Brainteasers

[Dgc1808]

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.