Simple probability calculation for dice game "Schock" I have a problem to calculate a probability within my programm I am currently writing on. I thought about it the whole day yesterday but I don't get it. Maybe it is quite simple for you guys.

It is a dice game without entire information. I have three or more player. I can calculate the possibility to win for each player against each other player. Using these numbers, I want to calculate the general possibility to win for each player.

Simple example: 3 player. Every player has 50% against each other. I used a tree to calculate the overall possibility, but that seems to fail. The possibility should be 33% for each player to have the best roll?! But using the tree would calculate 1 * 0.5 (beating villain 1) * 0.5 (beating villain 2) = 0.25.

Can someone please point me to the right direction?

Update: 
The game is called "Schock", a german tavern-game. Every player has 3 dices and up to 3 rolls, which will take place consecutively. When all player finished their roll, the player with the lowest rating roll gets minus points equal to the highest ranking rolls' value.

Player may choose to let some of their dices unrevealed.

Update 2, to make more clear what my problem is:
The numbers above are just example numbers to keep things simple. I already have a routine which is able to calculate odds between two players' dicecups, so lets take it as given that I can predict in any situation the odds between two players' cups even if they did not finish their rolls yet and have any number of dices unrevealed.
The core of my problem is using these numbers (1vs1) to calculate odds between more than two player, which is a pure mathematical problem.
My first approach was a tree, but this fails. As it was stated in the comments, this may be a complete false approach as players' rolls rankings are not independent of each other.
So the core question is: Is it possible to use all 1vs1 odds to calculate 1vsAll odds and how.

Maybe with a concrete situation:
Lets assume 3 players playing. Which all have made 3 rolls, so the round is finished. Before all cups are revealed, this is the situation:


Player 1: Has two 1s and one dice unrevealed.
Player 2: Has one 1 and two dices unrevealed.
Player 3: Has all dices unrevealed.


Given all 1 vs 1 odds to have the better roll:


P1 vs P2: 0.88426 (88%)
P1 vs P3: 0.97299 (97%)
P2 vs P3: 0.50759 (51%)


Without even knowing the rules of the game, is it possible to calculate the total chance for Player 1 to win or Player 3 not to loose using the given numbers?
The issue at hand is that the probabilities aren't independent - indeed, one result is necessarily defined by the other two - if A beats B and B beats C, then A necessarily beats C. And just knowing the probabilities for each of the three two-player matches isn't enough to define the system.

Allow me to demonstrate with a simple example of a similar game, but with easily-analysed properties. A bag holds four balls, each labelled between 1 and 4. Each of the three picks a ball out of the bag, and the one with the highest number wins. This is used to avoid issues of equal scores causing rematches, etc.

Ahead of time, it's easy to determine that the probability that each player wins is $\frac13$, and that the probability that any specific player beats any other player is $\frac12$. We can easily write that
$$
Pr(``\text{A wins''}) = Pr(A&gt;B) - Pr(C&gt;A&gt;B)
$$
Now, as we noted, $Pr(A&gt;B)=\frac12$ and $Pr(``\text{A wins''})=\frac13$, so we have that $Pr(C&gt;A&gt;B)=\frac16$. But this means that $Pr(C&gt;A|A&gt;B)=\frac13\neq Pr(C&gt;A)$, and thus the events aren't independent. To see why this is, let's look at the situation where A>B. In this case, there are six possible combinations.

--- NOTE: Below here, until noted otherwise, probabilities are for case in which $A&gt;B$, left off for notational simplicity ---

Specifically, if we denote the combinations as (A,B), then we have (2,1), (3,1), (3,2), (4,1), (4,2), and (4,3). Of these situations, $Pr(C&gt;A=4)=0$, $Pr(C&gt;A=3)=\frac12$, and $Pr(C&gt;A=2)=1$, where $Pr(C&gt;A=n)$ is shorthand for $Pr(C&gt;A|A=n)$. And given the probabilities that arise, we have
$$
Pr(C&gt;A) = \sum_{n=2}^4 Pr(C&gt;A=n)Pr(A=n)\\
= 0\cdot \frac12+\frac12\cdot \frac13+1\cdot\frac16 = \frac13
$$
Now suppose that we allow partial knowledge. We know that $B\neq2$ and $B\neq 3$. Now the set of combinations becomes (2,1), (3,1), (4,1). This means that we now have
$$
Pr(C&gt;A) = 0\cdot\frac13 + \frac12\cdot\frac13 + 1\cdot\frac13 = \frac12
$$

--- NOTE: Assumption that $A&gt;B$ is removed from here ---

When we have $B\not\in\{2,3\}$, we can see that $Pr(A&gt;B)=\frac12$. We also have that $Pr(C&gt;A|A&gt;B) = \frac12$, and so
$$
Pr(``\text{A wins''}) = \frac12 - \frac12\cdot\frac12 = \frac14
$$
Notice that $Pr(A&gt;B)$, $Pr(A&gt;C)$, and $Pr(B&gt;C)$ didn't change, but the final probabilities did? Notice that, in the latter case, $Pr(C&gt;A)=Pr(C&gt;A|A&gt;B)$?

So it's not enough to know the probabilities for each pair, you need to know how they interact, and that depends on the specifics of what each player shows (not just how many dice are shown, but number of dice+score on those dice).
If you make a probability tree for each player:
for Player 1 $P_1$
$$P_{win,win}=0.5*0.5=0.25$$
$$P_{win,loss}=0.5*0.5=0.25$$
$$P_{loss,win}=0.5*0.5=0.25$$
$$P_{loss,loss}=0.5*0.5=0.25$$

Summing all the probabilitites equals 1.
This is the same for all the other players ($2$ and $3$)