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Now say, the players aggree, that the second players may not play scissors. This looks like:
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):
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:
Update: Don Reble found an alternative solution to the above problem:
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:
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.