Answer this if ur high IQ.

[imontheloose]

How will u tell if it’s lighter or heavier as u don’t get the weight? you are all forgetting the second component of this question lmao. Ur not just working out which coin has a different weight, you are also working out if it’s lighter than the others or heavier

The scale tells you which side is heavier.

 

[CantStopTheMog]

Draw 24 leaves.

Sure, how will you work it out because u NEED the visual scale to decipher if it is heavier or lighter bro listen please

 

[BigBallsLarry]

Yes it is easier but it’s cos nobody other than imontheloose can get the answers to the others

Explain the method because yes it is POE but there’s a second step to find if it is heavier or lighter.

what i did is focus on doing the most each weight

first weight i use to group the coins up and eliminate atleast 2 groups:

group 1 - coins 1-4
group 2 - coins 5-8
group 3 - coins 9-12

i put group 1 on left scale and group 2 on right scale, so the outcome is simply:

- if group 1 pans, the counterfeit is in group 2 and vice versa
- if they stay leveled, the counterfait is in group 3

second weighting depends on the outcome of the first weighting but is pretty simple aswell

easy way - if it was balanced, simply weight the suspects vs what is confirmed to be real (group 3 vs group 1 or 2) but use 3 coins instead of 4 to instantly rule out the counterfeit

for example, i put coin 9, 10 and 11 next to 1, 2, 3. if they stayed balanced, the counterfeit is coin 12, if they weave, the counterfeit is either 9, 10 or 11 which can be tested in the third weight.

more complicated - if one side swayed, take 2 known real coins and 1 unknown coin (from lighter side) on one side (for example 9, 10 and 5) on the other side put 1 coin from heavy side, one from light and one from real (from group 3) [for example 1,6,11)

if it balances, fake is in one of the coins in group 1 or 2 that wasnt tested. (2, 3, 4 , 7, 8)

If it goes left:
either the coin from the lighter side on the left pan is heavy, or the coin from the heavier side on the right pan is light, or the coin from the heavier side on the left pan is heavy.
From the setup, that means the counterfeit is either 1, 5 or 6

if it goes right its basically the same but mirrored

third weighitng is just process of elimination with those

 

[BaldAutistV2]

You’re stupid. 6v6 does not work. 27>24 hence can be solved in three steps as I have detailed. Do not reply to me again.

You have 12 visually identical metal coins

 

[CantStopTheMog]

The scale tells you which side is heavier.

yes but how do u know if the counterfeit is heavier when it is stated it could be lighter or heavier and u have nothing to base the coins weight off of.

 

[imontheloose]

yes but how do u know if the counterfeit is heavier when it is stated it could be lighter or heavier and u have nothing to base the coins weight off of.

I addressed this already. I will have to write out a comprehensive guide for you.

A balance cares about relative weight, not absolute. The first balance determines one normal coin to use for the future. Compare a suspect set to a normal set. If S vs. N tips S, S is heavy. If it tips the other way, S is light. If it balances, S is normal.

By the time you name the coin, its heavy/light status has already been implied.

 

[BaldAutistV2]

these guys r literally not listening they’re doing the easy part like I’d expect but they’re not even answering the second part lmao

it's obvious when you are left with the pile of 4 coins with the weighted jew coin you permutate
coin 1 with coin 2
coin 2 with coin 3
coin 3 with coin 4

s P(A and B) = P(A) \* P(B|A), where P(A and B) is the probability of both dependent events A and B occurring, P(A) is the probability of event A, and P(B|A) is the probability of event B occurring given that event A has already occurred.

 

[CantStopTheMog]

what i did is focus on doing the most each weight

first weight i use to group the coins up and eliminate atleast 2 groups:

group 1 - coins 1-4
group 2 - coins 5-8
group 3 - coins 9-12

i put group 1 on left scale and group 2 on right scale, so the outcome is simply:

- if group 1 pans, the counterfeit is in group 2 and vice versa
- if they stay leveled, the counterfait is in group 3

second weighting depends on the outcome of the first weighting but is pretty simple aswell

easy way - if it was balanced, simply weight the suspects vs what is confirmed to be real (group 3 vs group 1 or 2) but use 3 coins instead of 4 to instantly rule out the counterfeit

for example, i put coin 9, 10 and 11 next to 1, 2, 3. if they stayed balanced, the counterfeit is coin 12, if they weave, the counterfeit is either 9, 10 or 11 which can be tested in the third weight.

more complicated - if one side swayed, take 2 known real coins and 1 unknown coin (from lighter side) on one side (for example 9, 10 and 5) on the other side put 1 coin from heavy side, one from light and one from real (from group 3) [for example 1,6,11)

if it balances, fake is in one of the coins in group 1 or 2 that wasnt tested. (2, 3, 4 , 7, 8)

If it goes left:
either the coin from the lighter side on the left pan is heavy, or the coin from the heavier side on the right pan is light, or the coin from the heavier side on the left pan is heavy.
From the setup, that means the counterfeit is either 1, 5 or 6

if it goes right its basically the same but mirrored

third weighitng is just process of elimination with those

u should explicitly pick the third weighing to resolve the remaining ambiguity

U also didn’t explain a way to show of the counterfeit is definitely heavier or lighter
BUT this is the closest to correct so far

 

[CantStopTheMog]

I addressed this already. I will have to write out a comprehensive guide for you.

A balance cares about relative weight, not absolute. The first balance determines one normal coin to use for the future. Compare a suspect set to a normal set. If S vs. N tips S, S is heavy. If it tips the other way, S is light. If it balances, S is normal.

By the time you name the coin, its heavy/light status has already been implied.

sorry ur correct I was reading it wrong the whole time lmao my bad

 

[imontheloose]

sorry ur correct I was reading it wrong the whole time lmao my bad

Chuddha forgives the lost lamb.

 

[CantStopTheMog]

it's obvious when you are left with the pile of 4 coins with the weighted jew coin you permutate
coin 1 with coin 2
coin 2 with coin 3
coin 3 with coin 4

s P(A and B) = P(A) \* P(B|A), where P(A and B) is the probability of both dependent events A and B occurring, P(A) is the probability of event A, and P(B|A) is the probability of event B occurring given that event A has already occurred.

U don’t even need to do allat

 

[CantStopTheMog]

Chuddha forgives the lost lamb.

View attachment 4182255

For some reason I thought u meant you’d compare their weights using a scale which would tell u their exact weights rather than 2 defined outcomes (different or balanced) not sure how I even came to the conclusion that that was what u were saying

 

[imontheloose]

For some reason I thought u meant you’d compare their weights using a scale which would tell u their exact weights rather than 2 defined outcomes (different or balanced) not sure how I even came to the conclusion that that was what u were saying

No brah. Chuddha uses le balance.

 

[CantStopTheMog]

That’s not what I’m saying. I’m saying it just makes it easier to group each step. Obviously you use your reasoning like @imontheloose said.

you guys were right lmao I was jus reading ur answers wrong because it’s hard to see what u guys r sayinh on a text as u have a different method to me

 

[Gaygymmaxx]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

well you weren't specific enough so I'm gonna have a little fun

How is one EXACTLY supposed to place the coins on the scale? you didn't say all at once or a few at a time

So i will use the scale ONCE But I will place coins two at a time (one on each side), until the scale shows them as uneven

then for my second use, I will compare one of the coins to another one which is not amongst the pair selected

If the coin i choose doesn't match, then it's the counterfeit, if it does match, the other of the pair is the counterfeit

 

[CantStopTheMog]

well you weren't specific enough so I'm gonna have a little fun

How is one EXACTLY supposed to place the coins on the scale? you didn't say all at once or a few at a time

So i will use the scale ONCE But I will place coins two at a time (one on each side), until the scale shows them as uneven

then for my second use, I will compare one of the coins to another one which is not amongst the pair selected

If the coin i choose doesn't match, then it's the counterfeit, if it does match, the other of the pair is the counterfeit

Blud that is using the scale up to 8 times as ur placing more and more coins on it which adds more measurements actually

 

[Gaygymmaxx]

@CantStopTheMog try to be more specific next time, my new goal with these threads is not to get the right answer, but instead to justify the wrong one

 

[Gaygymmaxx]

Blud that is using the scale up to 8 times as ur placing more and more coins on it which adds more measurements actually

Cope, you didn't outline how exactly I have to use the scale

Fine, I will be measuring 6v6 coins, the whole stack..

but I'll simply be doing it slowly

 

[CantStopTheMog]

Cope, you didn't outline how exactly I have to use the scale

Fine, I will be measuring 6v6 coins, the whole stack..

but I'll simply be doing it slowly

Fuck ur lowkey right I shoulda said all coins must be placed at the exact same time
Bro outsmarted the system

 

[gonnabehappy]

what i did is focus on doing the most each weight

first weight i use to group the coins up and eliminate atleast 2 groups:

group 1 - coins 1-4
group 2 - coins 5-8
group 3 - coins 9-12

i put group 1 on left scale and group 2 on right scale, so the outcome is simply:

- if group 1 pans, the counterfeit is in group 2 and vice versa
- if they stay leveled, the counterfait is in group 3

second weighting depends on the outcome of the first weighting but is pretty simple aswell

easy way - if it was balanced, simply weight the suspects vs what is confirmed to be real (group 3 vs group 1 or 2) but use 3 coins instead of 4 to instantly rule out the counterfeit

for example, i put coin 9, 10 and 11 next to 1, 2, 3. if they stayed balanced, the counterfeit is coin 12, if they weave, the counterfeit is either 9, 10 or 11 which can be tested in the third weight.

more complicated - if one side swayed, take 2 known real coins and 1 unknown coin (from lighter side) on one side (for example 9, 10 and 5) on the other side put 1 coin from heavy side, one from light and one from real (from group 3) [for example 1,6,11)

if it balances, fake is in one of the coins in group 1 or 2 that wasnt tested. (2, 3, 4 , 7, 8)

If it goes left:
either the coin from the lighter side on the left pan is heavy, or the coin from the heavier side on the right pan is light, or the coin from the heavier side on the left pan is heavy.
From the setup, that means the counterfeit is either 1, 5 or 6

if it goes right its basically the same but mirrored

third weighitng is just process of elimination with those

this don’t work, cuz if u left with 3 options with one weighing it’s unsolvable, you’ll be 50/50 in a remaining scenario

 

[gonnabehappy]

this don’t work, cuz if u left with 3 options with one weighing it’s unsolvable, you’ll be 50/50 in a remaining scenario

i will answer when I eat dinner but u can solve it if the 4 are equal fs, but idk if u can solve when the original 4v4 isn’t balanced

 

[5'7" 3/4s]

Im currently moving into a new room and been thinking about how to do this at the same time, even recited the question to my pops while we were changing the bed frame

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.


Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

my dad came up with an idea, i told him it wouldnt work but then i tweaked it a little bit, which i think i ended up making my own "regression" type algo.

--//grab two coins for control as to which one weighs regular, theres a 2/12 chance you select the counterfeit coin, if you do it makes the process simpler and can skip some steps.

if they weigh the same, you now have a control coin. the way to do this(whilst breaking rules) would then just to match each coin with the counterfeit coin, but this would take 9 more tries at most using the scale.
--SECOND ATTEMPT-- MY MODIFIED VERSION

the only way to simplify would be to do it in groups, so now we should try to create a control group instead of merely a control coin.
you take 3 groups of 4 and measure them against each other. if they weigh the same you have a control group, if not, you still do, because you know the third group would atleast weigh the same as one of the other two groups(through deduction, since only 1 coin).

Now obviously the issue with the groups is that we don't know which coin within the group is the coin.

So in the case that they weigh the same:
so how about this, we now split the leftover group into 2, and then compare it with two from our original control group, if they are the same, then you know the counterfeit is in the leftover of the leftover group,
and can just compare 1 of these coins(since only 2 left) with another control coin.

if they are different at any step of the process, then we can just regress in the same manner using control coins until we get to process of elimination, and it shouldnt exceed more than 3 steps(if my logic is correct)

if these are even the same, then use process of elimination, since only 1 coin left, using the same means of regression.

In the case that they weigh different:
take one group off and compare it with the leftover control group

if they weigh different -> you know that the group you didnt take off has a counterfeit, but we wasted a step in determining this, so our algo above wont work in some cases



@CantStopTheMog my midwit brain is just stuck here, i could solve it if i had the motivation, but i cant be fucked because its literally one step from being the solution. but idk. Can you give me a hint to get past this step?

 

[5'7" 3/4s]

i will answer when I eat dinner but u can solve it if the 4 are equal fs, but idk if u can solve when the original 4v4 isn’t balanced

Im currently moving into a new room and been thinking about how to do this at the same time, even recited the question to my pops while we were changing the bed frame

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.


Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

my dad came up with an idea, i told him it wouldnt work but then i tweaked it a little bit, which i think i ended up making my own "regression" type algo.

--//grab two coins for control as to which one weighs regular, theres a 2/12 chance you select the counterfeit coin, if you do it makes the process simpler and can skip some steps.

if they weigh the same, you now have a control coin. the way to do this(whilst breaking rules) would then just to match each coin with the counterfeit coin, but this would take 9 more tries at most using the scale.
--SECOND ATTEMPT-- MY MODIFIED VERSION

the only way to simplify would be to do it in groups, so now we should try to create a control group instead of merely a control coin.
you take 3 groups of 4 and measure them against each other. if they weigh the same you have a control group, if not, you still do, because you know the third group would atleast weigh the same as one of the other two groups(through deduction, since only 1 coin).

Now obviously the issue with the groups is that we don't know which coin within the group is the coin.

So in the case that they weigh the same:
so how about this, we now split the leftover group into 2, and then compare it with two from our original control group, if they are the same, then you know the counterfeit is in the leftover of the leftover group,
and can just compare 1 of these coins(since only 2 left) with another control coin.

if they are different at any step of the process, then we can just regress in the same manner using control coins until we get to process of elimination, and it shouldnt exceed more than 3 steps(if my logic is correct)

if these are even the same, then use process of elimination, since only 1 coin left, using the same means of regression.

In the case that they weigh different:
take one group off and compare it with the leftover control group

if they weigh different -> you know that the group you didnt take off has a counterfeit, but we wasted a step in determining this, so our algo above wont work in some cases



@CantStopTheMog my midwit brain is just stuck here, i could solve it if i had the motivation, but i cant be fucked because its literally one step from being the solution. but idk. Can you give me a hint to get past this step?

I didnt ask CHATGPT for the answer, I asked it for a hint and this is what it gave me:

So Ill try to solve it with this in mind holdon

 

[5'7" 3/4s]

I didnt ask CHATGPT for the answer, I asked it for a hint and this is what it gave me:

View attachment 4182702

So Ill try to solve it with this in mind holdon

holy fuck i didnt account for the scale tipping the opposite way

 

[Rothschild]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

weigh all the coins

see which one is idfferent

counterfiet found

 

[5'7" 3/4s]

holy fuck i didnt account for the scale tipping the opposite way

ye im not high iq enough or never learned how to conceptualize probabilities in this way, chat gpt just gave me the answer. good learning experience.

 

[gonnabehappy]

i think maybe the trick is to switch possibly fake coins into control groups during weighs 2-3 possibly

I’m still thinking about possibility of 4v4 unbalanced, 4 safe

Then what to do cuz 8 possible coins left in 2 balances

 

[ProBono]

1 4:4 coins 1-4 on the left, and 5-8 on the right.
2m If they match, take coins 9-11 and weigh against 1-3 (all safe).
3m In this case, if they match again you can simply weigh 12 against any other coin as it is the counterfeit and you will also find out if it's lighter or heavier.
2mb If the left side (9-11) is, say, lighter, the counterfeit is lighter., now weigh 9 against 10.
3mb If 9 is heavier, 10 is the counterfeit and it's lighter. If 10 is heavier, 9 is the counterfeit and it's lighter. if they match, 11 is the counterfeit and its lighter.

2h If the left side is heavier, 9-12 are safe and it's either 1-4 as heavier or 5-8 as lighter. Now, weigh 1-2, and 5 against 6 and 3-4.
3hm If they match, it's either 7 or 8 and you can weigh the two against eachother. Whichever is lighter (remember if the counterfeit is 5-8 it has to be lighter) is the counterfeit.
3hh If the left side is heavier again, it's between 1-2 and 6, as 1-2 can both be heavy counterfeits on the left, and 6 can be a light counterfeit on the right. Weigh 1 against 2, if 1 is heavier, it's the counterfeit. If 2 is heavier, it's the counterfeit. If they match, 6 is the counterfeit and it's lighter.
3hl, If the left side is lighter, you apply the exact same methodology but instead it's to 3, 4 and 5 and you weigh 3 against 4.

 

[ProBono]

more complicated - if one side swayed, take 2 known real coins and 1 unknown coin (from lighter side) on one side (for example 9, 10 and 5) on the other side put 1 coin from heavy side, one from light and one from real (from group 3) [for example 1,6,11)

if it balances, fake is in one of the coins in group 1 or 2 that wasnt tested. (2, 3, 4 , 7, 8)

If it goes left:
either the coin from the lighter side on the left pan is heavy, or the coin from the heavier side on the right pan is light, or the coin from the heavier side on the left pan is heavy.
From the setup, that means the counterfeit is either 1, 5 or 6

if it goes right its basically the same but mirrored

third weighitng is just process of elimination with those

Interesting I put 2 from the heavy and 1 from the light on one side, and an additional 2 from the heavy and 1 from the light on the other, instead of using any of the confirmed real coins. I don't think you are actually able to solve every case if you use real coins like that in the second test as they don't provide any data.