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Seeing the Pirate's game posted (thanks magicalcheese!) reminded me of my classes in polysci/game theory, so I thought I'd take a shot at making Pirates 3. So here's my spin:

So after some more looting and plundering and general pirating, the pirates have picked up an additional member (making for 7 pirates total) and now they each have 100 gold coins.

Somehow (I highly suspect there was rum involved ;) ), they agreed to allow the captain to levy a "pirate tax". The captain can choose tax each crew member a different amount, but the proposed tax scheme (like all proposed tax schemes) must be voted upon in the usual pirate format.

(For those of you who missed the previous pirates strings created by magicalcheese, here are the rules: The captain must obtain approval (yes votes) from at least half his crew (including himself) or else there will be a mutiny, the current captain will be forced to walk the plank, and the next highest ranked member of the crew will become the new captain, and the process will repeat until a proposal passes.

However there is one extra catch in this case: for the pirates who vote "no" on a proposal that passes, the amount of tax they have to pay is the proposed amount divided by (the number of pirates who vote no+1), rounded to the nearest whole coin. For example, if out of 7 pirates, if three of them are taxed 100 gold coins and vote no, then they each have to pay 100/(3+1)=25 gold coins to the captain.

As before, assuming they are all really smart (and relatively sober during the voting...;)), and really greedy, what is the maximum amount the captain can get out of taxes without risking his life?

Remember, the captain is not only trying to get enough yes votes to pass, he is trying to get the maximum number of gold coins as well.

I just tried working this out myself, but maybe someone will find a better answer than I did...

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I'm going to say he doesn't get anything. If the crew is really greedy, then they'll conspire to all say no to any tax amount, hence causing a dismal demise for the captain. The captain, knowing this could happen must set his taxes to zero. If anyone is stupid enough to say no, then the captain gets a bonus from the no-sayers.

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Also, the wording "The captain must obtain approval (yes votes) from at least half his crew (including himself)" implies that a tie goes to the captain if there are an even number of pirates. Is that the case?

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I say he taxes 4/7 pirates 100 coins each, and agrees to split the "revenue" with the remaining three, whom he taxes nothing, ensuring their 'aye' votes. He casts the tying vote. The four unlucky pirates hand over the loot and the three pirates plus captain split the 400 coins. Do I win a sticker?

- C money

Crap. There are 7 total pirates. So he taxes 3/7 pirates the full 100 coins, ensuring the votes of the remaining 4 (including himself). Do I still win a sticker?

Edited by cmoney
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Trying to examine things from bottom up, starting with 2 pirates, but what is a pirate's preference if there is a tie?

Pirates labeled 1 to 5, 1 being most senior(captain). In the case with only two pirates, only 4 and 5 remain. Four doesn't need 5 to agree for the vote to pass, so he will tax 100 coins from 5. Five will oppose the vote and keep 50 coins.

Case with 3 pirates:

Now pirate 3 is in charge and he needs the support of one other pirate and this is where I have a question. Pirate 5 knows that the most gold he can get is 50, so if pirate 3 offers to tax pirate 4 100 gold coins and pirate 5 only 50 gold coins, will pirate 5 accept? Does he care if pirate 3 lives? Either way, pirate 5 will only get 50 pieces of gold.

As I see it, whether ties matter or not changes the answer.

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Captain gets 285 coins.

Here's my reasoning: the captain wants to avoid getting hanged, and if any other pirate were captain he would as well. Every pirate that's not the captain wants to do better than they would if that captain were hanged, and the next pirate becomes captain. This means that you can reason up from a two pirate situation. Assign them numbers 1-7 with 1 being the current captain, and 7 being the lowest rank:

2 pirates: 7 will always vote no and lose the tie to 6, so he will be taxed 100 and pay 100/(1+1)=50 to maximize 6's revenue.

3 pirates: 5 will always vote yes, 7 will vote yes as long as he does better than he would if 6 were captain, so he votes yes on 49. To max the payment from 6 (as he is the only no vote) he will pay 50.

4 pirates: 4 votes yes, 7 votes yes if less than 49, and 6 votes yes if less than 50. The most 4 can get as captain is then 50 + 49 + 48 = 147.

Continuing this reason, the current captain maxes out at 50 + 49 + ... + 45 = 285, where everyone votes "yes" except 2.

Does that make sense? I feel like this would be odd because every pirate can improve their situation by voting no until there is enough for mutiny.

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keeping in mind that the pirates are greedy, there are 2 things each pirate will want. one is to keep/get the most coins. the second is to be captain or closest to being captain (so they can get more coins the next time).

work backwards (in each scenario, the higher number pirate is the lower ranking pirate):

2 pirates: captain will tax pirate 2 100 coins and will vote yes, pirate 2 will vote no. captain wins and gets 50 coins (100/(1+1)).

3 pirates: captain taxes both 100, both vote know captain dies. no good.

captain taxes pirate 2 100 coins and pirate 3 49 coins. pirate 3 is happy (keeps more coins than if there were 2 pirates) and votes yes along with captain. captain gets 50+49=99 coins.

4 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins and pirate 4 48 coins. captain and pirate 4 vote yes. captain can't tax pirate 4 49 coins because then pirate 4 has no reason to vote yes this time (they could vote no and now be 3rd in line for captaincy instead of 4th and keep the same number of coins). captain gets 33+33+48=114 coins

5 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins, pirate 4 32 coins and pirate 5 47 coins. captain, pirate 4 and 5 vote yes. captain gets 33 + 33 + 32 + 47 = 135 coins.

6 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 31 coins and pirate 6 46 coins. still gets 3 yes votes and gets 25+25+25+31+46=152 coins

7 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 24 coins, pirate 6 30 coins and pirate 7 45 coins. gets 4 yes votes and 25+25+25+24+30+45=174 coins.

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Captain gets 285 coins.

Here's my reasoning: the captain wants to avoid getting hanged, and if any other pirate were captain he would as well. Every pirate that's not the captain wants to do better than they would if that captain were hanged, and the next pirate becomes captain. This means that you can reason up from a two pirate situation. Assign them numbers 1-7 with 1 being the current captain, and 7 being the lowest rank:

2 pirates: 7 will always vote no and lose the tie to 6, so he will be taxed 100 and pay 100/(1+1)=50 to maximize 6's revenue.

3 pirates: 5 will always vote yes, 7 will vote yes as long as he does better than he would if 6 were captain, so he votes yes on 49. To max the payment from 6 (as he is the only no vote) he will pay 50.

4 pirates: 4 votes yes, 7 votes yes if less than 49, and 6 votes yes if less than 50. The most 4 can get as captain is then 50 + 49 + 48 = 147.

Continuing this reason, the current captain maxes out at 50 + 49 + ... + 45 = 285, where everyone votes "yes" except 2.

Does that make sense? I feel like this would be odd because every pirate can improve their situation by voting no until there is enough for mutiny.

chuck, in your 4 pirate example, pirate 6 could vote no and pay less so he isn't playing optimally.

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keeping in mind that the pirates are greedy, there are 2 things each pirate will want. one is to keep/get the most coins. the second is to be captain or closest to being captain (so they can get more coins the next time).

work backwards (in each scenario, the higher number pirate is the lower ranking pirate):

2 pirates: captain will tax pirate 2 100 coins and will vote yes, pirate 2 will vote no. captain wins and gets 50 coins (100/(1+1)).

3 pirates: captain taxes both 100, both vote know captain dies. no good.

captain taxes pirate 2 100 coins and pirate 3 49 coins. pirate 3 is happy (keeps more coins than if there were 2 pirates) and votes yes along with captain. captain gets 50+49=99 coins.

4 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins and pirate 4 48 coins. captain and pirate 4 vote yes. captain can't tax pirate 4 49 coins because then pirate 4 has no reason to vote yes this time (they could vote no and now be 3rd in line for captaincy instead of 4th and keep the same number of coins). captain gets 33+33+48=114 coins

5 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins, pirate 4 32 coins and pirate 5 47 coins. captain, pirate 4 and 5 vote yes. captain gets 33 + 33 + 32 + 47 = 135 coins.

6 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 31 coins and pirate 6 46 coins. still gets 3 yes votes and gets 25+25+25+31+46=152 coins

7 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 24 coins, pirate 6 30 coins and pirate 7 45 coins. gets 4 yes votes and 25+25+25+24+30+45=174 coins.

In line with the rest pirate riddles, it seems that the lowest ranked pirate (pirate 7)will always accept a tax rule that'll have him pay 50 or less. Hence, we never have to reduce it down to 49 to make it an optimal deal since 50 is optimal for him. I think this is a valid assumption since that's how the original pirate riddle was solved (or was it solved that way since you can't go below 1 coin, which is the amount that the last two pirates end up getting in that riddle). Anyways, since there is no desire to see other pirates die, we can safely assume the above is true.

So that said, there is still another ambiguity. The key to this puzzle is realizing that one guy will always say NO (the one ranked currently below the captain). To get the most coins out of this 2nd ranked captain, we have to formulate a strategy that will lead others to say YES, so that n+1 denominator stays at 2. In other words, every pirate besides the 2nd ranked one should be indifferent to saying yes or no to the tax rule. And here is where the ambiguity lies...best explained through example. Take 4 pirates.

Situation: Paid (taxed) by others

4 pirates: 50 (100), 33 (33) , 50 (50).

Now, pirate # 2 is taxed 33. If he says NO, he still pays 33. If he says YES, he still pays 33. It is not clear that whether the pirate will choose to act in favor of the Captain granted his benefit is maxed out.

If we assume he does, then the answer I believe is:

Captain gets 100 (his own) + 216 coins (tax) = 316.

Situation: Paid (taxed) (pirate 2,3,4...)

2 pirates: 50 (100)

3 pirates: 50 (100), 50 (50

4 pirates: 50 (100), 33 (33), 50 (50)

5 pirates: 50 (100), 33 (33), 33 (33), 50 (50)

6 pirates: 50 (100), 25 (25), 25 (25), 33 (33), 50 (50)

7 pirates: 50 (100), 25 (25), 25 (25), 33 (33), 33 (33), 50 (50)

If we assume that the pirate will not act in favor of the captain when his benefit is maxed out, then we need to give him an extra coin. Make his share 34 in the 4 pirate case. Adjust rest accordingly in rest of cases.... already too long a post. Note that only pirates 4 and 5 need to be adjusted. Pirates 2 and 3 are already paying the lowest.

* EDIT

Realized my mistake in logic. Pirates pay [whatever is proposed]/[n+1].

In that case, the answer is:

Captain gets 100 (his own) + 183 coins (tax) = 283.

Situation: Paid (taxed) (pirate 2,3,4...)

2 pirates: 50 (100)

3 pirates: 50 (100), 50 (50)

4 pirates: 33 (100), 33 (100), 50 (50)

5 pirates: 33 (100), 33 (100), 33 (33), 50 (50)

6 pirates: 25 (100), 25 (100), 25 (100), 33 (33), 50 (50)

7 pirates: 25 (100), 25 (100), 25 (100), 25 (25), 33 (33), 50 (50)

Edited by Suicide
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Also, the wording "The captain must obtain approval (yes votes) from at least half his crew (including himself)" implies that a tie goes to the captain if there are an even number of pirates. Is that the case?

Yep, ties go to the captain. Same voting rules as the previous pirates (if you haven't looked at them, you may want to since it will probably help you figure out the solution).

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Trying to examine things from bottom up, starting with 2 pirates, but what is a pirate's preference if there is a tie?

Pirates labeled 1 to 5, 1 being most senior(captain). In the case with only two pirates, only 4 and 5 remain. Four doesn't need 5 to agree for the vote to pass, so he will tax 100 coins from 5. Five will oppose the vote and keep 50 coins.

Case with 3 pirates:

Now pirate 3 is in charge and he needs the support of one other pirate and this is where I have a question. Pirate 5 knows that the most gold he can get is 50, so if pirate 3 offers to tax pirate 4 100 gold coins and pirate 5 only 50 gold coins, will pirate 5 accept? Does he care if pirate 3 lives? Either way, pirate 5 will only get 50 pieces of gold.

As I see it, whether ties matter or not changes the answer.

Same order of preference as pirates 2:

Living

Getting money (or in this case, losing less money)

Seeing someone else die (human nature...)

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Captain gets 285 coins.

Here's my reasoning: the captain wants to avoid getting hanged, and if any other pirate were captain he would as well. Every pirate that's not the captain wants to do better than they would if that captain were hanged, and the next pirate becomes captain. This means that you can reason up from a two pirate situation. Assign them numbers 1-7 with 1 being the current captain, and 7 being the lowest rank:

2 pirates: 7 will always vote no and lose the tie to 6, so he will be taxed 100 and pay 100/(1+1)=50 to maximize 6's revenue.

3 pirates: 5 will always vote yes, 7 will vote yes as long as he does better than he would if 6 were captain, so he votes yes on 49. To max the payment from 6 (as he is the only no vote) he will pay 50.

4 pirates: 4 votes yes, 7 votes yes if less than 49, and 6 votes yes if less than 50. The most 4 can get as captain is then 50 + 49 + 48 = 147.

Continuing this reason, the current captain maxes out at 50 + 49 + ... + 45 = 285, where everyone votes "yes" except 2.

Does that make sense? I feel like this would be odd because every pirate can improve their situation by voting no until there is enough for mutiny.

Well...right method of thinking about it (good), but...

For the 4 pirate case, if pirate 7 knows that pirates 4 and 6 are going to say yes, there is no reason for him to vote yes as well, since it only needs two votes to pass. Hence he would vote no, and in that case both his and pirate 5 only have to pay 1/3 of what the tax was.

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keeping in mind that the pirates are greedy, there are 2 things each pirate will want. one is to keep/get the most coins. the second is to be captain or closest to being captain (so they can get more coins the next time).

work backwards (in each scenario, the higher number pirate is the lower ranking pirate):

2 pirates: captain will tax pirate 2 100 coins and will vote yes, pirate 2 will vote no. captain wins and gets 50 coins (100/(1+1)).

3 pirates: captain taxes both 100, both vote know captain dies. no good.

captain taxes pirate 2 100 coins and pirate 3 49 coins. pirate 3 is happy (keeps more coins than if there were 2 pirates) and votes yes along with captain. captain gets 50+49=99 coins.

4 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins and pirate 4 48 coins. captain and pirate 4 vote yes. captain can't tax pirate 4 49 coins because then pirate 4 has no reason to vote yes this time (they could vote no and now be 3rd in line for captaincy instead of 4th and keep the same number of coins). captain gets 33+33+48=114 coins

5 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins, pirate 4 32 coins and pirate 5 47 coins. captain, pirate 4 and 5 vote yes. captain gets 33 + 33 + 32 + 47 = 135 coins.

6 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 31 coins and pirate 6 46 coins. still gets 3 yes votes and gets 25+25+25+31+46=152 coins

7 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 24 coins, pirate 6 30 coins and pirate 7 45 coins. gets 4 yes votes and 25+25+25+24+30+45=174 coins.

That's close to what I got, except in the 4 step, the captain would be better off taxing pirate 6(or what you call pirate 3 in that step) 49 coins (since pirate 6 would have to pay 50 coins if the vote failed and it went down to the 3 pirate case) then taxing pirate 7 (your pirate 4) 48. And then the rest of my reasoning goes from there.

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do all pirates vote simultaneously?....or do they vote by seniority?

I don't think it really matters, since they are all game theory geniuses who can figure out exactly how everyone else is going to vote based on the proposed tax scheme...

The way the vote is going to should already be known before the vote takes place. The voting itself is just a formality...

Actually I was thinking about that, too, but the way I worked it out, whether the vote is simultaneous or by seniority may affect who gets taxed what amount but it won't affect the amounts that are taxed and the total amount the captain gets, which is what the question is asking for.

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That's close to what I got, except in the 4 step, the captain would be better off taxing pirate 6(or what you call pirate 3 in that step) 49 coins (since pirate 6 would have to pay 50 coins if the vote failed and it went down to the 3 pirate case) then taxing pirate 7 (your pirate 4) 48. And then the rest of my reasoning goes from there.

in the 4 step, pirate 6 (3) knows that the captain and pirate 7 (4) are going to vote yes, so to save the most coins pirate 6 (3) will always vote no. so if the captain only taxes pirate 6 (3) 49 coins, pirate 6 (3) would still vote no and only owe 49/(2+1)=16 coins.

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in the 4 step, pirate 6 (3) knows that the captain and pirate 7 (4) are going to vote yes, so to save the most coins pirate 6 (3) will always vote no. so if the captain only taxes pirate 6 (3) 49 coins, pirate 6 (3) would still vote no and only owe 49/(2+1)=16 coins.

Actually, I meant that pirates 4 and 6 are going to vote yes, and pirates 5 and 7 are going to vote no. The tax scheme would be: 5 (100), 6 (49), and 7 (100). Pirates 5 and 7 vote no, and each have to pay 33. Pirate 6 votes yes, since he knows that if he votes no they would go down to the the 3 pirate case he would have to pay 50. So pirate 4 (the captain in this case) would get 33+49+33=115 instead of 33+33+48=114 coins.

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Actually, I meant that pirates 4 and 6 are going to vote yes, and pirates 5 and 7 are going to vote no. The tax scheme would be: 5 (100), 6 (49), and 7 (100). Pirates 5 and 7 vote no, and each have to pay 33. Pirate 6 votes yes, since he knows that if he votes no they would go down to the the 3 pirate case he would have to pay 50. So pirate 4 (the captain in this case) would get 33+49+33=115 instead of 33+33+48=114 coins.

so adjusting, i get 176 coins?

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Same order of preference as pirates 2:

Living

Getting money (or in this case, losing less money)

Seeing someone else die (human nature...)

[spoiler='

Well, this changes my previous answer']Apparently I was wrong in assuming that since no desire to see others die is mentioned, we can assume the last pirate will be happy with 50. I guess I didn't realize "seeing someone else die" is human nature...I thought it was in"human"e...haha, jk. Anyways, in that case my new answer:

Situation: Paid (taxed)

Order: [...pirate 5, pirate 6, pirate 7]

2 pirates(6&7): 50(100)

3 pirates(5-7): 50(100), 49(49)

4 pirates(4-7): 33(100), 49(49), 33(100)

5 pirates(3-7): 33(100), 33(100), 48(48), 32(32)

6 pirates: 25(100), 25(100), 25(100), 47(47), 31(31)

7 pirates: 25(100), 25(100), 25(100), 24(24), 46(46), 30(30)

Captain gets: 100 (own) + 175 = 275 coins...

4 pirate case is funny...if you tax pirate 7 48 coins, the captain gets 33+33+48. But if you tax pirate 6 49 instead of 100 and pirate 7 100, then the captain gets 33+49+33. One more coin...woo hoo...

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[spoiler='

Well, this changes my previous answer']Apparently I was wrong in assuming that since no desire to see others die is mentioned, we can assume the last pirate will be happy with 50. I guess I didn't realize "seeing someone else die" is human nature...I thought it was in"human"e...haha, jk. Anyways, in that case my new answer:

Situation: Paid (taxed)

Order: [...pirate 5, pirate 6, pirate 7]

2 pirates(6&7): 50(100)

3 pirates(5-7): 50(100), 49(49)

4 pirates(4-7): 33(100), 49(49), 33(100)

5 pirates(3-7): 33(100), 33(100), 48(48), 32(32)

6 pirates: 25(100), 25(100), 25(100), 47(47), 31(31)

7 pirates: 25(100), 25(100), 25(100), 24(24), 46(46), 30(30)

Captain gets: 100 (own) + 175 = 275 coins...

4 pirate case is funny...if you tax pirate 7 48 coins, the captain gets 33+33+48. But if you tax pirate 6 49 instead of 100 and pirate 7 100, then the captain gets 33+49+33. One more coin...woo hoo...

Well, actually, you can apply what you said about 4 pirate case to 6 pirate case as well...

And, well, it *is* human nature...hence the coliseums, public hangings/burnings/beheadings, reality shows... ;P

Edited by Yoruichi-san
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Yeah, that's what I got :) . Yay, I feel validated... ;)

Time to go make Pirates 4...(unless you're all sick of these...)

Without johnny depp it's just not a real prirate theme :D

Bring in jack sparrow

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Well, actually, you can apply what you said about 4 pirate case to 6 pirate case as well...

And, well, it *is* human nature...hence the coliseums, public hangings/burnings/beheadings, reality shows... ;P

Situation: Paid (taxed)

Order: [...pirate 5, pirate 6, pirate 7]

2 pirates(6&7): 50(100)

3 pirates(5-7): 50(100), 49(49)

4 pirates(4-7): 33(100), 49(49), 33(100)

5 pirates(3-7): 33(100), 33(100), 48(48), 32(32)

6 pirates: 25(100), 25(100), 32(32), 47(47), 25(100)

7 pirates: 25(100), 25(100), 25(100), 31(31), 46(46), 24(24)

Captain gets: 100 (own) + 176 = 276 coins...

Looking forward to pirates 4...

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