Game theory-surreal/dyadic/ordinal numbers

[roadrunner]

First of, a simple question. If we have 3 games with surreal equivalents, let's say *1/2 *-1/2 and *1/4, (where positive means in favor of player left) if left goes first he wants to keep that sum of those 3 games greater than or equal to 0 to win right?

Homework Statement

For every dyadic number q, find it's birthday

Homework Equations

A dyadic rational is a number of the form n/2^k where n and k are integers.


Day 0: On day zero, the number 0 is born, and we identify 0 with fjg which we think of as
a split of the empty set into two empty sets.
Day n + 1: On day n + 1 we take all the numbers that have been born on days 0; 1; : : : n,
and we consider all possible ways to split these numbers into two sets fL j Rg where every
x 2 L and y 2 R satisfy x < y (note that we also allow L or R to be empty). We call L
the Left set and R the Right set. For each such split, a new number is born. If both L
and R are nonempty, then this new number is halfway in between the largest element of L
and the smallest element of R. If R is empty, then the new number is the smallest counting
number (positive integer) greater than everything in L, and similarly, if L is empty, we get
the largest negative counting number smaller than everything in R.

The Attempt at a Solution

I really have no idea where to start..

 

[mighty2000]

I would first clarify some points in the forum post. It seems that the person is describing a game where players take turns choosing a number from a set of surreal numbers (numbers of the form n/2^k) and the goal is to keep the sum of the chosen numbers greater than or equal to 0. However, it is not clear how the players are choosing these numbers or what the rules are for the game. I would ask for more information on the game before attempting to answer the question.

As for finding the birthday of a dyadic number q, I would follow the given algorithm. On day 0, the number 0 is born. On day 1, we split 0 into two empty sets, which gives us the numbers -1/2 and 1/2. On day 2, we can split these numbers into two sets, giving us the numbers -3/4, -1/4, 1/4, and 3/4. We continue this process, splitting the numbers that have been born on previous days, until we reach the desired dyadic number q. The day on which q is born will be its birthday. For example, if q = 3/8, it will be born on day 4 since on day 4 we split the numbers 1/4 and 3/4 to get 3/8.