In this article I will explore the solvability of my game Theocracy through the light of the fundamentals of game theory.
PRELIMINARY STEPS
The first thing we must know is the concept of payoff on game theory. The payoff of a move, decision or play is sum of weighted value, towards the win of the game, of this different possible outcomes for this move/decision.
It have the form of
$$\sum_{j}p_j\cdot v_j$$
being $p_j$ the probability for an outcome j multiplied for it value towards the win $v_j$.
On Theocracy the value-towards-the-win are the points that you earn or lost with a decision. You have on Theocracy two potential sources of points: the auction (A) and the prophesy ($\chi$).
We will notate from here any probability as $p(x)$ and the complementary as $p(\bar x)$. E.g. the probability to win the auction A for the player J on the t turn will be notated as $p(A_J, t)$, and the probability to lose it will be notated as $p(\bar A_J, t)$.
In other words: what (win A), who (player J) and when (turn t).
Analogously we will notate from here any value as $v(x)$ for the win of an event or as $v(\bar x)$ for the lose of the same event.
CONCRETING PROBABILITIES
The probability to win an auction
The probability to win an auction will be in function of the turn that is being played (because you have less options to play or play-vs per turn) and the cards that were played before (the powers that arent on play any more from you and your opponent).
The probability to win (or lose) an auction is a fraction $\frac{n}{m}$ where the numerator n is related to the type of power played and the denominator m is related to the turn that is being played.
The powers in Theocracy are the five cards that each player have to bid on an auction. They are valued from 1 to 5 where the bigger number win to the lesser but 5 that lose vs 1 (this is named the Ouroboros rule). This is the basic mechanic on the game and is practically the same as in GOPS.
The numerator and denominator of the probability can change with the turn that is being played: the denominator, that starts on turn one as 5, decreases on one for each turn played; the numerator could decrease on one for each turn played or not decreases at all, it depends on the powers that the opponent have played on the previous turns.
We can write then
$$p(A_J,t)=\frac{n-d_n(t)}{6-t}\\ d_n(t)\in\{0,...,n\};\ t\in \{1,2,3,4\};\ n\in\{1,2,3\} $$
where $d_n(t)$ is a number that represent the winnable-discarded-powers of the opponent for the n-power played, where n is not the value of the power, it is the number of possible win vs all the others powers.
There is the $n_1$ type, i.e., the 1 and the 2 that just can win vs one power, the $n_2$ type that is the 3 that can win vs two powers, and the $n_3$ type, that are the 4 and the 5, that can win vs three powers.
We can write
$$p(A_J,1)=\frac{n}{5}\in\left\{\frac{1}{5},\frac{2}{5},\frac{3}{5}\right\}\\p(A_J,2)=\frac{n-d_n(2)}{4}\in\left\{0,\frac{1}{4},\frac{2}{4},\frac{3}{4}\right\};\ d_n(2)\in\{0,1\}\\ p(A_J,3)=\frac{n-d_n(3)}{3}\in\left\{0,\frac{1}{3},\frac{2}{3},1\right\};\ d_n(3)\in\{0,...,n\}\\ p(A_J,4)=\frac{n-d_n(4)}{2}\in\left\{0,\frac{1}{2},1\right\};\ d_n(4)\in\{0,...,n\}$$
The probabilities of 0 or 1 doesn't have any risk, any "real" probability is a fraction. Seeing this the last turn will be completely determined or completely random.
A table with all possible probabilities for each turn and each n-type for $p(A_J,t)$
| n\t | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $n_1$ | $\frac{1}{5}$ | 0,$\frac{1}{4}$ | 0,$\frac{1}{3}$ | 0,$\frac{1}{2}$ |
| $n_2$ | $\frac{2}{5}$ | $\frac{1}{4}$,$\frac{1}{2}$ | 0,$\frac{1}{3}$,$\frac{2}{3}$ | 0,$\frac{1}{2}$,1 |
| $n_3$ | $\frac{3}{5}$ | $\frac{1}{2}$,$\frac{3}{4}$ | $\frac{1}{3}$,$\frac{2}{3}$,1 | 0,$\frac{1}{2}$,1 |
The probability to win a prophesy
The probability to win a prophesy depends on the probability to win (or lose) the auction of the turn. If you dont need to match the win condition and you can choose for free prophesy a win or a lose then the probability can be written as
$$p(\dot\chi_J^J,t)=\max{p(A_J,t),p(\bar A_J,t)}=\frac{1}{2}+\left|\frac{1}{2}-p(A_J,t)\right|$$
If you prophesy yourself winning the auction the probability for this prophesy will be the same that for win the auction, i.e., $p(\chi_J^J,t)=p(A_J,t)$. If you prophesy for yourself a lose then $p(\bar\chi_J^J,t)=p(\bar A_J,t)$.
Prophesying a win for the opponent is notated as $p(\chi_J^O,t)=p(\bar A_J,t)$. Analogously the lose for the opponent $p(\bar\chi_J^O,t)=p(A_J,t)$. If needed you can notate the draw as $p(\ddot\chi_J,t)=p(A_J,t)\cdot p(\bar A_J,t)$.
The table for $p(\dot\chi_J^J,t)$ will be
| n\t | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $n_1$ | $\frac{4}{5}$ | 1,$\frac{3}{4}$ | $\frac{2}{3}$,1 | $\frac{1}{2}$,1 |
| $n_2$ | $\frac{3}{5}$ | $\frac{1}{2}$,$\frac{3}{4}$ | $\frac{2}{3}$,1 | $\frac{1}{2}$,1 |
| $n_3$ | $\frac{3}{5}$ | $\frac{1}{2}$,$\frac{3}{4}$ | $\frac{2}{3}$,1 | $\frac{1}{2}$,1 |
It obvious that, in many cases, it will be easier to win a independent prophesy that win the auction. All possible probabilities are the same but the cases of $n_1$ for turns 1 and 2.
Note that $p(\dot\chi_J^J,t)=p(\dot\chi_J^O,t)=p(\dot\chi_O^J,t)=p(\dot\chi_O^O,t)$ so it can be written just as $p(\dot\chi,t)$.
THE SOURCES OF POINTS
There are two main sources of points:
- Auctioned points
- Points that opponent holds
The is two sources of loses points:
- Losing a prophesy
- Opponent win a prophesy
The auction points only can be earned if the player holds at least a free believer or if it can earn a free believer of the opponent, so it depends on the existence of free believers for this turn.
The win or lose of points related so points on believers is related to the non-existence of free believers for the player that lose the points and the obvious existence of a believer holding points.
The balance of believers for a player for each turn goes from -2 to 1: you only can win at maximum one believer per turn if you win your prophesy, but you can lose at maximum two believers per turn, one losing your own prophesy and the other if the opponent win his prophesy.
Probabilities to win auction points
If you have no free believer to earn some point you must win auction AND prophesy... this means that the probability for this prophesy, in this case, will be the same that the probability for win the auction because you cant prophesy for your lose.
We will notate the number of free believers of player J as $F_J$ and for the opponent as $F_O$.
- For $F_J=0$ and $F_O\geq 1$ the probability to earn some points will be $p(A_J,t)^2$.
- For $F_J\geq 3$ the probability to win points from auction is independent of any prophesy so it will be just $p(A_J,t)$.
- For $F_J=2$ the probability to win points depends of prophesies: if you lose your prophesy about your win AND the opponent win his prophesy about his win, so it will be $p(A_J,t)\cdot (1-p(\bar A_J,t)^2)$.
- For $F_J=1$ and $F_O=0$ the probability to win points of the auction depends if the opponent win any prophesy, so it will be $p(A_J,t)p(A,t)$. We dont know what prophesy the opponent will do or if he will prophesy so we note his prophesy in the undefined form $p(A,t)\in\{p(A_J,t),p(\bar A_J,t),1\}$.
- For $F_J=1$ and $F_O\geq 1$ the probability will depends if opponent prophesy something and win and you dont. Analogously to above we define this probability as $p(A_J,t)p(A^*,t)$, where $p(A^*,t)\in\{p(A_J,t)^2,p(\bar A_J,t)p(A_J,t),1\}$.
We will name these probabilities to win the auction points as $p(AP_J,F_J,F_O,t)$.
Probabilities to win earned points
The earned points are the points that each players holds over a believer. We will name the believers of a player J as $B_J=F_J+P_J$ where $P_J$ are the earned points that this player holds.
To steal these points from the opponent you must win a prophesy and the opponent must have no free believers, i.e $F_O=0$, but he must have some earned points, i.e. $P_O\geq 1$.
If you meet these conditions then you probability to win points from this source is $p(\chi_J,t)$. And fot the best case the probability will be $p(\dot\chi,t)$.
Probabilities to lose earned points
This can happen only if $F_J=0$ and $P_J\geq 1$. Then if you lose a prophesy you lose points, and/or if opponent win his prophesy you lose points. You can lose, at maximum, two times points but just if you have $P_J\geq 2$.
The probabilities to lose some earned points will be $p(\bar\chi_J,t) + p(\chi_O,t)$. And for lose twice will be $p(\bar\chi_J,t)\cdot p(\chi_O,t)$.
This will be the first part of the analysis on this game.
