Assume that you are both human and computer unaided and therefore cannot choose truly random positions on the list.
Roll a die. Choose a number. (From the list, or based on order. For example, if you choose 7, the 7th number in the list is 96.)
You both know that if you pick the same number, a terrible fate looms.
Alas! You search you pockets desperately but you don’t have a die to roll, or even a coin to flip. What to do?
A solution procedure:
(It might help if both people know it advance—the key is “don’t avoid schelling points, just pick different ones”.)
Sort your names in alphabetical order (if that fails, use birth days).*
The person whose name comes first*, picks a number from the first part of the list. (11 numbers, leaves 5 numbers to choose from for the first part, excluding the middle.) The last person, chooses from the last part of the list. Taken to the extreme, of course, the first person chooses the first number from the list, and the second person chooses from the last number from the list.
*If you take turns choosing, and you know the order of the choices, that can be used to sort both of you (both players) instead.
“Randomization” without a “randomness generator”:
What if you don’t know anything about the other person?
The list has less than 26 elements. There are 26 possibilities for the first letter of your name. Is this risky?
Perhaps you have a favorite number. Maybe it’s not in the list—but is it less than 11? (And a positive number.) Just to be sure, this might be combined with another method—taking care to not increase convergence probability.
Using game theory, what logical strategy would you employ in two-player avoidance games similar to the one above?
Not sure how logic is relevant here—but in theory, if one can move, perhaps that will preclude meeting again. (The costs associated with moving may be a reason for not doing this in practice, but it seems like it would work in games with binary outcomes.)
*Donald Hobson observed that on a game show you might have a score. If your scores are different that might be used to differentiate the two of you.
This doesn’t even require both use the same randomization, just that both randomize without putting higher frequency on the same options (for example e being the most common letter, random letter in a book will push towards e/5)
Added: favorite number seems like it would be weighted to specific numbers. Unless you have a very distinctive, unique reason for that number it seems like a bad idea.
Roll a die. Choose a number. (From the list, or based on order. For example, if you choose 7, the 7th number in the list is 96.)
Alas! You search you pockets desperately but you don’t have a die to roll, or even a coin to flip. What to do?
A solution procedure:
(It might help if both people know it advance—the key is “don’t avoid schelling points, just pick different ones”.)
Sort your names in alphabetical order (if that fails, use birth days).*
The person whose name comes first*, picks a number from the first part of the list. (11 numbers, leaves 5 numbers to choose from for the first part, excluding the middle.) The last person, chooses from the last part of the list. Taken to the extreme, of course, the first person chooses the first number from the list, and the second person chooses from the last number from the list.
*If you take turns choosing, and you know the order of the choices, that can be used to sort both of you (both players) instead.
“Randomization” without a “randomness generator”:
What if you don’t know anything about the other person?
The list has less than 26 elements. There are 26 possibilities for the first letter of your name. Is this risky?
Perhaps you have a favorite number. Maybe it’s not in the list—but is it less than 11? (And a positive number.) Just to be sure, this might be combined with another method—taking care to not increase convergence probability.
Not sure how logic is relevant here—but in theory, if one can move, perhaps that will preclude meeting again. (The costs associated with moving may be a reason for not doing this in practice, but it seems like it would work in games with binary outcomes.)
*Donald Hobson observed that on a game show you might have a score. If your scores are different that might be used to differentiate the two of you.
This doesn’t even require both use the same randomization, just that both randomize without putting higher frequency on the same options (for example e being the most common letter, random letter in a book will push towards e/5)
Added: favorite number seems like it would be weighted to specific numbers. Unless you have a very distinctive, unique reason for that number it seems like a bad idea.