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.
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