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Home > Online Games > Numfield > Two Player Theory |

Rock | Paper | Scissors | |

Rock | 0 | +1 | -1 |

Paper | -1 | 0 | +1 |

Scissors | +1 | -1 | 0 |

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:

Rock | Paper | Scissors | |

Rock | 0 | +1 | -1 |

Paper | -1 | 0 | +1 |

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.

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):

pass | 1 | 2 | 3 | 4 | 5 | 6 | 7 | .. | |

pass | 0 | +1 | +1 | +1 | +1 | +1 | +1 | +1 | .. |

1 | -1 | 0 | 0 | 0 | -1 | -2 | -2 | -3 | .. |

2 | -1 | 0 | 0 | 1 | 0 | 0 | -1 | -2 | .. |

3 | -1 | 0 | -1 | 0 | +2 | +1 | 0 | 0 | .. |

4 | -1 | +1 | 0 | -2 | 0 | +2 | +2 | +1 | .. |

5 | -1 | +2 | 0 | -1 | -2 | 0 | +3 | +2 | .. |

6 | -1 | +2 | +1 | 0 | -2 | -3 | 0 | +4 | .. |

7 | -1 | +3 | +2 | 0 | -1 | -2 | -4 | 0 | .. |

.. | .. | .. | .. | .. | .. | .. | .. | .. | .. |

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:

pass | 1 | 2 | 3 | 4 | 5 | 6 | 7 | .. | |

pass | 0 | +2 | +2 | +2 | +2 | +2 | +2 | +2 | .. |

1 | -2 | 0 | 1 | 0 | -1 | -1 | -2 | -3 | .. |

2 | -2 | -1 | 0 | 1 | 1 | 0 | -1 | -1 | .. |

3 | -2 | 0 | -1 | 0 | +2 | +1 | +1 | 0 | .. |

4 | -2 | +1 | -1 | -2 | 0 | +3 | +2 | +1 | .. |

5 | -2 | +1 | 0 | -1 | +3 | 0 | +3 | +3 | .. |

6 | -2 | +2 | +1 | -1 | -2 | -3 | 0 | +4 | .. |

7 | -2 | +3 | +1 | 0 | -1 | -3 | -4 | 0 | .. |

.. | .. | .. | .. | .. | .. | .. | .. | .. | .. |

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: | 1 | 2 | 3 | 4 | 5 | 6 | other |

probability | 0.142857 | 0.404762 | 0.142857 | 0.095238 | 0.119048 | 0.095238 | 0.000000 |

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

bid: | 1 | 2 | 3 | 4 | 5 | 6 | other |

probability | 0.333333 | 0.166667 | 0.333333 | 0.000000 | 0.166667 | 0.000000 | 0.000000 |

bid: | 1 | 2 | 3 | 4 | 5 | 6 | other |

probability | 0.000000 | 0.583333 | 0.000000 | 0.166667 | 0.083333 | 0.166667 | 0.000000 |

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:

pass | 1 | 2 | 3 | 4 | 5 | 6 | 7 | .. | |

pass | 0 | +3 | +3 | +3 | +3 | +3 | +3 | +3 | .. |

1 | -3 | 0 | 1 | 0 | 0 | -1 | -2 | -2 | .. |

2 | -3 | -1 | 0 | 2 | 1 | 0 | 0 | -1 | .. |

3 | -3 | 0 | -2 | 0 | +2 | +2 | +1 | 0 | .. |

4 | -3 | 0 | -1 | -2 | 0 | +3 | +2 | +2 | .. |

5 | -3 | +1 | 0 | -2 | -3 | 0 | +4 | +3 | .. |

6 | -3 | +2 | 0 | -1 | -2 | -4 | 0 | +4 | .. |

7 | -3 | +2 | +1 | 0 | -2 | -3 | -4 | 0 | .. |

.. | .. | .. | .. | .. | .. | .. | .. | .. | .. |

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

bid: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | other |

probability | 0.285714 | 0.285714 | 0.071429 | 0.178571 | 0.071429 | 0.071429 | 0.035714 | 0.000000 |

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:

- As expected, passing is a mistake, since we give the others a chance to catch carry and or bonus...
- There seems to be a rational limit on the highest bid with two players. Maybe there is a higher rational limit with more players. Maybe depending on the number of players?
- The higher carry and bonus are, the higher the bidding might go.
- For certain numerical situations special numbers might not be very good. E.g. I was surprised about the high probabilites on 2s, I expected the 1s to be highest...

OK, so far the theoretical background.

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http://doris-frank.de/numthry.html changed 4/16/07, rendered 4/17/07

(c) 1997-2007 by Doris & Frank Site imprint

(c) 1997-2007 by Doris & Frank Site imprint