Hey there was a good thread yesterday with some hard riddles but im not able to solve some of them and i forgot to save...

Hey there was a good thread yesterday with some hard riddles but im not able to solve some of them and i forgot to save the solutions so can some of you pls post em? I post all riddles i've got.
General riddle thread

The minimum number possible is one. You put four balls on one side, four balls on the other, and if they weigh the same amount, you've got the light ball in your hand. Statistically, you stand a one in nine chance of that happening.

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I can't think of a solution that uses fewer than three checks. Maybe I'm just a moran.

First check - four balls on each side of scale, one left out. Take the lighter side to the second check. If neither side is lighter, the one ball you left out is the light one.

Second check - Two balls on each side, move the lighter side to the third check.

Third check - One ball on each side, you've found the light ball.

The question says minimum number to guarantee you've found it. 1 in 9 is not a guarantee.

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For the first one the answer is 2. Break into groups of 3. Weigh two of the groups against each other. If one is lighter it must contain the lighter ball. If they are balanced, the lighter ball is in the third set.

From this selected group of three, weigh two balls against each other. If one is lighter it is the desired ball. If they balance the lighter ball is the third one from the set.

Cut off the cop's ear with a razor blade.

312211 read the number of numbers on the previous line aloud. Three ones. Two twos. One one.

Cop? Which cop?

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2D>3D, easy

Shoot at himself.

Well, these two were easy...

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The A and the 7. Who cares if there's an even or odd on the back of the K. The 2 doesn't matter because if it is a vowel, it checks out.

I'm pretty sure survive doesn't mean to suicide still if they both start with an s

Much more elegant than my solution. To be fair, there was no rule saying you could go outside the box.

Try to draw a tree of your steps and look at its longest path. Think about how you can reduce it. Keyword non binary.
Ganbare, user

No, because he only stands a 25% chance of hitting. If Mr. Grey shoots at himself, he stands a lot greater chance of shooting himself in the head. Meanwhile, when Mr. Black shoots at himself, he's 100%. Tell me why it wouldn't work. Worst case scenario, Mr. White has to shoot at himself twice. By then, Mr. Grey's guaranteed to hit.

Doesn't matter cause you'll find the light one

Shoot mr Gray.
Have the cops shoot mr Black.

Yes.
If you get a son you throw it away and get as many kids as you want.

doesnt work.
doesnt matter how many children they have or dont have before stopping having children, the probability of the sex of the child is still 50/50 boy/girl so on average, the ratio remains 50/50

At first the chances do not add with each try and why the fzck should the other two shoot at themself?

Obviously it doesn't matter if it works or not because it's a completely retarded idea.

What if there's a vowel behind the K? Nobody ever said that each card has a number and a letter

gg wp

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one
weigh four and four
when the wight is equal: no. 9 is lighter
otherwise you need more than one step.

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Check 1 - three per side, three left off scales.
Check 2 - one per side, one left off scales.

Black is threatened by grey because he has the most accuracy out of the remaining two and grey is threatened by black because if he shoots white then he's shot by black, so grey will probably take a chance at black.

White shoots in the air. Thus reducing its chances of killing someone to zero.

By then, since two people are going to die, grey's accuracy raises up to 100. He shoots black and kills him. Black shoots while getiing hit and kills grey.

White is probably a jew.

By weighing 4x, you will always find the lighter ball using a basic two-sided scale.

Any less than that and you've found it earlier than would guarantee others to find it with a different configuration.

So what're you saying? Mr. White should just get on his knees and offer to give Mr. Grey and Mr. Black a blowwy-wowwy and that'll give him plenty of time to shoot both? Mr. White isn't gay. So he won't be doing none of that.

My solution is logically sound. Just imagine you're Mr. Grey. You've come to a gun fight, knowing Mr. Black is going to straight-up murder you because he doesn't miss. And the other guy just tries to shoot himself in the head. What're you going to do? You got no choice.

Now imagine you're Mr. Black. You're gonna pwn some noobs, and suddenly, they're trying to shoot themselves in the head. If all your friends go jump off a bridge, you bet your shit Mr. Black is gonna jump off the bridge too.

It is a way to outsmart the king, who made some kind of shit harem law where all the boy children have to kill two other people so there'll be more girls. Mr. White doesn't want to worry about the electricity and the water and the plumbing to his neighborhood houses. He just wants to go home and enjoy his 20 add two dots waifus.

See, it doesn't matter if the consonants have odds OR evens, because the question doesn't ask that. It is only worried about vowels and evens.

So while Mr. Black's last thoughts are about his lopsided balls, Mr. White walks away a free man, because he was brave enough to SHOOT AT HIMSELF.

smart and short, nice.
i declare the solution is yours!

*tips fedora*

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get lucky son, just pick up one and hope for the best

The secret to this one is there's no "You should be able to do this." Since there isn't that line, you shouldn't be able to do this, and thus, win.

Well if the scales are exactly equal after placing 4 balls on either side. Then you do indeed have the lighter one. So technically its just one use of the scale, and again you'd have to be incredible lucky to have picked the lighter ball, but its not impossible. And thus, after weighing them once, you've guaranteed that you have the lighter one, because the other 8 balls equaled out on the scale. Again extremely lucky scenario, but would be 1 weigh, guaranteed

See, this doesn't work, because Mr. White only stands a 25% chance of hitting the air. That'll make it a 75% chance that he'll hit either Mr. Grey or Mr. Black. Mr. Grey and Mr. Black both know this so they'll murder him immediately.

If you put it that way he also has a chance of shooting himself but at this point he just get on his knees and give both grey and black a good succ for being a dumb retarded slut with a feminine aim.

No, because the air is all around them, and Mr. Grey and Mr. Black are the only things that aren't air. But there's tons of things other than himself around them, like air, or cardboard boxes, or a cat.

I can't believe nobody has answered this simple as shit "riddle"

1) 4 balls on each side
2) divide the 4 lighter balls in pairs
3) divide the lighter pair

Three steps guarantee you find the lighter ball.

the problem can be simplified to V -> E (vowel implies even) = ¬V ∨ E (not vowel or even) building a truth table you get this
V E ¬V (¬V∨E)
1 1 0 1
0 1 1 1
1 0 0 0
0 0 1 1

so given a letter card the only time this statement can be false is if you have a vowel so you need to turn over all the cards that have a vowel on them to make sure that the back is even.
which is pretty simple we got that from the start.
going the other way (given the number side) the only time the statement can be false is if you are given an odd card, so you need to turn over all the odd cards to make sure that the back is a consonant

3

I know what you're thinking, there will be some families that will have eight girls before they have one boy, but all those families who only have one boy will counter-balance that. You see? It doesn't even matter if some people decide to stop after four girls and never have a boy, the ratio is still 50-50, because in that fourth generation, 50% of the children are still boys.

The only real harem solution is to force three of the boys to shoot each other until only one survives.

This

If he has a 25% of shooting air, he has a 75% chance of shooting non-air, including but not limited to:

>grey
>black
>himself
>his gun
>the riddle
>you
>me
>this thread
>the game

Grey and black would die by shooting at one another anyways, shooting himself makes him ascend to valhalla so that's a win, shooting his gun would make his gun ascend to valhalla and be reincarnated as a gatling gun with which you don't care about aim, shooting the riddle solves it automatically, shooting you and me ends the debate, which at this point might as well be considered a win, shooting the thread 404s it and this kills the riddle, and he lost so might as well shoot the game anyways.

You are assuming that each card has a letter on one side and a number on the other. The "riddle" does not state this is the case.

There could be a vowel on the back of the K. So you have to turn over three cards. A, K, and 7.

2 steps.

>Divide into 3 groups of 3 each.
>Weigh 2 groups
>If the groups being weight are equal, ball is in the third group.
>If they aren't equal, you will know which group contains the lighter ball

Now you are down to 3 balls

>Weigh 2 of them.
>If they are equal, the third ball is the lightest.
>If not, you have the lighter ball.

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Thank god someone is neither dumb nor hiding its dumbness behind retarded sarcasm.

No, because if you shoot AT anything else, the bullet will hit air, on its way, to hitting the target. And the only things Mr. White can put his gun directly against, so there's no air between the gun and his target, is Mr. Grey or Mr. Black or himself. The riddle, you, me, this thread, the game, the cardboard boxes, the cat, they're all too far away...he'll hit AIR on its way.

fair point the riddle doesn't specify that, typically it does and i just assumed that the given riddle did

2, you weigh 3 against 3 and if they weigh the same then the ball is within the 3 you didn't weigh, if ones side weighs less than the other side then those 3 contain the light one. continuing: when you know which 3 contain the ball you weigh 1 against 1 and if they weigh the same then it's the one you didn't weigh and if one weighs less than the other then, well, you got it

He'll only have a 75 percent chance of hitting the air if he shoots at the cat.

WINRAR

Yes, but because he has a 75% of not hitting stuff, then there's a chance the bullet will just phase through air and break open new dimensions for aforementioned targets to be hit without the bullet entering into contact with air.

Other dimensions don't matter. Take the first problem, for example. If it took place in another dimension, one of the balls would be heavier. The solution is still the same.

Or a dimension where a big-breasted queen decrees that all families should stop bearing children when they have a girl wouldn't make her harem of shota-flavored boytoys any larger.

There's only one riddle here concerning different dimensions, but 2D>3D is true in all dimensions. So it is a moot point.

As far as i see mr white could overtalk them to play russian rullet so he would prabably survive?

That's what I've been saying all along.

White shoots at the air. Mr. Grey shoots at Mr. Black, but misses (25% chance), and hits Mr. White instead. Mr. Black shoots at Mr. Grey. Two people have died, and one has lived, and it isn't Mr. White. You've lost.

What is relevant is what missing the shot implies in terms of consequences, as and i are discussing, since it applies not only to white but also to grey, what matters is to correctly define the domain of all possibilities, and it should be noted that in the big breasted mother queen dimension, the riddle wouldn't involve guns but feminine dicks, hence the whole air problem still being a relevant problem in need of solving, since sperm oxydation is a thing.

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False.

Well this one was easy.
>You should be able to solve this