How to avoid Schelling points?

post by cwg1g16 · 2020-05-14T19:37:27.467Z · LW · GW · 5 comments

This is a question post.

Contents

  The Scenario
  A real life example:
  The point of this post:
  The apology in advance:
None
  Answers
    16 Charlie Steiner
    4 Dagon
    3 Purplehermann
None
5 comments

After reading an introductory post on game theory, specifically regarding Schelling Points, I noticed a paradox and I wonder if anyone here can shed some light on the subject.

The scenario I will use to illustrate this is the inverse of a frequently used example to describe popular Schelling points.

The Scenario

Suppose, in a dystopian future, you are on a game show with a partner. A list of numbers is presented to you:

[2, 5, 9, 25, 69, 73, 82, 96, 100, 126, 150]

You and your partner have to pick one number from the list each, without communicating with each other. You both know that if you pick the same number, a terrible fate looms. If you pick different numbers, you are both set free and live long and happy lives.

Assume that you are both human and computer unaided and therefore cannot choose truly random positions on the list.

Naturally, you would want to avoid Schelling points (special numbers or numbers in special positions in the list) to minimise the chance of picking matching numbers. In this case, the Schelling points are numbers which you would think your partner would be more likely to pick, for whatever reason. However, if you both rule out Schelling points, you make the list of numbers to choose from smaller, thus increasing the chance of you both picking the same number significantly. Therefore, if you both actively pick numbers which you think your partner is least likely to pick, assuming you both think rationally, you inadvertently increase the chance of picking the same number. Thus the notion of the Schelling point has become the numbers that are especially insignificant, and the cycle continues. This is the paradox.

A real life example:

Which bar do i choose to drink at on a Friday night if I want to avoid my ex (assuming she's actively avoiding me too)?

The point of this post:

Using game theory, what logical strategy would you employ in two-player avoidance games similar to the one above?

The apology in advance:

I'm new to the site and if any of this is convoluted or has been covered before, I apologise in advance.

Answers

answer by Charlie Steiner · 2020-05-14T23:56:54.773Z · LW(p) · GW(p)

Since you and the other player are cooperating, rather than thinking of the "Schelling point number," think of the Schelling point strategy that you expect each other to implement to try to win.

If I'm avoiding my ex, I might go to a bar that I like more than her, while she goes to a bar that she likes more than me. In the case of the list of numbers, I might pick a number that I think is more significant to me than to the other player.

This assumes that the two players have information that distinguishes them. But not only is that how it is in real life, it's also easy to show that it's necessary for any kind of nontrivial answer: if the two players are identical copies of the same physical system, and they don't have access to any source of randomness like an internet connection or a Geiger counter, then they're going to give the same answer.

answer by Dagon · 2020-05-15T21:31:43.158Z · LW(p) · GW(p)

Human brains suck at this sort of thing. You allude to the correct strategy - just randomize. Even without computer aids, you can probably find a reasonably random proxy, or if you have time a procedure to take non-computer random source and map it to the choices. In this case, I'd flip a coin 4 times to generate a 0-15 binary number, and just re-do it if it's over 10 (as there are only 11 choices). This would completely bypass the idea of Schelling points.

You can probably do better by making use of Schelling's insight rather than the specific of commonly-attractive points. Schelling observed that shared culture is actual information, and that what you know of your partners can be used without further communication. For this example, if I knew my partner well, I might think the younger partner would tend to choose lower, and I'd pick from the other half of the distribution. This is more a Schelling position-in-strategy-space than a Schelling point in the solution space, but it's the same concept.

answer by Purplehermann · 2020-05-19T20:37:15.447Z · LW(p) · GW(p)

The taller/bigger person should pick the bigger number.

You and your ex should a) actually coordinate b)not go to each other's favorite place or places very close to their home.

Anything simple, not reversible, obvious. The big guy gets a big number.

The closer person goes to the bar.

comment by Purplehermann · 2020-05-19T20:40:47.865Z · LW(p) · GW(p)

Or actually randomize I suppose. Number all options 1-N, multiply random numbers in your head until you get stuck (eg 2×6×9×95×34 in order, then when you miltiply that by 37 you get stuck. Use the last number,) though obviously getting it wrong is fine. Modulo N, +1. Walla, your choice, randomized.

5 comments

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comment by Donald Hobson (donald-hobson) · 2020-05-14T20:29:11.197Z · LW(p) · GW(p)

Almost all game theory assumes that you have access to random numbers for problems like this.

Although if you have a distinction making schelling point, you could use that too.

If I had 69 points, and my opponent had 73, or those numbers are our ages or something, then I choose 69, my opponent chooses 73 is an obvious schelling point.

comment by shminux · 2020-05-14T23:39:43.203Z · LW(p) · GW(p)

Seems related to the "guess 2/3 of the average" game.

comment by Ben Pace (Benito) · 2020-05-15T00:32:17.966Z · LW(p) · GW(p)

Mod note: Nice question. I changed the title to be an actual question, and changed a bit of your formatting in a way I think improves it. You're perfectly welcome to change the formatting of course, though if you'd like to change the title please keep it concise and make sure it's a question.

(Let the record show that the former title was "Schelling paradox" and the updated title is "How to avoid Schelling points?")

comment by Pattern · 2020-05-14T23:39:01.775Z · LW(p) · GW(p)
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 [LW(p) · GW(p)] 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.

Replies from: Purplehermann
comment by Purplehermann · 2020-05-19T20:44:23.522Z · LW(p) · GW(p)

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.