distribution until the problem is solved (the paradox), but by putting the conclusion as the premise or given, the iteration now can converge (the circular fallacy).
But it converges to the conclusion that was given (circularity) but could not have been known (paradox).

Anonymous Sat Dec 17 19:14:33 2022 No.15060275

>>15060258
Now you have changed the problem.

Now you have equivocated the distribution of all the states with the reduced set of outcomes states that could occur of simply being gold, nothing or a chest, which itself would have the distribution of 1/3 1/3 1/3 contradicting the distribution of 30% 50% and 20%.

Anonymous Sat Dec 17 19:20:58 2022 No.15060284

>>15060271
>which is the conclusion the problem is trying to find out.
No, the problem is to find the chance of an event not to establish a probability distribution.
By that logic any question that establishes a probability distribution and asks you to figure out the chance of an event is circular.
>>15060275
That's what the problem has been from the start.
>30% chance of containing a gold bar
Nothing about that implies that it is a chance of opening up more chests inside of it to contain a gold bar.
>which itself would have the distribution of 1/3 1/3 1/3
No, those events don't occur with equal probability why would you assume that they do?
We need further information, which the question provides, to weigh each event.
If I ran a dice with 6 sides but renumbered them to only have 1, 2 and 3 there's no reason to assume they would occur equally as I very well could have renumbered every other number into a 3 biasing it in the distribution.

>>	Anonymous Sat Dec 17 19:29:18 2022 No.15060301 >>15043983 36%

>>	Anonymous Sat Dec 17 19:40:34 2022 No.15060322 >>15043983 37.5%

Anonymous Sat Dec 17 19:42:22 2022 No.15060328

>>15060284
Still don't get it...
>No, the problem is to find the chance of an event not to establish a probability distribution.
Say that to yourself again and see the circularity. The probability distribution IS the chance of the event. If the only chance of gold was a set percent then problem would be over, but the problem says that there could be a chest that also had gold, and so right from the start any given chance of gold that is given has to be wrong.

Now to your dice.
Use your dice example of a dice with three numbers on in but instead of knowing how many sides there are, what if you don't know how many sides the dice has?
If I started off saying saying the probability of anything, I would be wrong because I can't know the probability until I know how many sides there are.
But if I stated as a given that three of the sides were numbered and had a 10% chance each, and then further stated that there were other sides that had the same distribution as the dice, even though I could not know how many sides the dice had, then, by the definition of a Random Variable, you could conclude that all of those sides must have a probability of 70% ONLY because dice have a finite number of sides, and ONLY because I GAVE you the probabilities to begin with that the conclusion would now prove!
However, since you can’t know how many sides there are, you could not have known that the three sides would have a probability of 10% each; therefore you could not know that the remaining sides added to 70% OR that they were finite!
I would have forced that probability by the given that it was a finite die AND by my givens of the percentages, and circularly proven it by the result.

Anonymous Sat Dec 17 19:50:17 2022 No.15060348
File: 40 KB, 600x450, 1614543158934.jpg [View same] [iqdb] [saucenao] [google]

>>15043983
>>15043985
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>>15060328
The fact that this simple riddle has exploded into this rambunctious debate just goes to show you're not as intelligent as you think you are, just because you like pop science.

Anonymous Sat Dec 17 19:57:25 2022 No.15060369

>>15060328
>The probability distribution IS the chance of the event.
No, the probability distribution is the chance of all events in the space.
The question is giving us that and asking us to work out a specific event.

> I would be wrong because I can't know the probability until I know how many sides there are.
No, a question can give you the probability distribution and ask you to work backward to find how many sides there are.
Even in physical systems, you don't need to know how many sides they are to know the distribution, someone else could have rolled the dice measure the chance of all outcomes over many rolls, and given you the measured distribution as a result.
>and ONLY because I GAVE you the probabilities to begin with that the conclusion would now prove!
That's how hypothetical problems work they establish the necessary information to prove details about the system.

>>	Anonymous Sat Dec 17 20:01:37 2022 No.15060384 >>15060348 >huehue look I tag everyone Kill yourself retard idek what debate you're talking about. I posted in this thread once and never revisited it till now.

>>	Anonymous Sat Dec 17 20:03:52 2022 No.15060390 >>15060348 >mass reply faggot Why are you a screen door on my submarine?

>>	Anonymous Sat Dec 17 20:06:28 2022 No.15060398 >>15060384 >I posted in this thread once no one is gonna know or care, dork.

Anonymous Sat Dec 17 20:18:10 2022 No.15060432

>>15060369
>No, the probability distribution is the chance of all events in the space.
>The question is giving us that and asking us to work out a specific event.
The probability of any one specific event is 100%

>No, a question can give you the probability distribution and ask you to work backward to find how many sides there are.
That is exactly what I said you were doing circularly by assuming that it was a die and had FINITE sides! AND by assuming the probabilities were right, which could not be known until you had the number of sides, AND SO ANY calculation would force the number of sides by the probabilities given. That can’t work with infinte sides, and changes depending upon the how many sides there are on the die.

Now to your chest. Did you get the probability from trying infinite chests? That experiment is not doable.
You also cannot determine the number of iterations of chests within chests because the problem gives you that they are infinite.

I am going to give up on you. These kinds of problem aren’t invented by internet trolls. They are invented by large corporations as employment tests to see if the recent college graduate has an intuitive sense of what the fuck is going on. They are tricks to see if you understand the problem so that you don’t encode stupidity into their algorithms. You have to learn to recognize paradox and fallacies so that you don’t bankrupt their companies.

If you don’t understand this, spend some time working this out, because if this is the kind of answer you give, I would not hire you.

Anonymous Sat Dec 17 20:22:54 2022 No.15060447

>>15060348
I'm not answering the question in the image, I'm answering OP's question as to why people and 4chan in general are usually bad at probabilty. Even some of the greatest mathematicians get stumped by the most basic probability logic problems because we're not designed to view the world in the understanding of a random walk but rather we're designed to view everything with the potential for an underlying pattern. It's really cool when you look into it.

>>	Anonymous Sat Dec 17 20:36:34 2022 No.15060471 >>15060447 Not the mass replier, but what is the answer the average person is gonna give vs the mathematician? I refuse to scroll back up.

Anonymous Sat Dec 17 21:14:25 2022 No.15060527

>>15060471
Your average person is going to give up, your average smart person who isn't trained in probability is going to try and give some answer similar to a limit, for example .3 + (.2)*.3 +(.2^2)*2 and so on. But your mathematician will find the likelihood of getting no gold bars at all p(0) and subtract that from 1. This particular example isn't great but the book I mentioned earlier is called "The Drunkard's Walk"

>>	Anonymous Sat Dec 17 21:19:46 2022 No.15060540 >>15060527 But the answer here is 3/8ths

>>	Anonymous Sat Dec 17 21:20:47 2022 No.15060544 >>15060348 >>15043983 The answer is 3/8ths.>>15043983

>>	Anonymous Sat Dec 17 23:24:57 2022 No.15060804 >>15060220 There are two points at which the game ends: win or lose. The probably of losing is 50%. What’s the probability of winning? Everyone on this website is so retarded it makes me sick

>>	Anonymous Sat Dec 17 23:27:47 2022 No.15060814 >>15060275 Please explain, how can you "get nothing" simultaneously with getting a gold bar or two chests? The problem is describing mutually exclusive outcomes. Retard.

Anonymous Sun Dec 18 01:40:49 2022 No.15060956

>>15060804
>There are two points at which the game ends: win or lose.
Technically the game could go on forever.

>The probably of losing is 50%.
No, the probability of not getting gold or two chests immediately is 50%. You can also get two chests and no gold in either of them, so the probability of losing is greater than 50%.

>>	Anonymous Sun Dec 18 01:46:38 2022 No.15060967 >>15060540 Proof?

>>	Anonymous Sun Dec 18 01:57:38 2022 No.15060989 >>15060967 Nvm, I goofed I thought it was just one treasure box and not two. But the guy who said roughly .436 is right I just ran it in python.

>>	Anonymous Sun Dec 18 01:58:53 2022 No.15060992 >>15060527 That doesn't really expiration anything, finding p(0) is not much easier than finding p(1). The key is seeing that there is a recursive function for these probabilities.

>>	Anonymous Sun Dec 18 02:01:12 2022 No.15060998 >>15060956 50% You either get it, or you don't

>>	Anonymous Sun Dec 18 02:23:38 2022 No.15061016 >>15060998 Hahaha what a funny meme

>>	Anonymous Sun Dec 18 02:40:41 2022 No.15061030 >>15060992 Yeah, I goofed. My instincts from taking probability years ago was to find p(0) and subtract that from 1. That's a hell of a mess.

>>	Anonymous Sun Dec 18 03:00:40 2022 No.15061066 >>15060956 > the probability of not getting gold or two chests immediately = 50% what’s the probability of getting gold or two chests? checkmate, atheist

>>	Anonymous Sun Dec 18 03:04:36 2022 No.15061069 >>15060956 > the game could go on forever 0.2^infinity = 0. fuck off with this retarded nonsense

Anonymous Sun Dec 18 04:37:34 2022 No.15061186

>>15060956
>Technically the game could go on forever.
No, if it could then there could then summing all probabilities where the game ends wouldn't sum to one, as we'd be discounting some non-zero probability of the game going forever.
But as it does >>15060010 implying there is no chance the game goes on forever.

>>	Anonymous Sun Dec 18 04:53:45 2022 No.15061204 File: 869 KB, 1000x1024, Untitled.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Sun Dec 18 05:29:11 2022 No.15061239 >>15061066 >what’s the probability of getting gold or two chests? Two chests are not guaranteed to have gold, so this is a non sequitur.

>>	Anonymous Sun Dec 18 05:31:47 2022 No.15061243 >>15061069 0 probability does not mean it can't occur. You know nothing about probability. https://en.wikipedia.org/wiki/Almost_surely

Anonymous Sun Dec 18 05:36:03 2022 No.15061246

>>15061186
>No
Yes, it could. You could just keep getting chests. There is nothing preventing this from occurring.

>if it could then there could then summing all probabilities where the game ends wouldn't sum to one
Wrong. They sum to 1 because the probability of the game not ending is 0. However, 0 probability does not mean it can't occur.

>as we'd be discounting some non-zero probability of the game going forever.
Wrong.

>>	Anonymous Sun Dec 18 05:48:16 2022 No.15061254 >>15061243 >0 probability does not mean it can't occur For all practical purposes in this situation it does.

>>	Anonymous Sun Dec 18 12:45:01 2022 No.15061728 >>15043983 The chance is zero let's be real

>>	Anonymous Sun Dec 18 16:01:34 2022 No.15062059 >>15061254 No, it certainly doesn't in this situation because there is nothing stopping you from getting infinite chests. When infinities are involved, probability 0 doesn't mean impossible.

Anonymous Sun Dec 18 16:41:27 2022 No.15062141

>>15061246
>However, 0 probability does not mean it can't occur.
It means that as you increase the game size it becomes more and more unlikely to be reached.
You'd never really consider the game going forever to be a thing as it becomes less and less likely.

In contrast, if the chance of getting two boxes was higher you have a non-0 chance of the game going on forever.

Anonymous Sun Dec 18 17:21:17 2022 No.15062184

>>15062141
>It means that as you increase the game size it becomes more and more unlikely to be reached.
No, you can have a monotonically decreasing probability without it approaching 0.

>You'd never really consider the game going forever to be a thing as it becomes less and less likely.
Non sequitur. The question was whether it can occur, not how likely it is to occur.

>In contrast, if the chance of getting two boxes was higher you have a non-0 chance of the game going on forever.
??? x^n approaches 0 as long as x<1.

It's pretty funny how you rely on intuition to make these claims instead of thinking about them, and every single one turns out to be wrong.

>>	Anonymous Sun Dec 18 17:27:36 2022 No.15062198 >>15062184 The probability of the other outcomes dominate the asymptotically infinitesimal chance of continually getting 2 chests.

>>	Anonymous Sun Dec 18 17:41:15 2022 No.15062219 >>15062059 What real world situation would you encounter where you would take into consideration something with probability zero occuring (besides pure math puzzle crap)?

>>	Anonymous Sun Dec 18 17:54:06 2022 No.15062235 >>15043983 1.33 repeating ofcourse

>>	Anonymous Sun Dec 18 18:02:21 2022 No.15062249 >>15052127 The CSfag arrives

>>	Anonymous Sun Dec 18 18:13:26 2022 No.15062277 >>15062198 And? Again, non sequitur.

>>	Anonymous Sun Dec 18 18:15:16 2022 No.15062284 >>15062219 If this problem is a "real world situation" then this problem. If this problem is not a "real world situation" then I don't see how "real world situations" are relevant to this thread.

Anonymous Sun Dec 18 18:19:04 2022 No.15062290

>>15060328
>>15060271
You're acting retarded. You're not given a distribution then asking to solve for it, you're given a distribution of a set of outcomes and asking to find the probability of a collection of them. You can claim its circular only by limiting the sample space to having gold or not.

These kinds of questions come up often in real life. Say there's a soccer championship of n length. A soccer team has a 40% chance of winning, if they lose, they cannot proceed. If they win, they have another 40% chance of winning again. It is a perfectly valid question to say for a given n, how likely are they to always win.

Your dice example is ridiculous. It doesn't work because the probability of the outcomes is not known.

>>	Anonymous Sun Dec 18 18:38:47 2022 No.15062328 >>15044040 Abstraction. >>15043990 Yes, obviously. Real geniuses don't do anything and sit around on 4chan.

Anonymous Sun Dec 18 18:58:04 2022 No.15062368

>>15062184
>??? x^n approaches 0 as long as x<1.
That's not at all how it's done.
While the individual chance of every event decreases the number of ways you can maintain opening chests forever grows fast enough that the overall chance of going through ANY way to open chests forever increases.
>'s pretty funny how you rely on intuition to make these claims instead of thinking about them,
I already calculated this, and showed you that, I'm not relying on intuition for these claims.
>The question was whether it can occur, not how likely it is to occur.
Without it approaching 0.
It approaches 0 as I already showed.

>>	Anonymous Sun Dec 18 20:30:13 2022 No.15062544 File: 1.12 MB, 1440x2880, Screenshot_20221219-015854_WhatsApp.png [View same] [iqdb] [saucenao] [google] >>15043983 >>15044016 >>15044029 >>15044037 >>15044465 >>15044687 I have the right answer, suckers

Anonymous Sun Dec 18 21:44:02 2022 No.15062684

>>15062368
>That's not at all how it's done.
That is how it's done. Since the chance of getting two boxes at each stage is less than 1 regardless of how many chests you have, the chance of this process going on forever is always 0.

>I already calculated this, and showed you that
Where?

>Without it approaching 0.
No, the probability of a finite number of chests approaches zero as the number of chests increases. The probability of infinite chests is 0, yet can occur. You can't "calculate" anything to show this is not the case because it's a logical necessity. There is nothing stopping you from continuously getting chests. It's perfectly possible, but almost impossible.

>>	Anonymous Sun Dec 18 21:59:35 2022 No.15062720 File: 53 KB, 803x554, file.png [View same] [iqdb] [saucenao] [google] well /sci/?

>>	Anonymous Sun Dec 18 22:04:26 2022 No.15062730 >>15062720 The second paragraph is where it fails. It multiplied probabilities for events that aren't independent.

Anonymous Sun Dec 18 23:09:37 2022 No.15062857

>>15062684
>Since the chance of getting two boxes at each stage is less than 1 regardless of how many chests you have, the chance of this process going on forever is always 0.
No.
The number of ways you can get to x number of boxes increases the bigger x is, so there is a growing factor that can offset the individual chance becoming smaller depending on how fast the individual chance shrinks.
So it's a question of if the individual chance shrinks fast enough compared to that growing factor
>Where
>>15060010
>You can't "calculate" anything to show this is not the case
You can.
I calculated and added the probabilities of every way the game can end, the only way they don't sum to 1 is if there is a non-zero probability of the game going forever.
For the one in OP they add up to 1 implying the game must end but if we change the probabilities to bias getting more chests then we can see the sum doesn't become 1.

Anonymous Sun Dec 18 23:14:36 2022 No.15062866

>>15062720
>The probability of getting no gold bar from any of the treasure chests inside it i equal to the probability of gettting no gold bars from the first chets multiplied by itself for each chest.
Not true.
It is equating the chance of getting gold from a single chest not counting any chests within to the chance of getting gold from a single chests via any means including chests from within.

Anonymous Sun Dec 18 23:23:41 2022 No.15062887

>>15062857
>>Where
>>>15060010 #
Everything there is correct except the last line. Probability 1 doesn't mean guaranteed when you're dealing with infinite possible outcomes. Our means almost guaranteed. I don't know why I have to keep explaining this to you.

>>	Anonymous Sun Dec 18 23:27:14 2022 No.15062892 >>15043983 How many chests do I need to get to guarantee infinite gold?

>>	Anonymous Sun Dec 18 23:50:00 2022 No.15062966 >>15043983 That question is dumb cause there is a .00001% chance of opening infinite nesting chests until the end of time. It needs clarification

Anonymous Mon Dec 19 00:47:10 2022 No.15063126
File: 9 KB, 480x360, AreYouEvenTryinm8.jpg [View same] [iqdb] [saucenao] [google]

>>15043983
The third chest have 20% chance of containing two chests with the same odds as "this one" referring to the third chest itself, it is never mentionned that there is a bar of gold in these.
Therefore the infinity of boxes does net even enter the problem, it is sufficient to define the last box as no chance of finding a gold bar.
The lottery is also not properly defined, assuming randomization and only one selection the answer is quite obvious.
The chance of have at least one gold bar is 1-(1/3*70%+1/3*50%+1/3*100%)~0.27
This problem is a pure bait.
/thread

>>	Anonymous Mon Dec 19 15:12:40 2022 No.15064330 >>15043983 idk how to do math equations here so bear with me is it 0.3 + (sigma, n=1 -> n=infinity) 0.2^n * 0.3 ?

>>	Aether king of sci Mon Dec 19 15:19:03 2022 No.15064340 50% /Thread

>>	Anonymous Mon Dec 19 16:18:25 2022 No.15064420 >>15064340 Literal retard This is the right answer >>15062544 Never come back to sci again

>>	Aether king of sci Mon Dec 19 16:52:53 2022 No.15064494 >>15064420 Fuck off stage rat You're looking for the mean ratio It is 1 bar not the mean Therefore I know it's 50% Move back along to /b/ now tardcake

>>	Anonymous Mon Dec 19 17:56:58 2022 No.15064618 >>15043983 This is trivial.

>>	Anonymous Mon Dec 19 19:34:18 2022 No.15064850 >>15062544 Nice answer m8. I overcomplicated it by trying to incorporate series.

>>	Anonymous Mon Dec 19 20:31:16 2022 No.15064997 >>15064340 it is not 50%

Anonymous Mon Dec 19 20:38:55 2022 No.15065024

>>15044044
Not worth wasting 15 seconds on, midwit. You do because you need unceasing distraction consuming your waking life. Every smart person I've ever known is content with doing nothing. Midwits love to whip out their phones, scroll, listen to podcasts, and do sudoko. Try being still for a few minutes, I doubt you can manage.

Anonymous Mon Dec 19 20:59:39 2022 No.15065102

>>15043983
Not 100% sure but here's what I think:
Let p denote the probability in question. Either you get a gold bar from the original treasure chest or you get it from one that was nested inside of it. The former case accounts for 0.3 of p, so p - 0.3 = 0.5 * q, where q is the chance that you end up getting a gold bar from either of the chests in the original chest if it has two chests inside it. That happens to be 1 - (1 - p)^2, which can be rewritten as 2p - p^2. So p - 0.3 = p - (1/2)p^2. Rewrite as p^2 = 3/5. So p = sqrt(3/5) = approx. 77.46%.

>>	Anonymous Mon Dec 19 21:01:53 2022 No.15065106 >>15065102 Oops, mixed up the 20% and the 50%.

>>	Aether king of sci Mon Dec 19 21:06:04 2022 No.15065118 >>15064997 Yes it is kek

>>	Anonymous Mon Dec 19 21:08:22 2022 No.15065124 >>15043983 You can't get any bars before opening it faggot, 0%

>>	Aether king of sci Mon Dec 19 21:19:43 2022 No.15065156 File: 36 KB, 340x178, 0fd.jpg [View same] [iqdb] [saucenao] [google] >>15065124 LOL it is this actually how dumb are these posters hah

>>	Anonymous Tue Dec 20 00:42:28 2022 No.15065516 lel this shitpost thread is still alive?

>>	Anonymous Tue Dec 20 00:47:03 2022 No.15065522 >>15062720 Even though it failed, I'm honestly a bit impressed that it managed to spit out the first paragraph at all. I would have expected the recurrence to trip it up completely.

>>	Anonymous Tue Dec 20 00:52:07 2022 No.15065525 >>15065522 >it failed It didn't fail. It's a statistical regurgitator. You failed at telling apart a statistical regurgitator from an artificla intelligence.

>>	Anonymous Tue Dec 20 01:06:12 2022 No.15065547 I can't get the zelda chest opening music out of my head now.

>>	Anonymous Tue Dec 20 11:22:27 2022 No.15066247 >>15043990 >>15048518

>>	Anonymous Tue Dec 20 11:36:43 2022 No.15066264 >>15043983 [Math]E[X] = 0.31 + 0.50 +0.2*2x[/math] [Math]0.6E[X] = 0.3[/math] [Math]E[X] = 0.5[/math]

>>	Anonymous Tue Dec 20 11:46:46 2022 No.15066278 >>15044016 49.999 recouring retard

>>	Anonymous Tue Dec 20 22:38:51 2022 No.15067313 >>15046288 heads up, "chest() or chest()" would be much faster since you return as soon as you find the first gold bar

>>	Anonymous Tue Dec 20 23:02:35 2022 No.15067366 >>15043983 3/8?

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