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Are you smart enough to save your waifu /bant/?

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Thread replies: 110
Thread images: 21

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Are you smart enough to save your waifu /bant/?
>>
>>1141002
Are you fucking serious?
1 golden key with 75% vs. 3 silver keys with 75%. Anyway silver! It could make some sense if golden key was 75% and 3 silver keys was 25%.
>>
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>>1141019
>>
>>1141002
I'd choose the golden key. The silver keys have LOWER chances of opening the door than the gold key.
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>>1141019
u dumb
>>
>>1141019
3 silver keys - 75% => 1 silver key - 25%
2 silver keys - 50%
1 golden key - 75%
>>
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>>1141019
>>
>>1141057
EACH Silver Key, not all silver keys as a whole.
>>
>the word "each"
>>
>>1141019
1 gold key is better than 3 silver dude gold has 25% chance to fail. you need two silver which have 25% chance to fail both. 0,75 times 0,75 is 56,25% chance with 2 keys. although i'm not completely sure i believe it is times 1,25 for the extra key and that would make a 70% chance of the silver route working while gold has 75%
>>
>>1141002
Golden key change of success: 0.75
Silver keys chance of success: 0.5625

Golden key, every time.
>>
>>1141132
you forgot the extra silver key
>>
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>>1141019
>>1141032
>>1141051
>>1141054
>>1141057
>>1141063
>>1141068
>>1141098
>>1141131
>>1141132
>>1141140
3 good silver keys - 0.75^3 - one possible permutation
2 good silver keys and one bad silver key - 0.75^2 * 0.25 - three possible permutations

0.75^3 + (0.75^2 * 0.25)*3 = 0.84375

Easy-tier. Silver keys win.
>>
silvers should have ~84% success rate
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>>1141002
>these threads
>answer ususally obvious
>hoop doop ima tard hurrrrr i'll just do it wrong and call people retards
ironic shitposting is STILL shitpostin
>>
The chance of the golden key = 3/4

The chance of getting 2 working silver keys and one wrong key = (3/4)^2 * 1/4 = 9/64
The chance of getting 3 working silver keys = (3/4)^3 = 27/64
(9/64 + 27/64)= 36/64 < 3/4

You should always go for the golden key
Or am I retarded?
>>
>>1141164
This is the correct math.
>>
>>1141164
>Or am I retarded?
Yes, you are, because there are multiple ways to get 2 working silver keys.
>>
Silver keys, duh
>>
>>1141002
Take all the keys and sell them for profit
>>
>>1141002
silver keys A,B,C
x. chance A doesn't work with B: .25 * .25
y. chance A doesn't work with C: .25 * .25
z. chance B doesn't work with C: .25 * .25

Chance silver keys don't work: x or y or z = x + y + z = 0.1875
Chance silver keys work: 1 - 0.1875 = 0.8125
>>
>>1141146
Why the third case of 2 bad keys and 1 good key which also have 3 permutations was not mentioned? The answer is more than .84375
>>
>>1141264
nvm i did not read the first part of the question
>>
>>1141002
Drill a gloryhole
>>
>>1141002
The chance of getting 2 or more bad silver keys is less than 10%
The chance of getting 1 bad gold get is 25%
However, the obvious choice is to simple blow up the door.
>>
>>1141179

Ah yes, I just saw your earlier post.

I suppose you're right.
>>
I stand corrected. I choose the silver key.

You see, there's a better chance of opening the door with two silver keys than one golden key.

Let's say that the challenge can only let us choose once. If we choose the golden key, there's a 25% chance that we might not be able to open the door. If it fails, then there's no other chance of opening the door.

However, if we choose the silver key, there's a chance that we might be able to replace the wrong key (since the question did not say we cannot change the silver key).

I'm not a math person, but I am (sort of) good with logic (since I'm a programmer).
>>
>>1141051
My post earlier is wrong, Please refer to >>1141343
>>
>>1141343
>I'm not a math person, but I am (sort of) good with logic (since I'm a programmer).
Unfortunately, your logic is a non-sequitur.
>>
>>1141366
Actually, my logic is right. The question didn't say I can't replace the bad silver key if it fails, which means there's a chance I can replace it.
>>
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>>1141667
>The question didn't say I can't replace the bad silver key if it fails
The question also didn't say you can't break the door down or threaten the keymaster with a gun. However, it says that your option are to either take 1 gold key or 3 silver keys. Nothing about replacements.
>>
>>1141355
Stop replying to yourself and reply to me.

>>1141245
My math is correct if you can't tell which of the two silver keys don't work on each attempt. This would apply to a lock with a single mechanism connecting two keyholes.
>>
>>1141690
Read this again:
I have three silver keys that I can use. If one fails, there's another key I can use just in case.

It's called maximizing your resources.
>>
>>1141709
Also forgot to add: I only need two keys and I have three keys, two of which actually works.
>>
>>1141709
Your logic is a non-sequitur. There's nothing to argue about here.
>>
>>1141723
Do I have to fucking explain it again?

>Choose between one key and three keys
>Choose three keys
>Only need two for it to work
>If one fails, use the other one
>You can also look at the keys closely and look for the odd one out.
>>
>>1141690
Have you ever used a key before? Ever one that didn't open the lock you put it in?
>>
>>1141742
>being this retarded
Okay. Let's try something else. The silver keys have a 67.4% chance of working. Silver or gold?
>>
>>1141752
>being this retarded
You're invited to use this """logic""" to solve an alternative version of the problem: >>1141753
>>
>>1141742
It says the silver key lock is a SINGLE LOCK that needs two keys to work. It doesn't say that you can tell which of the two keys is a dud. Stop arguing with the wrong person.
>>
>>1141770
> It doesn't say that you can tell which of the two keys is a dud.
Okay, but he can try ever combination until he finds one that works, assuming he's got at least two working keys. However, this is irrelevant. His logic is a non-sequitur.
>>
>>1141781
>Okay, but he can try ever combination until he finds one that works, assuming he's got at least two working keys. However, this is irrelevant. His logic is a non-sequitur.


Even if you do the math, it's still logical to choose the three keys since it gives me the chance to change the key. Choosing gold, however, won't give me that chance.

Mind explaining how my logic is null?
>>
golden key=
75%, plain and simple

silver keys=
75% each
0,75*0,75*0,25+0,75*0,75*0,75= 84,4% of 2 or 3 keys working. I think.

I'll take my chances with the silver keys.
>>
>>1141804
(0,75*0,75*0,25)+(0,75*0,75*0,75) just to be clear
>>
This is just a question about if you have reading comprehension or not. Most people here are retarded though so it might be hard for some.
>>
>>1141817
it's more a math question
>>
>>1141803
>Mind explaining how my logic is null?
Okay. Let's try something else. The silver keys have a 67.4% chance of working. Silver or gold? Use your "logic".
>>
>>1141875
You didn't answer my question.
>>
>>1141890
I did answer your question: your logic is a non-sequitur. This is clearly demonstrated by your inability to apply this "logic" to the exact same problem if the numbers change a bit.
>>
The chance of getting a bad gold key is 25%.
The chance of getting two or more bad silver keys is 3*(32.6%*32.6%*67.4%)+(32.6%*32.6%*32.6%) which equals about 24.95%.

You have a better chance of failing if you take the gold key. Therefore, take the silver keys.
>>
>>1141928
That was directed at >>1141916
>>
>>1141742
But there is a chance every one of them is dud.
>>
>>1141938
Yes, but the chance that two or more are duds is 3*(25%*25%*75%)+(25%*25%*25%) which equals about 15%. The chance that the gold key is a dud is 25%, which is much higher. So take the silver keys.
>>
>>1141766
I did and the chances seem equal
>>
>>1141928
>>1141936
>>1141992
You niggers are missing the point. His "logic" doesn't provide a solution. You had to do explicit calculations to get there, similar to the ones I did.
>>
>>1141002
working golden key = 3/4
working silver key = 3/4
2 working silver key = 3 * 3/4* 4 = 9/16
3 working silver key = 3 * 3 * 3/4 * 4 * 4 = 27/64
>>
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>>1142013
Ohhhhh I see what you're arguing about now.
>>
>>1142316
The point is that 67.4% favors silver keys, while 67.36% favors gold keys. If it wasn't blatantly obvious to you that his logic was a non-sequitur in the first place, then it should be blatantly obvious now because it doesn't differentiate between the two cases above.
>>
>>1142026
See
>>1141146

Your answer is correct, but there are more correct silver key combinations
>>
>>1142407
But there is just 1 lock, not 2
>>
>>1142461
You still need 2 working silver keys for some reasons, ergo, two locks if you're going the silver route.
>>
>>1142407
>there are more correct silver key combinations
There are only three permutations:
BGG
GBG
GGB
>>
>>1142478
The lock is magic.
>>
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I'll just stick my dick into it they all have a chance to fail so what if they all fail? Its the best to just fuck the keyholes
>>
>>1142013
I can't believe you just replied without saying non sequitur again
>>
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>>1141002
The gold key has a 0.75 chance of working.
The silver key part follow a Binomial Distribution that can be used to work out the probability of at least 2 keys working.
Here is the proof that the silver keys are better:

Let X be the number of working keys
there is a 0.75 chance of *each* silver key working, and there are 3 silver keys.

Binomial Distributions work such that the number of trials and the probability of each trial succeeding is taken into account.
for any binomial distribution, X~(n,p) [X being distributed with n number of trials and p probability]
this is where things get a little complicated... ( i wonder if anyone will actually read this )
P(X=x) = (nCx)*(p)^x*(1-p)^(n-x) [The probability that the number of working keys "X" is equal to "x" is n Choose x (search up mathematical combinations if you dont know what that is) multiplied by the probability to the power of x, multiplied by 1-p to the power of x]

so lets put this into the context of our problem:

X~B(3,0.75) [X is Binomially Distributed with 3 being the number of trials, and 0.75 being the chance that each trial succeeds]
P(X>=2) = P(X=2) OR P(X=3) [The probability that at least 2 keys are working is the same as the probability that 2 or 3 keys are working]

P(X=2) = 3C2*(0.75)^2*(0.25)^1 = 0.421875
P(X=3) = 3C3*(0.75)^3*(0.25)^0 = 0.421875 [they are the same due to the symmetrical nature of the binomial curve]

therefore the probability that the silver keys work is 0.84375, which is more than the probability of the gold key working.
>>
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>>1141002
>test V4 is more trivial compared to V1
What did he mean by this?
>>
>>1141146
Alternative Solution:
You have 3 silver keys, but you need 2 of them to work. Multiply .75 by .75 to get the chances of 2 silver keys working. Since you have an extra silver key, multiply your previous answer by 1.5, since you technically have "50% more keys." You get the same answer.
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>>1142615
>all this unnecessary verbiage
Literally all you need is the addition and multiplication laws of probability...
>>
>>1142635
That's not an alternative solution. It's the same calculation, only your reasoning is... TUN DUN DUN... a non-sequitur.
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>>1142617
That version is poorly worded, here is the more precisely worded version.
>>
>>1142775
Also easy-tier.
p=0.3*1 + 0.5*0 + 0.2*2*p
p=0.3+0.4*p
p-0.4*p=0.3
0.6p = 0.3
p=0.3/0.6 = 0.5
>>
>>1142920
Pr(1 chest has some gold) = 0.3*1 + 0.5*0 + 0.2*Pr(2 chests has some gold)
Pr(2 chests has some gold) = 1 - Pr(2 chests has no gold)
Pr(2 chests has no gold) = Pr(1 chest has no gold)^2 = (1 - Pr(1 chest has some gold))^2

If p = Pr(1 chest has some gold):
p = 0.3*1 + 0.5*0 + 0.2*(1 - (1 - p)^2)

At this point you can solve for valid values of p.

Also, on a side note, intuitively the value of p has to be strictly larger than 0.3 and strictly smaller than 0.5 because if you get the two chests, you may or may not get some gold.
>>
>>1141146
I'm taking this one as a thing, assuming we have 200 couples, with a 50/50 birth ratio, we get 100 males and 100 females, from this we can cut the couple count in half, with the next set following the same probability, we get to a total of 150 males and 150 females, using the recursive function we will eventually conclude that there is still a near 50/50 ratio of men to women, as for each boy, there is a girl, to be precise, we will have a X/X+1 ratio with X being equal to the couple count
>>
>>1141002
Take the golden one as it has 75% of working. If you need two working Silver keys, and each one has a 75% that means you have a 56,25% of a working combination as shown here: 0,75*0,75=0,5625.
>>
>>1143200
I honestly have no idea what you were trying to say, so I'm not going to call it wrong, but please elaborate.
>>
>>1143528
the 50/50 ratio predicts that each girl born in a set should be paired with a girl, as the rule states that after a boy is born, the couple will stop reproducing, the amount of reproduction is halved, rounding up in this case, we eventually get to the point of having only one couple left, so at that point they can produce one more child, if it is female, this leads to the X/X+1 ratio, otherwise the inverse ratio of X+1/X is true
>>
>>1143933
damnit, boy, not girl, I was that close to not fucking up
>>
>>1143933
>the 50/50 ratio predicts that each girl born in a set should be paired with a [boy]
Okay.
>as the rule states that after a boy is born, the couple will stop reproducing, the amount of reproduction is halved
Now that doesn't make sense.
>>
>>1143975
sorry for my explaination making as much sense as the rule the test is based on, but what I'm trying to say is their law essentially makes it a 50% chance for a couple not to be allowed to reproduce, the quote I will place here is a rundown of it
>The family can continue to have children if they have a female child, but must stop as soon as they have a male child.
>>
>>1144012
>their law essentially makes it a 50% chance for a couple not to be allowed to reproduce
That's not true... it's a 50% chance that a boy will be their only child.
>>
>>1143474
If you think that, you should read the last sentence which essentially proves that your answer is wrong:

"Also, on a side note, intuitively the value of p has to be strictly larger than 0.3 and strictly smaller than 0.5 because if you get the two chests, you may or may not get some gold."
>>
>>1144060
>If you think that, you should read the last sentence which essentially proves that your answer is wrong:
Your logic makes no sense and solving your equation produces an objectively wrong answer. Let me see if I can figure out where your error is.
>>
>>1144045
I apologise for keeping it on a each stage ratio, I also apologise for apologising like a leaf, but with each stage, that would be true, say on if they had two kids, it'd be 75%, not 50%, I may have assumed that people would take the issue as a case by case thing.
>>
>>1144092

Here is a break down of the logic:

p = Pr(1 chest has some gold) = 0.3*1 + 0.5*0 + 0.2*Pr(2 chests has some gold)

Now your answer, p = 0.5, can only happen if Pr(2 chests has some gold) = 1, which implies that if you have two chests, you are guaranteed to get some gold, which is not true at all.
>>
>>1141002
Everyone saying that the golden key is correct you are NOT correct. It clearly says that each silver key has a 75% chance, not 25% each.
>>1141019 You were right Dimitri, you are a smart Russian.
>>
>>1144114
>Now your answer, p = 0.5, can only happen if Pr(2 chests has some gold) = 1, which implies that if you have two chests, you are guaranteed to get some gold, which is not true at all.
This logic is based based on:
>p = Pr(1 chest has some gold) = 0.3*1 + 0.5*0 + 0.2*Pr(2 chests has some gold)
And I suspect that it's inherently invalid, which makes your argument invalid as well. Your logic is circular.
>>
>>1141002
nice try professor, I'm not installing mathlab again
>>
>>1144191
>And I suspect that it's inherently invalid
Can you actually prove it is invalid? It's a very simple statement:

"1 chest has some gold" is the event that the chest you are looking at contains some gold inside at some point

"2 chests has some gold" is the event that either of the two chests has some gold inside at some point

It is the same as the first line in your solution:
p=0.3*1 + 0.5*0 + 0.2*2*p

Where you defined Pr(2 chests has some gold) = 2*p.
>>
>>1144238
>Can you actually prove it is invalid?

Yes. The possible scenarios that leave you with one or more pieces of gold are:

E1: None from self AND one or more from children
E2: One from self AND one or more from children
E3: One from self AND none from children

Your equation does not reflect this, and it produces a wrong answer.
>>
>>1144324
Actually that was total nonsense. I don't know what I was thinking.
>>
>>1144324
There are three distinct events when you open up the first chest:

E1: The chest contains a gold bar, probability = 0.3
E2: The chest is empty, probability = 0.5
E3: The chest contains two chests, probability = 0.2

The way you define your events is flawed, consider the following:
>E2: One from self AND one or more from children.

If a chest contains a gold bar, you are done with that chest, there are no children of that chest to consider. In fact, if you run a simulation of this scenario and draw out the resultant tree structure, you will find out that every non-child node is a chest and every child node is either a gold bar or nothing.
>>
>>1144434
I'm not sure why your answer is wrong, but I know for sure that it is wrong, because I know for sure that mine is correct.
>>
>>1144445
Actually I guess you're right.

E1: You get a gold bar directly
E2: You get at least one gold bar from inner chests

p = P(E1) + P(E2)
P(E1) = 0.3
P(E2) = 0.2 * (p*p + 2*p*(1 - p))

This yields p = ~0.436. What I calculated was the expected number of gold bars per chest.
>>
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>>1144748
Yea, the problem is not as easy as it appears on the surface. V10 is another problem in a similar vein.
>>
>>1144777
knives
>>
>>1144777
In order to kill the beast with the grenade launcher, you need to get 4 out of 5 hits (at least). The probability of that happening is 5(.75^4)(.25)+(.75^5), which is 63.28%.
In order to kill the beast with the AK47, you need to get 25 out of 30 hits (at least). The probability of that happening is 142506(.80^25)(.20^5)+27405(.80^26)(.20^4)+4060(.80^27)(.20^3)+435(.80^28)(.20^2)+30(.80^29)(.20)+(.80^30), which is 42.75%.
In order to kill the beast with the throwing knives, you need to get 2 out of 3 hits (at least). The probability of that happening is 3(.60^2)(.40)+(.60^3), which is 64.8%.

So you would have the best chance of killing the beast with the throwing knives (followed by the grenade launcher, then the AK47).
>>
>>1144777
I guess there's always the dumb way of doing it, which is the same as the keys reasoning:

Grenade launcher requires at least 4/5 hits:
0.75^5 + 0.25*0.75^4*5 = 0.6328125

Knife requires at least 2/3 hits:
0.6^3 + 0.4*0.6^2*3= 0.648

Now the AK requires at least 25/30 hits, so that one gets pretty ugly:
(0.8^29)*(0.2^1)*30!/(29!*1!)+
(0.8^28)*(0.2^2)*30!/(28!*2!)+
(0.8^27)*(0.2^3)*30!/(27!*3!)+
(0.8^26)*(0.2^4)*30!/(26!*4!)+
(0.8^25)*(0.2^5)*30!/(25!*5!) = 0.426

I'm sure there's a cleaner way to do it using binomial distribution formulas which I just don't remember anymore.
>>
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>>1141002
Fucking hell, are y'all retarted?

Three keys, we need two.
So:
- if first key doesn't work (0.25 chance), two other keys must work. So 0.25*0.75*0.75
- if first key works, only one key must work. Chance of one-key-of-two-working can be calculated two ways, either (0.75 + 0.25*0.75) or (1-0.25*0.25). So 0.75*(0.75 + 0.25*0.75)

Final answer:

In [1]: 0.75*(0.75 + 0.25*0.75) + 0.25*(0.75*0.75)
Out[1]: 0.84375
>>
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>>1141146
>>1142407
>>1142615
Ok, some of you are correct
>>
>>1144777
Actually the formula for cumulative binomial probabilities isn't much better than (>>1145451). Is there some trick to this that makes it easy, or is it intended to be tedious? (If there is one, don't give it away just yet)
>>
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>>1144445
oFF BY

Wun


n
>>
>>1145887
>oFF BY Wun
And the one right before it, too...
>>
>>1144777

I got schooled the last time but did my best on this one.

First of all, the grenade launcher.

4 or 5 must hit

Amount of possibilities: if 4 hit = 5 (5C1), if 5 hit =1

The chance of killing the beast: 4((3^4/4^4) * 1/4)+(3^5/4^5) = 0.5537109375

The AK

25,26,27,28,29 or 30 bullets must hit:

25 (30C5) = 142506 possibilities
26 (30C4) = 27405
27 (30C3) = 4060
28 (30C2) = 435
29 (30C1) = 30
30 = 1

The chance to kill the beast:

142506((4^25/5^25)*1^5/5^5)+27405((4^26/5^26)*1^4/5^4)+4060((4^27/5^27)*1^3/5^3)+435((4^28/5^28)*1^2/5^2)+30((4^29/5^29)*1/5)+(4^30/5^30)= 0.42751243759

Knives:

2 or 3 must hit

Amount of possibilities:

2 (3C1) = 3
3 = 1

The chance to kill the beast:

2((3^2/5^2)* 2/3)+(3^3/5^3) = 0.696

Knives>Grenade Launcher >AK
>>
>>1141002
Silver. The probability of at least 2 keys working is 84% while your gold key is just a flat 75%. Now give me my waifu.
>>
>>1146042
>Amount of possibilities: if 4 hit = 5 (5C1), if 5 hit =1
>The chance of killing the beast: 4((3^4/4^4) * 1/4)+(3^5/4^5) = 0.5537109375
I think you meant: 5((3^4/4^4) * 1/4)

>2((3^2/5^2)* 2/3)+(3^3/5^3) = 0.696
I think you meant: 3((3^2/5^2)* 2/5)+(3^3/5^3)

Check the last one yourself.
>>
>>1146193

yeah, those are pretty bad errors.

Thanks for sorting me out.
>>
>>1141002
just let her die
>>
>>1146311
Just remembered one I saw on Omegle once: on average, how many random numbers between 0 and 100 do you have to sum together to get a result divisible by 400?
>>
>>1141002
1 gold key = 0.75
3 silver keys = 0.75(0.75+0.25*0.75)+0.25(0.75^2) = 0.84375
Pick the silver keys.
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