Pirate democracy at its finestPirates and gold coinsThe coolest checkerboard magic trickThe Trickster's GamePirate Puzzle Double TwistIf 27 x 147 = 3969 What is its square root WITHOUT using complex / long calculationx
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Pirate democracy at its finest
Pirates and gold coinsThe coolest checkerboard magic trickThe Trickster's GamePirate Puzzle Double TwistIf 27 x 147 = 3969 What is its square root WITHOUT using complex / long calculationx
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited May 23 at 12:26
Glorfindel
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
$endgroup$
add a comment |
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited May 23 at 12:26
Glorfindel
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
$endgroup$
add a comment |
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited May 23 at 12:26
Glorfindel
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
$endgroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
strategy combinatorics
edited May 23 at 12:26
Glorfindel
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
edited May 23 at 12:26
Glorfindel
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
edited May 23 at 12:26
Glorfindel
16.7k46294
edited May 23 at 12:26
Glorfindel
16.7k46294
edited May 23 at 12:26
Glorfindel
16.7k46294
16.7k46294
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
asked May 23 at 12:07
Thomas BlueThomas Blue
2,8951851
2,8951851
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
$endgroup$
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
add a comment |
$begingroup$
If you believe @Glorfindel's answer:
489 men have been killed and the man with #490 is next up. Before the votes are cast, the man holding #745 (the # before the next stable point) walks up to the Captain. He says, "If you guarantee me you won't vote to kill me when my turn comes, I'll vote to kill off #490 here." The Captain thinks for a second and says, "If I agree, what makes you think I'll keep my word when your number comes?" #745 replies, "whatever you stand to make from killing me, I will pay you one more. I'd even double it." The captain replies, "That's... actually a really good deal for me, isn't it?"
But then...
The man holding #744 overhears this conversation and chimes in, "No, wait... I'll give you more than what you stand to make in that deal with #745!" Before long most of the crew is shouting at each other trying to cut deals with the captain and everyone else in earshot. The captain fires his pistol in the air and everybody goes quiet. "Hey #490," the Captain calls out. #490 is currently standing on the plank awaiting the vote. "Aren't you the guy that cleans the toilets?" "Oh, aye, Captain! The smell doesn't bother me so much since I lost most of my nose off the coast of Zanzibar." "Right," said the Captain, "let's just call it here, then. Break out the mead, everybody's drinking for 1.957 people!"
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
$endgroup$
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
add a comment |
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
$endgroup$
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
add a comment |
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
$endgroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
answered May 23 at 12:25
GlorfindelGlorfindel
16.7k46294
16.7k46294
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
add a comment |
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
$begingroup$
This is the same exact problem (actually off-by one due to the voting rules) of Pirates and gold coins with 0 coins. This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
$endgroup$
– 12345ieee
May 23 at 20:39
10
10
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
$begingroup$
It's faster to break out yer musket than compute that...
$endgroup$
– smci
May 24 at 23:55
add a comment |
$begingroup$
If you believe @Glorfindel's answer:
489 men have been killed and the man with #490 is next up. Before the votes are cast, the man holding #745 (the # before the next stable point) walks up to the Captain. He says, "If you guarantee me you won't vote to kill me when my turn comes, I'll vote to kill off #490 here." The Captain thinks for a second and says, "If I agree, what makes you think I'll keep my word when your number comes?" #745 replies, "whatever you stand to make from killing me, I will pay you one more. I'd even double it." The captain replies, "That's... actually a really good deal for me, isn't it?"
But then...
The man holding #744 overhears this conversation and chimes in, "No, wait... I'll give you more than what you stand to make in that deal with #745!" Before long most of the crew is shouting at each other trying to cut deals with the captain and everyone else in earshot. The captain fires his pistol in the air and everybody goes quiet. "Hey #490," the Captain calls out. #490 is currently standing on the plank awaiting the vote. "Aren't you the guy that cleans the toilets?" "Oh, aye, Captain! The smell doesn't bother me so much since I lost most of my nose off the coast of Zanzibar." "Right," said the Captain, "let's just call it here, then. Break out the mead, everybody's drinking for 1.957 people!"
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
$endgroup$
add a comment |
$begingroup$
If you believe @Glorfindel's answer:
489 men have been killed and the man with #490 is next up. Before the votes are cast, the man holding #745 (the # before the next stable point) walks up to the Captain. He says, "If you guarantee me you won't vote to kill me when my turn comes, I'll vote to kill off #490 here." The Captain thinks for a second and says, "If I agree, what makes you think I'll keep my word when your number comes?" #745 replies, "whatever you stand to make from killing me, I will pay you one more. I'd even double it." The captain replies, "That's... actually a really good deal for me, isn't it?"
But then...
The man holding #744 overhears this conversation and chimes in, "No, wait... I'll give you more than what you stand to make in that deal with #745!" Before long most of the crew is shouting at each other trying to cut deals with the captain and everyone else in earshot. The captain fires his pistol in the air and everybody goes quiet. "Hey #490," the Captain calls out. #490 is currently standing on the plank awaiting the vote. "Aren't you the guy that cleans the toilets?" "Oh, aye, Captain! The smell doesn't bother me so much since I lost most of my nose off the coast of Zanzibar." "Right," said the Captain, "let's just call it here, then. Break out the mead, everybody's drinking for 1.957 people!"
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
$endgroup$
add a comment |
$begingroup$
If you believe @Glorfindel's answer:
489 men have been killed and the man with #490 is next up. Before the votes are cast, the man holding #745 (the # before the next stable point) walks up to the Captain. He says, "If you guarantee me you won't vote to kill me when my turn comes, I'll vote to kill off #490 here." The Captain thinks for a second and says, "If I agree, what makes you think I'll keep my word when your number comes?" #745 replies, "whatever you stand to make from killing me, I will pay you one more. I'd even double it." The captain replies, "That's... actually a really good deal for me, isn't it?"
But then...
The man holding #744 overhears this conversation and chimes in, "No, wait... I'll give you more than what you stand to make in that deal with #745!" Before long most of the crew is shouting at each other trying to cut deals with the captain and everyone else in earshot. The captain fires his pistol in the air and everybody goes quiet. "Hey #490," the Captain calls out. #490 is currently standing on the plank awaiting the vote. "Aren't you the guy that cleans the toilets?" "Oh, aye, Captain! The smell doesn't bother me so much since I lost most of my nose off the coast of Zanzibar." "Right," said the Captain, "let's just call it here, then. Break out the mead, everybody's drinking for 1.957 people!"
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
$endgroup$
If you believe @Glorfindel's answer:
489 men have been killed and the man with #490 is next up. Before the votes are cast, the man holding #745 (the # before the next stable point) walks up to the Captain. He says, "If you guarantee me you won't vote to kill me when my turn comes, I'll vote to kill off #490 here." The Captain thinks for a second and says, "If I agree, what makes you think I'll keep my word when your number comes?" #745 replies, "whatever you stand to make from killing me, I will pay you one more. I'd even double it." The captain replies, "That's... actually a really good deal for me, isn't it?"
But then...
The man holding #744 overhears this conversation and chimes in, "No, wait... I'll give you more than what you stand to make in that deal with #745!" Before long most of the crew is shouting at each other trying to cut deals with the captain and everyone else in earshot. The captain fires his pistol in the air and everybody goes quiet. "Hey #490," the Captain calls out. #490 is currently standing on the plank awaiting the vote. "Aren't you the guy that cleans the toilets?" "Oh, aye, Captain! The smell doesn't bother me so much since I lost most of my nose off the coast of Zanzibar." "Right," said the Captain, "let's just call it here, then. Break out the mead, everybody's drinking for 1.957 people!"
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
answered May 24 at 19:08
Dark ThunderDark Thunder
773114
773114
add a comment |
add a comment |
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