Royal Flamingo Casino
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6 members have voted
Wizard
Administrator
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. Administrator
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?
It's not whether you win or lose; it's whether or not you had a good bet.
rudeboyoi
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%.
beachbumbabs
Administrator
Administrator
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.
unJon
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.
unJon
Flamingo Hotel
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.
Babs, I think 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.
beachbumbabs
Administrator
Administrator
Quote: beachbumbabs
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.
Babs, I think 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.
Wizard
Administrator
Administrator
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.
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.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Administrator
Administrator
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%.
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.
It's not whether you win or lose; it's whether or not you had a good bet.
EvenBob
It's raining, I would call Uber Eats andset up tomorrow.
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'It's not enough to succeed, your friends must fail.' Gore Vidal
unJon
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Quote: unJon
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.
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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.
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.
Royal Flamingo Casino
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.