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