Three hot dog vendors
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6 members have voted
January 16th, 2019 at 3:14:07 AM
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There is a beach one mile long with people distributed uniformly along the mile. When anyone on the beach gets hungry, he goes to the nearest hot dog stand.
There are three vendors who are all perfect logicians. Each wishes to maximize his own sales. They don't trust each other, so no colluding.
It is known that A will arrive first, then B, and then C. B and C will note the locations of the vendors who arrived before them and locate their stand accordingly.
Where should A set up his stand?
There are three vendors who are all perfect logicians. Each wishes to maximize his own sales. They don't trust each other, so no colluding.
It is known that A will arrive first, then B, and then C. B and C will note the locations of the vendors who arrived before them and locate their stand accordingly.
Where should A set up his stand?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 5:00:22 AM
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I don't think this is the answer you're looking for but this reminded me of an interesting video on the economics of why competitors group close together. And they use a beach as an example.
https://youtu.be/jILgxeNBK_8
If A picks the middle though the most he is going to end up with is only 25% while B and C end up with 37.5%.
https://youtu.be/jILgxeNBK_8
If A picks the middle though the most he is going to end up with is only 25% while B and C end up with 37.5%.
January 16th, 2019 at 5:24:57 AM
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I think the one assumption is that all are perfect logicians. So A should set up 34% (or 33.4, or 33.34, as far as you want to run it out, but just slightly more than 1/3) of the way from the end of the beach in either direction.
If B sets up 35% from the opposite side or more, that forces C to set up to the far end of him, right next to him, so B can set up no closer to the middle than 33%.
C will then set up at the mid-point between them to get both sides rather than an end run, as that will be slightly more advantageous to him than setting up on either end. So A will get the "most" by a very slight amount, but it will be the most.
Not sure how much room you take up for the actual width of the stand, but it could matter in measuring "most" once all 3 are there.
If B sets up 35% from the opposite side or more, that forces C to set up to the far end of him, right next to him, so B can set up no closer to the middle than 33%.
C will then set up at the mid-point between them to get both sides rather than an end run, as that will be slightly more advantageous to him than setting up on either end. So A will get the "most" by a very slight amount, but it will be the most.
Not sure how much room you take up for the actual width of the stand, but it could matter in measuring "most" once all 3 are there.
If the House lost every hand, they wouldn't deal the game.
January 16th, 2019 at 5:34:25 AM
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So if A goes in the middle, B will set up either just to the left or right of him and C just on the other side, leaving A with territory approaching 0. So A needs to go to one side to make B and C move away from him towards the larger territory. This is definitely an economic problem I’ve solved in college (I forget the name, but it’s different if the oligopolies move sequentially or simultaneously.
Let’s say A sets up to the left of center (closer to 0). B will want to make C indifferent about moving just to the left or right of him So B sets up at a distance where 1-B = 0.5(B-A). That’s because C captures all the territory between B and the 1 mile mark and only half of the territory between B and C. Oh but there should also be a boundary condition where 1-B > A or else B would move slightly to the left of A to steal all of that territory between 0 and A. Since A is going to want to make B indifferent about moving to the left and right of him, the boundary condition yields 1-B=A.
So we have two equations and two unknowns.
Manipulating the first equation gets to:
A = 3B - 2
Plug into above for B = 1-A
A = 3(1 - A) - 2
A = 3 - 3A - 2
4A = 1
A = 1/4
I think that works. B then sets up at 3/4. And C is indifferent between just to the right of B, anywhere between A and B or just to the left of A.
Hmm I guess there was another hidden boundary condition where C could cut off A, which would happen if A moved to the right of 1/4.
ETA: I guess A is risking C taking the territory just to his left, which would be bad, so I think technically A should move a “smidge” to the left of 1/4 so that C is disincentivized to do that. Then B should move a smidge to the right of 3/4, making C want to land in the middle of A and B.
Let’s say A sets up to the left of center (closer to 0). B will want to make C indifferent about moving just to the left or right of him So B sets up at a distance where 1-B = 0.5(B-A). That’s because C captures all the territory between B and the 1 mile mark and only half of the territory between B and C. Oh but there should also be a boundary condition where 1-B > A or else B would move slightly to the left of A to steal all of that territory between 0 and A. Since A is going to want to make B indifferent about moving to the left and right of him, the boundary condition yields 1-B=A.
So we have two equations and two unknowns.
Manipulating the first equation gets to:
A = 3B - 2
Plug into above for B = 1-A
A = 3(1 - A) - 2
A = 3 - 3A - 2
4A = 1
A = 1/4
I think that works. B then sets up at 3/4. And C is indifferent between just to the right of B, anywhere between A and B or just to the left of A.
Hmm I guess there was another hidden boundary condition where C could cut off A, which would happen if A moved to the right of 1/4.
ETA: I guess A is risking C taking the territory just to his left, which would be bad, so I think technically A should move a “smidge” to the left of 1/4 so that C is disincentivized to do that. Then B should move a smidge to the right of 3/4, making C want to land in the middle of A and B.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 16th, 2019 at 6:08:48 AM
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Babs, I thinkQuote: beachbumbabsI think the one assumption is that all are perfect logicians. So A should set up 34% (or 33.4, or 33.34, as far as you want to run it out, but just slightly more than 1/3) of the way from the end of the beach in either direction.
If B sets up 35% from the opposite side or more, that forces C to set up to the far end of him, right next to him, so B can set up no closer to the middle than 33%.
C will then set up at the mid-point between them to get both sides rather than an end run, as that will be slightly more advantageous to him than setting up on either end. So A will get the "most" by a very slight amount, but it will be the most.
Not sure how much room you take up for the actual width of the stand, but it could matter in measuring "most" once all 3 are there.
At A=1/3 and B=2/3 then C does best avoiding the middle (where he only gets half of the middle so 1/6) and taking one of the sides, which screws over A, since B will make his side slightly smaller than As
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 16th, 2019 at 7:20:22 AM
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Quote: unJonBabs, I thinkQuote: beachbumbabsI think the one assumption is that all are perfect logicians. So A should set up 34% (or 33.4, or 33.34, as far as you want to run it out, but just slightly more than 1/3) of the way from the end of the beach in either direction.
If B sets up 35% from the opposite side or more, that forces C to set up to the far end of him, right next to him, so B can set up no closer to the middle than 33%.
C will then set up at the mid-point between them to get both sides rather than an end run, as that will be slightly more advantageous to him than setting up on either end. So A will get the "most" by a very slight amount, but it will be the most.
Not sure how much room you take up for the actual width of the stand, but it could matter in measuring "most" once all 3 are there..At A=1/3 and B=2/3 then C does best avoiding the middle (where he only gets half of the middle so 1/6) and taking one of the sides, which screws over A, since B will make his side slightly smaller than As
That makes sense, thanks!
If the House lost every hand, they wouldn't deal the game.
January 16th, 2019 at 7:27:16 AM
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Quote: unJonSo if A goes in the middle, B will set up either just to the left or right of him and C just on the other side, leaving A with territory approaching 0. So A needs to go to one side to make B and C move away from him towards the larger territory. This is definitely an economic problem I’ve solved in college (I forget the name, but it’s different if the oligopolies move sequentially or simultaneously.
Let’s say A sets up to the left of center (closer to 0). B will want to make C indifferent about moving just to the left or right of him So B sets up at a distance where 1-B = 0.5(B-A). That’s because C captures all the territory between B and the 1 mile mark and only half of the territory between B and C. Oh but there should also be a boundary condition where 1-B > A or else B would move slightly to the left of A to steal all of that territory between 0 and A. Since A is going to want to make B indifferent about moving to the left and right of him, the boundary condition yields 1-B=A.
So we have two equations and two unknowns.
Manipulating the first equation gets to:
A = 3B - 2
Plug into above for B = 1-A
A = 3(1 - A) - 2
A = 3 - 3A - 2
4A = 1
A = 1/4
I think that works. B then sets up at 3/4. And C is indifferent between just to the right of B, anywhere between A and B or just to the left of A.
Hmm I guess there was another hidden boundary condition where C could cut off A, which would happen if A moved to the right of 1/4.
ETA: I guess A is risking C taking the territory just to his left, which would be bad, so I think technically A should move a “smidge” to the left of 1/4 so that C is disincentivized to do that. Then B should move a smidge to the right of 3/4, making C want to land in the middle of A and B.
Yes, this is what I get too, and for the same reasons. Congratulations! I now owe you two beers.
However, I think I'm going to put you on the list of math wizards who are on a 24-hour delay for future problems, to give others a chance.
However, I think I'm going to put you on the list of math wizards who are on a 24-hour delay for future problems, to give others a chance.
General comment #1 -- If one of the vendors is indifferent between multiple locations, you may assume he picks randomly.
General comment #2 -- Once a vendor picks a spot, he must stay there the rest of the day.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 7:33:13 AM
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Quote: rudeboyoiI don't think this is the answer you're looking for but this reminded me of an interesting video on the economics of why competitors group close together. And they use a beach as an example.
https://youtu.be/jILgxeNBK_8
If A picks the middle though the most he is going to end up with is only 25% while B and C end up with 37.5%.
Yes, this a classic game theory exercise. However, it a process of many steps. I contend if the three vendors were allowed to relocate, then they would all end up exactly at the half way point, forcing customers to pick a vendor arbitrarily.
I probably should have made a comment that the cart had to remain in the same place once the location was chosen for the rest of the day.
I probably should have made a comment that the cart had to remain in the same place once the location was chosen for the rest of the day.
Last edited by: Wizard on Jan 16, 2019
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 10:05:05 AM
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It's raining, I would call Uber Eats and
set up tomorrow.
set up tomorrow.
"It's not called gambling if the math is on your side."
January 16th, 2019 at 10:07:30 AM
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Quote: WizardQuote: unJonSo if A goes in the middle, B will set up either just to the left or right of him and C just on the other side, leaving A with territory approaching 0. So A needs to go to one side to make B and C move away from him towards the larger territory. This is definitely an economic problem I’ve solved in college (I forget the name, but it’s different if the oligopolies move sequentially or simultaneously.
Let’s say A sets up to the left of center (closer to 0). B will want to make C indifferent about moving just to the left or right of him So B sets up at a distance where 1-B = 0.5(B-A). That’s because C captures all the territory between B and the 1 mile mark and only half of the territory between B and C. Oh but there should also be a boundary condition where 1-B > A or else B would move slightly to the left of A to steal all of that territory between 0 and A. Since A is going to want to make B indifferent about moving to the left and right of him, the boundary condition yields 1-B=A.
So we have two equations and two unknowns.
Manipulating the first equation gets to:
A = 3B - 2
Plug into above for B = 1-A
A = 3(1 - A) - 2
A = 3 - 3A - 2
4A = 1
A = 1/4
I think that works. B then sets up at 3/4. And C is indifferent between just to the right of B, anywhere between A and B or just to the left of A.
Hmm I guess there was another hidden boundary condition where C could cut off A, which would happen if A moved to the right of 1/4.
ETA: I guess A is risking C taking the territory just to his left, which would be bad, so I think technically A should move a “smidge” to the left of 1/4 so that C is disincentivized to do that. Then B should move a smidge to the right of 3/4, making C want to land in the middle of A and B.
That compliment (however unfounded) is worth at least the two beers. Thanks, Wizard!Yes, this is what I get too, and for the same reasons. Congratulations! I now owe you two beers.
However, I think I'm going to put you on the list of math wizards who are on a 24-hour delay for future problems, to give others a chance.
General comment #1 -- If one of the vendors is indifferent between multiple locations, you may assume he picks randomly.
General comment #2 -- Once a vendor picks a spot, he must stay there the rest of the day.
Thanks, Wizard! That compliment (however unfounded) is worth at least two beers. (Side note: it’s three beers)
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 16th, 2019 at 10:48:08 AM
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No matter their locations, there will be some customers equidistant from two or three vendors. How do they decide which stand to visit?
If you are vendor C, wouldn't the perfectly logical tactic be to "bookend" one or both vendors who are already there by setting up your stand just inches from theirs on the longer open side? This would allow you to intercept all of the customers coming from that side since folks only go to the closest stand. This tactic is similar to the last player to bid on "The Price is Right"'s "Contestant's Row" bidding a dollar more than another player.
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 16th, 2019 at 11:38:11 AM
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Extra credit (no prize and I don’t know answer).
Same problem but the beach is a 4 dimensional 1x1x1x1 hypercube and there are 7 vendors.
Same problem but the beach is a 4 dimensional 1x1x1x1 hypercube and there are 7 vendors.
January 16th, 2019 at 3:15:08 PM
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You left out the nationality of the
vendors. For instance, if they're
Egyptian, Iranian, and Jewish, there
might be other problems..
vendors. For instance, if they're
Egyptian, Iranian, and Jewish, there
might be other problems..
"It's not called gambling if the math is on your side."
January 16th, 2019 at 3:44:45 PM
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Quote: AyecarumbaNo matter their locations, there will be some customers equidistant from two or three vendors. How do they decide which stand to visit?
They would pick randomly. Same as a vendor who was indifferent between multiple spots.
Quote: AyecarumbaIf you are vendor C, wouldn't the perfectly logical tactic be to "bookend" one or both vendors who are already there by setting up your stand just inches from theirs on the longer open side? This would allow you to intercept all of the customers coming from that side since folks only go to the closest stand. This tactic is similar to the last player to bid on "The Price is Right"'s "Contestant's Row" bidding a dollar more than another player.
The correct answers assumes they will do that.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 3:47:36 PM
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Quote: RSExtra credit (no prize and I don’t know answer).
Same problem but the beach is a 4 dimensional 1x1x1x1 hypercube and there are 7 vendors.
You are quite the masochist with that one. I was actually thinking of just doing a 4-vendor case on the same mile.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 4:22:48 PM
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Rather than repeat it I with Union's answer but used trial and error assuming the beach was 100m long and you could only setup at a (different) 1m posts and then logic to fine tune it. If A starts in the middle (50) then B goes one side of him (49) and C goes the other (51). As A moves it's interesting how the various decisions change.
January 16th, 2019 at 4:38:08 PM
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Quote: WizardThey would pick randomly. Same as a vendor who was indifferent between multiple spots.
Quote: AyecarumbaIf you are vendor C, wouldn't the perfectly logical tactic be to "bookend" one or both vendors who are already there by setting up your stand just inches from theirs on the longer open side? This would allow you to intercept all of the customers coming from that side since folks only go to the closest stand. This tactic is similar to the last player to bid on "The Price is Right"'s "Contestant's Row" bidding a dollar more than another player.The correct answers assumes they will do that.
But B doesn't. Does B know that C is going to show up later?
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 16th, 2019 at 4:46:30 PM
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Quote: AyecarumbaDoes B know that C is going to show up later?
Yes.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 4:52:03 PM
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This could easily be wrong, but here is where I'm at with the four-vendor case:
Vendor 1 picks a spot a smidge to the left of 0.1
Vendor 2 picks a spot a smidge to the right of 0.9
Vendor 3 will be indifferent between 0.4 and 0.6. Let's just say he picks 0.4
Vendor 4 will pick 0.65.
This would give space as follows:
Vendor 1 0.25
Vendor 2 0.275
Vendor 3 0.25
Vendor 4 0.225
However, Vendor 3 could pick the spot closer to vendor 2. It would average out to:
Vendor 1 0.2375
Vendor 2 0.2750
Vendor 3 0.2500
Vendor 4 0.2375
Thoughts?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2019 at 6:06:28 PM
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Quote: Wizard
This could easily be wrong, but here is where I'm at with the four-vendor case:
Vendor 1 picks a spot a smidge to the left of 0.1
Vendor 2 picks a spot a smidge to the right of 0.9
Vendor 3 will be indifferent between 0.4 and 0.6. Let's just say he picks 0.4
Vendor 4 will pick 0.65.
This would give space as follows:
Vendor 1 0.25
Vendor 2 0.275
Vendor 3 0.25
Vendor 4 0.225
However, Vendor 3 could pick the spot closer to vendor 2. It would average out to:
Vendor 1 0.2375
Vendor 2 0.2750
Vendor 3 0.2500
Vendor 4 0.2375
Thoughts?
Note it's 2am here so I know there a fault in my logic as if C doesn't pick 50% then this makes it worse for A.
If C is nearer then the average is between 1/6 and 1/3; if C is further away then it's between 1/6 and 1/4 (but not linearly).
Using the same idea of having 100m beach with 1m spots then moving A from 50.
A at 50 gets cut off by B and C who eiher get 49 to themselves or share it with D.
At the solution for 3 vendors A goes to 25, B goes to 76, C goes between them and D goes to 24.
Similarly A=20, B=80, C=50 D = 19.
So the cut over is where it's better for D to go in the middle.
When A = 1/6, B = 5/6 then C's probably better off going 1/2. From A's perspective it has to be better for D to go randomly between A and C rather than left of A.
If it lands up
A 16.666%
C 50% (I think this is where my logic isn't complete ##!! as if C shows any bias then it may help A or hinder)
B 83.334%
then D is indifferent whether between A and C or C and B, and gets just over 1/6.
A's average catchment area...when D is beyond C then A has first 1/3, when D is nearer than C then A has first 1/4 (average of 1/6 thru 1/3). These are equally likely so A gets 7/24 (29.1666%)
January 16th, 2019 at 8:30:51 PM
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Quote: Wizard
This could easily be wrong, but here is where I'm at with the four-vendor case:
Vendor 1 picks a spot a smidge to the left of 0.1
Vendor 2 picks a spot a smidge to the right of 0.9
Vendor 3 will be indifferent between 0.4 and 0.6. Let's just say he picks 0.4
Vendor 4 will pick 0.65.
This would give space as follows:
Vendor 1 0.25
Vendor 2 0.275
Vendor 3 0.25
Vendor 4 0.225
However, Vendor 3 could pick the spot closer to vendor 2. It would average out to:
Vendor 1 0.2375
Vendor 2 0.2750
Vendor 3 0.2500
Vendor 4 0.2375
Thoughts?
Let’s try a simplifying assumption. A and B will want to be at the lowest and highest spots because tending too far away from an end leave them open to C and D grabbing that territory that is unshared to the beach ends. A and B will pick spots such that the distance to the island end is half the distance that C and D could claim in the middle (since middle beach territory only runs to the midpoint to the next vendor).
If that holds let’s run a test where A goes to 1/5 mile mark. If 3 vendors means A goes to 1/4, then maybe it’s as simple as A going to 1/(1+n)? And B responds by going to 4/5 mile marker.
C will want to make D indifferent about going to the left or right of him. That means C will go to 1/2 and D will land a smidge to the right or left of C. This is generally true of a two vendor problem, which is exactly what the four vendor problem reduces to (since A and B simply “shrink” the useable end points of the beach). C and D capture territory of (1/2 - 1/5) / 2 = 3/20.
Since 3/20 territory is worse than D could have done by cutting A or B off from the end (capturing 1/5 or 4/20), that means A and B are too far from the island edge.
The indifference equation then is A - 0 = (1/2 - A)/2. So A should equal 1/6.
So A goes to 1/6 (or a smidge less), B goes to 5/6 (or a smidge more), C goes a smidge to the left (or right) of 1/2 and D takes the middle.
If that holds let’s run a test where A goes to 1/5 mile mark. If 3 vendors means A goes to 1/4, then maybe it’s as simple as A going to 1/(1+n)? And B responds by going to 4/5 mile marker.
C will want to make D indifferent about going to the left or right of him. That means C will go to 1/2 and D will land a smidge to the right or left of C. This is generally true of a two vendor problem, which is exactly what the four vendor problem reduces to (since A and B simply “shrink” the useable end points of the beach). C and D capture territory of (1/2 - 1/5) / 2 = 3/20.
Since 3/20 territory is worse than D could have done by cutting A or B off from the end (capturing 1/5 or 4/20), that means A and B are too far from the island edge.
The indifference equation then is A - 0 = (1/2 - A)/2. So A should equal 1/6.
So A goes to 1/6 (or a smidge less), B goes to 5/6 (or a smidge more), C goes a smidge to the left (or right) of 1/2 and D takes the middle.
Looks like my answer agrees with Charliepatrick’s.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 17th, 2019 at 3:45:55 AM
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Quote: unJon...C goes a smidge....
Last night I wasn't sure why C would go to the middle. In the two vendor case whereever C goes D goes next door. However in this case D is indifferent where he goes on the segment, except it will be the larger segment remaining, as the other end is shared with A or B. In fact D gets half the segment between C and either A or B.
Thus if C leaves a larger segment on one side then D gets a larger slice than C going in the middle. It didn't seem obvious this would be to the detriment of C.
For argument's sake (since we're looking at C) look at various points where C is nearer to A. C will share that segment so has a AC/2 slice.
D picks a point randomly between C and B, so on average CD = CB/2. Thus C's slice is CB/4. The total of the two slices is (AC/4+AC/4)+CB/4=AC/4+AB/4. Thus C should maximise AC and pick the middle.
Thus if C leaves a larger segment on one side then D gets a larger slice than C going in the middle. It didn't seem obvious this would be to the detriment of C.
For argument's sake (since we're looking at C) look at various points where C is nearer to A. C will share that segment so has a AC/2 slice.
D picks a point randomly between C and B, so on average CD = CB/2. Thus C's slice is CB/4. The total of the two slices is (AC/4+AC/4)+CB/4=AC/4+AB/4. Thus C should maximise AC and pick the middle.
January 17th, 2019 at 4:02:07 AM
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That makes sense to me.Quote: charliepatrickLast night I wasn't sure why C would go to the middle. In the two vendor case whereever C goes D goes next door. However in this case D is indifferent where he goes on the segment, except it will be the larger segment remaining, as the other end is shared with A or B. In fact D gets half the segment between C and either A or B.
Thus if C leaves a larger segment on one side then D gets a larger slice than C going in the middle. It didn't seem obvious this would be to the detriment of C.
For argument's sake (since we're looking at C) look at various points where C is nearer to A. C will share that segment so has a AC/2 slice.
D picks a point randomly between C and B, so on average CD = CB/2. Thus C's slice is CB/4. The total of the two slices is (AC/4+AC/4)+CB/4=AC/4+AB/4. Thus C should maximise AC and pick the middle.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 17th, 2019 at 8:34:58 AM
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Quote: unJonLooks like my answer agrees with Charliepatrick’s.
I agree and stand corrected.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 17th, 2019 at 7:46:22 PM
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How about the case of n pirates? Let's think of it in terms of units of space, as opposed to 1 mile.
The first two vendors to act will pick spots a smidge less than 1 unit from either edge.
Vendors 3 to n-2 will pick spots 2 minus a smidge units away from already established vendor
Vendor n-1 will pick a spot exactly between the closest vendors in the progressions from each end.
Vendor n will be indifferent between the spot exactly between n-1 and one of the vendors to his side, because he will get a full unit, as opposed to one unit minus a smidge.
At the end of the day:
Vendors 1 and 2 will get 1.5 units of space
Vendors 3 to n-4 will get 1 unit each
Vendors n-2 and n-3 will get 1 and 1.5 units each. Who gets how much will depend on n's action.
Vendor n-1 will get 1.5 units
Vendor n will get 1 unit.
p.s. Why doesn't my spell checker like "smidge?"
p.p.s Do I put the question mark inside or outside the right quote?
The first two vendors to act will pick spots a smidge less than 1 unit from either edge.
Vendors 3 to n-2 will pick spots 2 minus a smidge units away from already established vendor
Vendor n-1 will pick a spot exactly between the closest vendors in the progressions from each end.
Vendor n will be indifferent between the spot exactly between n-1 and one of the vendors to his side, because he will get a full unit, as opposed to one unit minus a smidge.
At the end of the day:
Vendors 1 and 2 will get 1.5 units of space
Vendors 3 to n-4 will get 1 unit each
Vendors n-2 and n-3 will get 1 and 1.5 units each. Who gets how much will depend on n's action.
Vendor n-1 will get 1.5 units
Vendor n will get 1 unit.
p.s. Why doesn't my spell checker like "smidge?"
p.p.s Do I put the question mark inside or outside the right quote?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 17th, 2019 at 9:39:46 PM
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Quote: WizardHow about the case of n pirates. Let's think of it in terms of units of space, as opposed to 1 mile.
The first two vendors to act will pick spots a smidge less than 1 unit from either edge.
Vendors 3 to n-2 will pick spots 2 minus a smidge units away from already established vendor
Vendor n-1 will pick a spot exactly between the closest vendors in the progressions from each end.
Vendor n will be indifferent between the spot exactly between n-1 and one of the vendors to his side, because he will get a full unit, as opposed to one unit minus a smidge.
At the end of the day:
Vendors 1 and 2 will get 1.5 units of space
Vendors 3 to n-4 will get 1 unit each
Vendors n-2 and n-3 will get 1 and 1.5 units each. Who gets how much will depend on n's action.
Vendor n-1 will get 1.5 units
Vendor n will get 1 unit.
p.s. Why doesn't my spell checker like "smidge?"
p.p.s Do I put the question mark inside or outside the right quote?
1) Pirates on the brain! :-)
2) I’d put the ? Outside the quotes there.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 17th, 2019 at 11:56:24 PM
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Quote: WizardHow about the case of n pirates? Let's think of it in terms of units of space, as opposed to 1 mile.
The first two vendors to act will pick spots a smidge less than 1 unit from either edge.
Vendors 3 to n-2 will pick spots 2 minus a smidge units away from already established vendor
Vendor n-1 will pick a spot exactly between the closest vendors in the progressions from each end.
Vendor n will be indifferent between the spot exactly between n-1 and one of the vendors to his side, because he will get a full unit, as opposed to one unit minus a smidge.
At the end of the day:
Vendors 1 and 2 will get 1.5 units of space
Vendors 3 to n-4 will get 1 unit each
Vendors n-2 and n-3 will get 1 and 1.5 units each. Who gets how much will depend on n's action.
Vendor n-1 will get 1.5 units
Vendor n will get 1 unit.
p.s. Why doesn't my spell checker like "smidge?"
p.p.s Do I put the question mark inside or outside the right quote?
What if none of the vendors knows n? Will vendor 1 set up in the middle of the strand?
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 18th, 2019 at 6:48:24 AM
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Quote: AyecarumbaWhat if none of the vendors knows n? Will vendor 1 set up in the middle of the strand?
I'd guess the correct logic is the first person goes in the middle and the rest just go in the middle of the largest gap left. Thus three people would be at 25% 50% and 75%, seven at 1/8 2/8... 7/8 etc.
January 18th, 2019 at 7:14:59 AM
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Quote: AyecarumbaWhat if none of the vendors knows n? Will vendor 1 set up in the middle of the strand?
They know what n is.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 22nd, 2019 at 2:27:03 AM
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1/(2n-2)?
February 22nd, 2019 at 3:39:21 AM
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Quote: 99kv991/(2n-2)?
Especially when n = 1, ; )
edit:
sorry, I don't know the answer, but this thread looked interesting
February 22nd, 2019 at 8:38:42 AM
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Anywhere to the left or right of the middle.
If he sets up in the middle, that will leave too many people to the left or right of him for the other vendors to nab.
If he sets up off-center, because the people are distributed evenly, he is guaranteed to cut out customers for one of the vendors since they won't be able to set up their stands evenly.
If he sets up in the middle, that will leave too many people to the left or right of him for the other vendors to nab.
If he sets up off-center, because the people are distributed evenly, he is guaranteed to cut out customers for one of the vendors since they won't be able to set up their stands evenly.
