fun math problem!

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OK, this problem is about bandits.

There are 10 bandits with 100 pieces of gold. The bandits are marked one through ten. The bandits have priorities: 1)stay alive 2) get the most gold The bandits have to make up a way of splitting up money keeping in mind the two priorities. the first bandit makes a plan first, when the bandits vote on the plan (including himself) if he gets less than half the votes (he can get a tie and the plan would still work) he would die. Concluding that if he gets more than half or ties with the votes the plan is passed and he lives. if the plan is voted down, the other bandits kill the first bandit. then the second bandit has to make a plan. Keep in mind that all the bandits are logical.

Have fun with this, and if you think you have the answer show your work please! and dont answer it if you have already done a problem similar to this, let people have fun! Smile

I will answer questions too. Good Luck!

What is the plan the first bandit can make that will give him the most gold but not die?

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What's the question?

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lol, what can i say, im a blonde.... Smile

the question is what is the plan the first bandit will make to earn the most money but not die?

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I think the question is what is the best strategy for each bandit?-- or maybe assuming that all bandits are logical (as specified) what will the outcome be?

I need a clarification.

I assume that dead bandits don't get a vote, but you say the term "half the votes", does this mean "half the votes of remaining (living) bandits" or "half the votes of the original bandits? I think the first option makes the problem more interesting, but when you specified "<5" it makes it unclear.

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This problem is the same one as this one.

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editing... lol sry

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If that is the question, the first guy wants to stay alive.

I assume that if he took none of the money, and split it between guys 2-6 (the next guys whose life will be at stake) he would get 5 votes to stay alive, but he wouldn;t get any money.

But all bandits being logical, I am pretty sure the first guy doesn't die (although I am not 100% sure yet that the first guy isn't doomed from the start).

Assuming I am correct, the question is whether the first guy can safely keep any money.

I also assume that the 10th bandit will vote down any plan, since if everyone dies before him, he will get all the money.

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you have a very good start!

and sry JoeFX, i dont usually go looking around the whole site for some1 with the same riddle lol, and i can guarantee you i didn't see it and then repost it, i just came home from an after school math club, and learned it there, anyways, some people that only go to the math and science forum may want to figure this one out Smile

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Nah, the tenth pirate will vote for any plan ig he gets as much as a single piece of gold, because if it comes down to him and number nine the vote ties, and number nine takes all the money.

I'd give one gold to number 10, one vote for me, and none to number 9, one vote against me. Not sure about what to give the rest, athough i would probably barter for the support of pirates two three and four.

Edit: or maybe not, number 10 might vote nay and expect a larger cut by number 2.

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you all are thinking well... ur all on the right track.... keep going!

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Pirate 2 might not want to end up in the hotseat, as any majority with an odd number of pirates would tie with one of them dead.

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nothing to be sorry about, latilatilas... I'm sorry if I sounded rude..

anyways my answer is the next one:

nobody dies and the distribution of the money is the next one (#1 being the highest ranked bandit):

#1=8
#2=6
#3=4
#4=2
#5=0
#6=0
#7=0
#8=0
#9=0
#10=80

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nope... why would the second pirate vote for that plan when he knows if he doesn't vote for him, he would die, then he could make a plan and get more money? and im not looking for the rank, but dont give the answer, bcuz u've already figured it out once!! lol gj by the way

u didn't sound rude, and i didn't mean to either... im jsut saying i didn't see it Smile

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nebody want a hint? (i dont want to give one if youd rather figure it out alone)

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I think I got it! (sorry I posted in two places, but I am excited and we haven't chosen one yet)

The first bandit can keep 96 gold coins.

Assuming the other bandits are all perfectly logical and greedy (that is they would prefer 1 coin to 0 coins) the first pirate needs to give 1 coin each to bandit #3, #5, #7 and #9.

She will get a vote from each bandit that received a coin which (with her own vote) will be 5, enough to get the plan ratified.

Does anyone want a proof?

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very good!! do you want me to explain it or would you rather?

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ebrown_p wrote:

I think I got it! (sorry I posted in two places, but I am excited and we haven't chosen one yet)

The first bandit can keep 96 gold coins.

Assuming the other bandits are all perfectly logical and greedy (that is they would prefer 1 coin to 0 coins) the first pirate needs to give 1 coin each to bandit #3, #5, #7 and #9.

She will get a vote from each bandit that received a coin which (with her own vote) will be 5, enough to get the plan ratified.

Does anyone want a proof?

Lets see it.

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I presented a proof in the pirate thread.

http://www.able2know.com/forums/viewtopic.php?p=1054251#1054251

What a great problem!

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very good!! you got it!! that took me awhile... lol and sry i was so slow at coming back, im practing and i got kinda intrigued, and 4get... lol gj!

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fun math problem!

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