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Home  >  Online Games  >  Numfield  >  Two Player Theory

Numfield, Two Player Theory

This page is a devoted to the 2 player zerosum case of Numfield. This case is not the most interesting, but on the other hand it has a theoretical kind of solution.

The setting

Lets assume 2 players play a (infinite) sequence of numfield games and try to maximise their total score advantage over the other one. Any fixed bid is a immediate looser since the opponent can switch to a bid, which beats that particular bid. In fact mathematical game theory suggest, in those games optimal behaviour is randomized. By an appropriately randomized behaviour for a large number of games the laws of stochastics make it happen, that both players can approximately achieve the fair result of a draw.

A simpler game

As game theory teaches the point is to adapt the randomized behaviour to the situation. In the very simple Rock, Paper, Scissors the optimal behaviour in that sense is, to roll a perfect die secretly and play Rock when the die shows 1,2, Paper for 3,4 and Scissors for 5,6.

This reads as follows: In the each Column a player a can see his result, against a play of his oppenent given by the row. E.g. when I play Rock against Paper, I can see from the entry in the column Rock and the row Paper, I suffer a loss of one. In Rock, Paper, Scissors there is only win or loss or draw, so outcome is -1, 0 or +1. And as stated, the optimal behaviour is to play all strategies randomly with the same probability.

Now say, the players aggree, that the second players may not play scissors. This looks like:


When our opponent decides to play randomly both of his options with probability 0.5, then we can play Paper all the time, for a average result of 0.5. But theory tells the second player, to play paper (p=0.666..) twice as often as Rock, our result drops to 0.33. And this is in fact his optimal strategy. The point is that we might have very special probabilites to take to achieve an optimal result.

Rock, Paper, Scissors is a very special class, for which strategies can be calculated by the so called Simplex Algorithm: Its a symmetrical game for two players. A game is called symmetrical, when its tableau is square and when it is reflected at the main diagonal (from left upper corner to right lower) all entries change their sign. In symmetrical games the optimal strategies for both players are identical.

Numfield Again

The mathematical theory can in fact solve a larger class of non symmetrical two players games, like the above modified Rock, Paper, Scissors. But this is not necessary her, for this page, since Numfield for two players is symmetrical. In fact numfield is symmtrical for any number of players, as a simple result of the symmetrical rules.

Lets see what happens when two players play numfield: If they both bid the same, no one gains anything. If one of us passes, the other one gets the whole bonus and carry, independent his bid. Now lets say it is carry plus bonus equals 1, I bid 5 and you bid 1. The pot is 7. I get two thirds of seven, that is 4. You get one third of 7, i.e. 2. So the overall effect is I loost two against you, since I payed one more than I gained and you gained one more than you payed (This is the zero-sum principle at work).

By this we can denote Numfield in the above tableau form. But the point is, we need a different tableau, depending on the number of additional points which get into the pot. These additional points come from the bonus per round and from the carry from the previous round.
For carry + bonus = 1 we get (the turns are the bid numbers):


Of course this is truncated since the total system would allow for all 99 numbers for both players. This reads as, when I bid 2 (column 2) and he bids a 3, then I would suffer a loss of 1. This is so, cause we had 3+2+1=6 in the pot, he gets 2 parts and I get 1, so I get back one third of 6, which gives two and he gets back two thirds of 6 which is four. So I my opponent made a gain of one and I didn't, so I've lost one relative to him...

The solution in this case can also seen by common sense and needs no high math: To play a one in this case is better or equal than any other number against all answers of the opponent. So both players bid ones and neither gain or loose. But in this case the one bonus point carries over and in the next round we have the next case:

For carry + bonus = 2 we get:


This is again the truncated tableau. The fun thing is, even if you let your program do the calculations with the full (100x100) tableau, you get the following optimal probabilities for the bid:

bid: 123456other
Thus the optimal strategy involves only bids from 1 to 6, no passing in this case and no higher bids. And one should bild a lot of 2s!

Update: Don Reble found an alternative solution to the above problem:

bid: 123456other
This solution is much easier, it can be described this way: Roll a perfect die, if it shows 1 or 2 then bid 1, did you roll 3 bid 2, did you roll 4 and 5 then bid 3, with a roll of 6 bid 5. Note that this solution does never use bids of 4 or 6. In general a second solution is no real surprise, since the above problem may have different solutions. Note, that any linear combination of above perfect strategies is also a perfect strategy... E.g.
bid: 123456other
Is another solution.

Higher values of the additional points are unlikely in practical play since the maximum number of points carried of from one round to the next is 1. But when we started with a higher number of players and some of them just died away, higher carries might at least once in the larger game. Lets look at this:

In the case carry + bonus = 3 we get:


Again the numerical solution shows, that we need not more of the tableau:

bid: 1234567other
So we should never bid more than a 7 in this case and never pass again. In fact we should bid a lot of 1s and 2s and even nearly no 7s...

The special case of 2 players above seems to give a few indications for the general case of more players, all not very spectacular, but not all theories must contradict intuition:

OK, so far the theoretical background.


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