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solution
0, 50, 50, 0, 0
as described in edit above. 4&3 are smart and realize that 50 is the most they can get.
Alternate answer, put all the coins in a pillow case and beat the other 4 pirates with death with them. Hey, I saw Sean Penn do it in Bad Boys!
Number 2: 16 is the minimum
First you make 5 races. You pick the first 3 of each race(because the 3 fastest may be in the same race). Now you pick one winner and race against the 4 guys who were 3rd.And you do this with every winner.if a 3rd beats a 1st you switch them.you do the same with the 2nds and one final race with all the 1sts.
EDIT:i dont really get number 1.whats the point there?get all the money?if not, just give 20 coins to each pirate and thats it.
nope
Races
Round 1 - 5 races - take 3 fastest from each leaves 15 runners.
Round 2 - 3 Races - take 3 fastest runners leaves 9 runners.
Round 3 - 2 Races - take 3 fastest runners leaves 6 runners.
Round 4 - 1 Race with 5 of the 6
Round 5 - 1 Race with 3 from the previous round plus the other man.
Gives 12 Races.
It has to be worked backwards (rewording of previous post)
If there are 2 left, P2 will definately get all the money leaving p1 with nothing.
Because of this, if there are 3 left, P1 will vote even if he has just one coin (leaving p2 with nothing).
Because of this, if there are 4 left, p2 will vote even if he has just coin (leaving p1 and p3 with nothing).
Therefore, with 5 left, all p5 needs to do is give P1 and p3 a coin each as if you die, they will definately get nothing anyway (it will be settled in the scenario above)
@J'Pol, that is the same as my first post but in the next one I got it down to 11 races by a "top 3 stay in and the last 2 go out and repeat" method.
you are getting close on solving the first question, work back wards
no one is close on the second :D
Chevy, if 3 votes "no", 5 will die.Quote:
Originally posted by Chevy@24 April 2004 - 22:23
It has to be worked backwards (rewording of previous post)
If there are 2 left, P2 will definately get all the money leaving p1 with nothing.
Because of this, if there are 3 left, P1 will vote even if he has just one coin (leaving p2 with nothing).
Because of this, if there are 4 left, p2 will vote even if he has just coin (leaving p1 and p3 with nothing).
Therefore, with 5 left, all p5 needs to do is give P1 and p3 a coin each as if you die, they will definately get nothing anyway (it will be settled in the scenario above)
@J'Pol, that is the same as my first post but in the next one I got it down to 11 races by a "top 3 stay in and the last 2 go out and repeat" method.
In the next round, 3 and 4 will split the pot 50/50.
So, I think that it is 98, 0, 0, 1, 1.
Because if 5 dies 4&3 will take all the coins.
Getting 1 is all 1&2 can hope for.
Read the scenarios for the amount of pirate left I mentioned. You always get the vote of the person who would be guarenteed nothing before. THis is the case for however many are left and the others can buy just that one pirate with a coin. As the 5th, you have to buy 2 but they are the 2 who would get nothing anyway if you die.Quote:
Originally posted by hobbes@24 April 2004 - 21:33
Getting 1 is all 1&2 can hope for.
I got the final answer for number 2,and dont tell me its wrong,because it isnt :P
5+3+1=9
5 races again.pick 3 from each.now make 3 races:1sts vs 1sts;2nds vs 2nds and 3rds vs 3rds.from here you get the top 5,and besides you know who is the fastest guy.he doesnt even have to run the last race.
what if the 3 fastest happened to be in the same race at the beginning?Quote:
Originally posted by yonki@24 April 2004 - 21:37
I got the final answer for number 2,and dont tell me its wrong,because it isnt :P
5+3+1=9
5 races again.pick 3 from each.now make 3 races:1sts vs 1sts;2nds vs 2nds and 3rds vs 3rds.from here you get the top 5,and besides you know who is the fastest guy.he doesnt even have to run the last race.