Math Problem

6/4 ?

i dont have any logic, im bad for riddles, but mike said 'i didnt know it at FIRST' so he might be telling him a clue that its on january

ah it´s 9/1. misread something

1. statement: month 3 and 9 are left
2. statement: days 1, 4 and 8 are left
3. statement: day 1 is left => 9/1

John says: "If I don't know the date, then Mike can't either"

Means that it cant be either in the month 6 since thn Mike could have known the date 6/7.
It cant be any in the month 12 either for the same reason.


Mike says: "I didn't know it at first, but now I know it"

Means it is one of the number which are single days 3/4 3/8 and 9/1

John says: "Now I know it too."

means it is 9/1

If mike's number was D = 8, john would have M = 12. If john had M = 12, he would not know that mike didn't know it, because 12/2 is the only date with 2 as the day. John's first statement shows that he knows it is not a date that has only 1 date with the given day. This means his month cannot be 12 and it cannot be 6.

I would've understood it better if John just would've said: "Mike can't know the date". Because the if-clause is totally confusing and useless (John can't know the date in the first place with no information about Mike's number)

3/4

John says: "If I don't know the date, then Mike can't either"

(Mike, knowing he has 4 would deduce that it's either 3/4 or 6/4 at the beginning. John, having 3, knows it's 3/4, 3/5, or 3/8.)

Mike says: "I didn't know it at first, but now I know it"

if it was 6/4, John's statement would be false because John would have 6 which gives
the possibility of Mike knowing the answer right from the beginning

John says: "Now I know it too."

(John can use the same logic as Mike here to figure it out)

month
3-- | -- | -- | 4 | 5 | -- | 8
6-- | -- | -- | 4 | -- | 7 |-
9-- | 1 | -- | -- | 5 | -- |-
12 | 1 | 2 | -- | -- | -- | 8

then you clearly see which month have which days i common and you just have to cross out whole rows/colmns

I must be missing something because I don't see how it is anything but 50%. I'm terrible at these types of logic problems anyways, I always miss the point.

there's 3 socks, one is black. Number 3 isn't black. What's the chance Number 1 is black. What am I missing here?

We have a winner! Congratulations to lololol!
http://en.wikipedia.org/wiki/Three_Prisoners_Problem

17 mins

I'm pretty sure this is ceiling(log base (y+1) (x)).
In a less technical format, (y+1)^n = x, solve for n, rounding up, gives the smallest possible worst case number of measurements.

Label the coins 1-12. Then do the following weighings:
1 4 6 10 against 5 7 9 12
2 5 4 11 against 6 8 7 10
3 6 5 12 against 4 9 8 11

Listing out all the cases shows that these weighings uniquely determine the false coin and whether it is lighter or heavier.

To answer the question as worded the answer to me would be 16 mins. However, the 1 min person will still be on the same side of the bridge, having crossed the bridge with the 10min person, returning which takes 1 min, and then brought the torch back allowing the 5 min (and 2 min) person to cross with the torch.

If the 4 people are to remain on the opposite side of the bridge, would not the answer be 19 mins, as the 1 min person has to come back twice taking an additional 2 mins, to accomany the others on their respective trips. Please explain how 17 min is correct

they each ask the guy below them what color their hat is. then that guy tells them and that's what they right down. everyone lives except for the first guy.

4 people are on, say, the left side of the bridge, and we want to bring all of them over to the right side.
They take 10, 5, 2, 1 minutes to cross, and only 1 or 2 can cross at once, and they need to carry the torch.

Person 1 and Person 2 cross to right side. +2 minutes have elapsed.

Person 1 crosses back to left side. +1 minutes cost.

Person 1 hands the torch over to Person 10 and Person 5, who cross the bridge together. +10 minutes cost.

Person 5 hands the torch over to Person 2, who crosses the bridge again (to the left). +2 minute cost.

Person 2 and Person 1 cross the bridge together to the right. +2 minute cost.

Adding them gives us 17 minutes.

http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle
1. basically what it says is, there is a finite number of ways to sequence the hats.
2. the first person looks to ppl above him and make a best guess what the sequence is.
3. the second person knowing the result of the first person deduces what the remaining possible sequences are, and takes his best guess at it.
4. repeat the result and but deduction, eventually the entire sequencing of hats have been revealed. after n prisoners.
5. n+1'st prisoner and onwards, now knowing the correct sequence, will be able to correctly determine his hat.
6. thus a finite number (even though this number can approach infinite) of prisoners will die, and an infinite number will survive.

There are three doors (or cups, or boxes, or whatever) that you can choose from. You can only choose one. One of the doors, when opened, will give you a prize. The other two give you nothing.

Once you have chosen a door, one of the two remaining doors is opened to reveal that it contains nothing.

You can now choose to keep your first pick, or switch your pick to the other remaining closed door.

Should you keep your first pick, or swap? Why?

You should swap.

The probability that you correctly chose on your first pick is 1/3. Obvious; there are three doors, and only one of them is correct.

Now, the probability that the prize lies behind the remaining two doors is 2/3. This probability does not change when one of those doors is opened. P(revealed door) + P(remaining door) = 2/3. Except that we know P(revealed door) = 0, so P(remaining door) = 2/3. And P(selected door) = 1/3, still. So swapping gives you 2/3, and keeping gives you 1/3 chance to win.

Ok here is the solution. I hope it is satisfactory . It's supposed to be more of a demonstration of a cool paradox than a puzzle.

Represent a sequence of hats by 1s and 0s, with 1s being black and 0s being white. Then the judges input is going to be an infinite sequence of 1s and 0s. Call two sequences "almost the same" if, after a finite number of initial entries, they become the same. For example, two almost the same sequences could be different in the billionth place but never differ afterwards.

Now, all of the possible infinite sequences can be partitioned into classes, where any two elements of the same class are almost the same. The prisoners all get together and choose one special sequence out of each class.

When a given prisoner sees the hats in front of him, he can immediately tell the class of the sequence the judge has chosen, because that does not depend on the hats before him. He then guesses that his hat is the one from the chosen sequence out of that class.

Now, the chosen sequence must eventually agree entirely with the judges sequence, so from a certain point onwards all prisoners will be set free.

The solution requires the prisoners to be infinitely fast and have infinite memory, which I can imagine.

I'm having a problem with dividing infinity into a finite number of classes, though, which is what the solution seems to require. Maybe that's why I fail at math

If there are infinitely many classes, how do the prisoners all choose the correct one? There must be some relation (function) that maps each sequence into exactly one class. Is such a function definable? I'm guessing it will also take an infinite amount of memory and time to solve, too, but that's no problem here... I guess if we could define some such function, this problem would be solved. I can't quite grasp it though. Any examples to make it easier to imagine?

Ok here is the solution. I hope it is satisfactory . It's supposed to be more of a demonstration of a cool paradox than a puzzle.

Represent a sequence of hats by 1s and 0s, with 1s being black and 0s being white. Then the judges input is going to be an infinite sequence of 1s and 0s. Call two sequences "almost the same" if, after a finite number of initial entries, they become the same. For example, two almost the same sequences could be different in the billionth place but never differ afterwards.

Now, all of the possible infinite sequences can be partitioned into classes, where any two elements of the same class are almost the same. The prisoners all get together and choose one special sequence out of each class.

When a given prisoner sees the hats in front of him, he can immediately tell the class of the sequence the judge has chosen, because that does not depend on the hats before him. He then guesses that his hat is the one from the chosen sequence out of that class.

Now, the chosen sequence must eventually agree entirely with the judges sequence, so from a certain point onwards all prisoners will be set free.

Split the coins into 3 groups of 4 as you normally would. I label them a1, a2, a3, a4, b1, etc. Start by weighing a1-4 against b1-4.

If the scale balances: weigh c1:c2, then c1:c3. If both balance the oddball is c4, if both tip it's c1, if just one of them does it's c2 or c3 as appropriate.

If it doesn't balance on the initial weighing: weigh a1a2b1:a3a4b2. If it balances now, the odd coin is either b3 or b4, by weighing them against each other and taking into account which direction the scale tilted earlier, you can find which one.

If on the a1a2b1:a3a4b2 weighing it tips the same direction as the first time, the odd coin is a1, a2, or b2. Weigh a1:a2, if they balance it's b2, otherwise you can decide by the way it leaned earlier as with b3/4 above. If a1a2b1-a3a4b2 tips the opposite way from the original a1-4:b1-4, the odd coin is one of a3, a4, b2, so you weigh a3:a4 and proceed as above.

Hopefully I actually typed that all up correctly, it worked out when I did it on paper.

Each sequence representing its class is identical to every other sequence in its class, except perhaps for the first m elements. The first m people may therefore die. However, every person after person m will definitely live.

Where m is, is irrelevant, because it's a finite number.

But then again, I keep running into this idea:

Every ith person can correctly choose the class that represents the sequence correctly ("correctly enough"), except perhaps his own. Iterating this over i from 0 to N will still give us 1/2 survival? I can't reconcile it.