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unJon
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?
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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.
charliepatrick
...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.
unJonThus 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.
Thanks for this post from:
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.
That makes sense to me.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.
Wizard
Administrator
Administrator
Looks like my answer agrees with Charliepatrick’s.
I agree and stand corrected.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Administrator
How about the case of n pirates? Let's think of it in terms of units of space, as opposed to 1 mile.Administrator
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?
It's not whether you win or lose; it's whether or not you had a good bet.
unJon
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?
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.
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Ayecarumba
How about the case of n pirates? Let's think of it in terms of units of space, as opposed to 1 mile.
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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?
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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
charliepatrick
What 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.
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Wizard
Administrator
Administrator
What if none of the vendors knows n? Will vendor 1 set up in the middle of the strand?
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They know what n is.It's not whether you win or lose; it's whether or not you had a good bet.
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1/(2n-2)?