Brainteasers

[Johann]

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...