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TruckOSaurus said:
Okay I think I've got something :

We all know what happens if we have only two pirates (Scenario 1) :

4 - 100
5 - 0

It passes on the tie breaker

Now for 3 pirates (Scenario 2) :

3 - 100, 4 - 0, 5 - 0 would get 3 killed but if 3 is killed we go back to scenario 1 so at this time pirate 5 would accept anything over 0. So at 3 pirates the only possible outcome is :

3 - 99
4 - 0
5 - 1

This gets accept by 3 and 5.

Now if we have 4 pirates (scenario 3):

Pirate 2 knows pirate 5 gets 1 piece in scenario 2 and 0 in scenario 1 so he'll be willing to take the 1 piece again . So we get :

2 - 99
3 - 0
4 - 0
5 - 1

This gets accepted in the tie breaker.

Now we get to our actual scenario, 5 pirates :

Since pirate 4 gets nothing in scenarios 2 & 3, he'll be willing to take anything over 0. So we get this offer :

1 - 98
2 - 0
3 - 0
4 - 1
5 - 1

Final answer?

Am starting to think that there is no real answer, but in fact multiple out comes to this. I think this was made for everyone to lose some sleep. XD

Nah, am just joking, but that's how it feels.

My: answer, Pirate 4 get's everything.

 



4 ≈ One

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TruckOSaurus said:
Okay I think I've got something :

We all know what happens if we have only two pirates (Scenario 1) :

4 - 100
5 - 0

It passes on the tie breaker

Now for 3 pirates (Scenario 2) :

3 - 100, 4 - 0, 5 - 0 would get 3 killed but if 3 is killed we go back to scenario 1 so at this time pirate 5 would accept anything over 0. So at 3 pirates the only possible outcome is :

3 - 99
4 - 0
5 - 1

This gets accept by 3 and 5.

Now if we have 4 pirates (scenario 3):

Pirate 2 knows pirate 5 gets 1 piece in scenario 2 and 0 in scenario 1 so he'll be willing to take the 1 piece again . So we get :

2 - 99
3 - 0
4 - 0
5 - 1

This gets accepted in the tie breaker.

Now we get to our actual scenario, 5 pirates :

Since pirate 4 gets nothing in scenarios 2 & 3, he'll be willing to take anything over 0. So we get this offer :

1 - 98
2 - 0
3 - 0
4 - 1
5 - 1

Final answer?

Yeah, that seems probable, it's even better than mine.

 



http://www.vgchartz.com/games/userreviewdisp.php?id=261

That is VGChartz LONGEST review. And it's NOT Cute Kitten DS

TruckOSaurus said:
Okay I think I've got something :

We all know what happens if we have only two pirates (Scenario 1) :

4 - 100
5 - 0

It passes on the tie breaker

Now for 3 pirates (Scenario 2) :

3 - 100, 4 - 0, 5 - 0 would get 3 killed but if 3 is killed we go back to scenario 1 so at this time pirate 5 would accept anything over 0. So at 3 pirates the only possible outcome is :

3 - 99
4 - 0
5 - 1

This gets accept by 3 and 5.

Now if we have 4 pirates (scenario 3):

Pirate 2 knows pirate 5 gets 1 piece in scenario 2 and 0 in scenario 1 so he'll be willing to take the 1 piece again . So we get :

2 - 99
3 - 0
4 - 0
5 - 1


This gets accepted in the tie breaker.

Now we get to our actual scenario, 5 pirates :

Since pirate 4 gets nothing in scenarios 2 & 3, he'll be willing to take anything over 0. So we get this offer :

1 - 98
2 - 0
3 - 0
4 - 1
5 - 1

Final answer?

 

Dude, you almost had it...

Pirate 5 wouldn`t agree with the bolded part `cause he can get the same ammount of money from the next most senior pirate (Pirate 3) and still throw Pirate 2 overboard (remember, they all hate each other). Keep trying `cause you`re almost there!



Quem disse que a boca é tua?

Qual é, Dadinho...?

Dadinho é o caralho! Meu nome agora é Zé Pequeno!

Yeah I didn't think of that, let's adjust then, I'll copy/paste and bold changes :

We all know what happens if we have only two pirates (Scenario 1) :

4 - 100
5 - 0

It passes on the tie breaker

Now for 3 pirates (Scenario 2) :

3 - 100, 4 - 0, 5 - 0 would get 3 killed but if 3 is killed we go back to scenario 1 so at this time pirate 5 would accept anything over 0. So at 3 pirates the only possible outcome is :

3 - 99
4 - 0
5 - 1

This gets accept by 3 and 5.

Now if we have 4 pirates (scenario 3):

Pirate 2 knows pirate 4 gets nothing in scenario 2 so he'll be willing to take 1 piece. So we get :

2 - 99
3 - 0
4 - 1
5 - 0

This gets accepted in the tie breaker.

Now we get to our actual scenario, 5 pirates :

Since pirate 3 & 5 gets nothing in scenario 3, Pirate 1 only has to offer money to these two. So we get this offer :

1 - 98
2 - 0
3 - 1
4 - 0
5 - 1

I seriously hope I haven't forgotten something this time too.



Signature goes here!

TruckOSaurus said:

Yeah I didn't think of that, let's adjust then, I'll copy/paste and bold changes :

We all know what happens if we have only two pirates (Scenario 1) :

4 - 100
5 - 0

It passes on the tie breaker

Now for 3 pirates (Scenario 2) :

3 - 100, 4 - 0, 5 - 0 would get 3 killed but if 3 is killed we go back to scenario 1 so at this time pirate 5 would accept anything over 0. So at 3 pirates the only possible outcome is :

3 - 99
4 - 0
5 - 1

This gets accept by 3 and 5.

Now if we have 4 pirates (scenario 3):

Pirate 2 knows pirate 4 gets nothing in scenario 2 so he'll be willing to take 1 piece. So we get :

2 - 99
3 - 0
4 - 1
5 - 0

This gets accepted in the tie breaker.

Now we get to our actual scenario, 5 pirates :

Since pirate 3 & 5 gets nothing in scenario 3, Pirate 1 only has to offer money to these two. So we get this offer :

1 - 98
2 - 0
3 - 1
4 - 0
5 - 1

I seriously hope I haven't forgotten something this time too.

 

YES! That`s it! Any other answer isn`t good enough `cause it either they rather throw him overboard or they know they can get more money!\

Okay I still have one more problem but well... is too damn hard!

I know the answer but not the logic of it. A friend of mine (that has an actual degree in math) tried to explain to me once but I just didn`t get it! I`ll post it later for you guys to see if maybe one of you can explain it to me...



Quem disse que a boca é tua?

Qual é, Dadinho...?

Dadinho é o caralho! Meu nome agora é Zé Pequeno!

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I've got a new one :

You've got four people that need to cross a bridge : A sprinter, a regular guy, a fat guy and an old grandma. However the bridge is not in good shape so only two people can cross it at once. It is also pitch black and you need a torch to cross the bridge. Each pair can only go as fast as the slowest of the two.

The sprinter can cross the bridge in 1 minute
The regular guy can cross the bridge in 2 minutes
The fat guy can cross the bridge in 5 minutes
The old grandma can cross the bridge in 10 minutes

What's the quickest way to get everyone across?



Signature goes here!

^ Does the need of a torch have anything to do with the answer?? Because I'm trying tot hink...But I think my answer is toooo easy...



 

I seriously think there is no true answer now.... this puzzle will continue until the puzzle's poster answers us....



4 ≈ One

supermariogalaxy said:
^ Does the need of a torch have anything to do with the answer?? Because I'm trying tot hink...But I think my answer is toooo easy...

 

Yes because if the sprinter crosses the bridge with the old lady it's gonna take 10 minutes because they share the torch.



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OK...hmm...The sprinter and regular go...Then the fat guy and grandma go....That's a total of 12 minutes...I doubt I'm right though...