Csdrhrt

I would like to find a probability between opponents that have never played with each other.

I would like to explain what I mean in an example.

Player A won m times against Player B and never lost a match against Player B.
Player B won n times against Player C and never lost a match against Player C.

If Player A and Player C play a match, As a sense, we can easily say that Player A has big advantage against Player C even though they have never played a match before.

I would like to find a formula to define the probability for such events that we need to formalize such kind of probability events.

Firstly, I want to define limit possibilities

A won all matches against B. $P_{A/B}=1$

B won all matches against C. $P_{B/C}=1$

Let's define probability before a real match between A and C
the probability of winning A against C: $$P_{A/C}=1$$

A lost all matches against B . $P_{A/B}=0$

B lost all matches against C . $P_{B/C}=0$

Let's define the probability before a real match between A and C

The probability of winning A against C: $$P_{A/C}=0$$

Other limit possibilities:

A won all matches against B. $P_{A/B}=1$

B lost all matches against C. $P_{B/C}=0$

We can not say anything about a real match between A and C. We can consider their chances are same as limit definition for this condition.

$$P_{A/C}=1/2$$

A lost all matches against B. $P_{A/B}=0$

B won all matches against C . $P_{B/C}=1$

We can not say anything about a real match between A and C. We can consider their chances are same as limit definition for this condition.

$$P_{A/C}=1/2$$

More Generally

A won m times against B in n games. $P_{A/B}=frac{n}{m}$

B won x times against C in y games. $P_{B/C}=frac{x}{y}$

I would like to find an estimated probability before a real match between A and C
$$P_{A/C} =f(m,n,x,y)$$

Some properties of the function $f(m,n,x,y)$ must satisfy that:
The limit conditions:

• 
  $f(0,n,0,y)=0 tag{1}$


• $f(0,n,y,y)=frac{1}{2} tag{2}$


• $f(n,n,0,y)=frac{1}{2} tag{3}$


• $f(n,n,y,y)=1 tag{4}$

Other important property:

A won x times against B in y games. $P_{A/B}=frac{x}{y}$

B won m times against C in n games. $P_{B/C}=frac{m}{n}$

We should have the same probability with previous result. $$P_{A/C} =f(x,y,m,n)=f(m,n,x,y)$$

Thus,

• $f(m,n,x,y)=f(x,y,m,n) tag{5}$

Other property is:

$$P_{C/A} =1-P_{A/C}$$

• $f(y-x,y,n-m,n)=1-f(m,n,x,y)$

Because of property 5 above, we can write that

$f(y-x,y,n-m,n)=f(n-m,n,y-x,y)$

• $f(n-m,n,y-x,y)=1-f(m,n,x,y) tag{6}$

I have found the candidate function that satisfies these 6 conditions:

$$f(m,n,x,y)=frac{1}{2} (frac{m}{n}+frac{x}{y}) =frac{P_{A/B}+P_{B/C}}{2} tag{7}$$
$$P_{A/C} =frac{P_{A/B}+P_{B/C}}{2} tag{8}$$

I want to extend the idea for 4 players and
I tested the candidate function(7) for 4 players conditions but the candidate function did not work

Let's see the status for 4 players.

A won m times against B in n games. $P_{A/B}=frac{n}{m}$

B won x times against C in y games. $P_{B/C}=frac{x}{y}$

C won k times against D in z games. $P_{C/D}=frac{k}{z}$

I want to find $$P_{A/D} =g(m,n,x,y,k,z)$$

Limit conditions:

1. 
   $P_{A/D}=g(0,n,0,y,0,z)=0 $


2. 
   $P_{A/D}=g(0,n,0,y,z,z)=frac{1}{2}$
   
   /Absolutely, C is the most powerful player but We cannot say anything between A and D before a real match, so we need to give same chance for A and D before a real match. (C>[A,B,D])


3. 
   $P_{A/D}=g(0,n,y,y,0,z)=frac{1}{2}$
   /Absolutely, B is the most powerful player
   but we do not know the level between A and D. (B>[A,C,D])


4. 
   $P_{A/D}=g(0,n,y,y,z,z)=frac{1}{2}$
   /Absolutely, B is the most powerful player but we do not know the level between A and D.(B>[A,C,D])


5. 
   $P_{A/D}=g(n,n,0,y,0,z)=frac{1}{2}$
   /Absolutely, The worst player is B but we do not know the level between A and D.([A,C,D]>B)


6. 
   $P_{A/D}=g(n,n,0,y,z,z)=frac{1}{2}$
   /Absolutely, The worst player is B but we do not know the level between A and D.([A,C,D]>B)


7. 
   $P_{A/D}=g(n,n,y,y,0,z)=frac{1}{2}$
   /Absolutely, The worst player is C but we do not know the level between A and D.([A,B,D]>C)


8. $P_{A/D}=g(n,n,y,y,z,z)=1$

other properties:

9. $g(m,n,x,y,k,z)=g(m,n,k,z,x,y)=g(x,y,m,n,k,z)=g(x,y,k,z,m,n)=g(k,z,x,y,m,n)=g(k,z,m,n,x,y)$


10. 
    $P_{D/A} =1-P_{A/D}$ thus $g(n-m,n,y-x,y,z-k,z)=1-g(m,n,x,y,k,z)$
    

EDIT:
I have found an candidate function $P_{A/D}$ for 4 players : It satisfies all 10 conditions above .

$$P_{A/D}=frac{1}{2}(P_{A/B}+P_{B/C}+P_{C/D})-frac{1}{2}(P_{A/B}P_{B/C}+P_{B/C}P_{C/D}+P_{A/B}P_{C/D})+P_{A/B}P_{B/C}P_{C/D}$$

My candidate function for 5 players ${{A,B,C,D,E}}$ and known values $P_{A/B},P_{B/C},P_{C/D},P_{D/E}$
$$P_{A/E}=frac{1}{2}(P_{A/B}+P_{B/C}+P_{C/D}+P_{D/E})-frac{1}{2}(P_{A/B}P_{B/C}+P_{A/B}P_{C/D}+P_{A/B}P_{D/E}+P_{B/C}P_{C/D}+P_{B/C}P_{D/E}+P_{C/D}P_{D/E})+frac{1}{2}(P_{A/B}P_{B/C}P_{C/D}+P_{A/B}P_{B/C}P_{D/E}+P_{B/C}P_{C/D}P_{D/E}+P_{A/B}P_{C/D}P_{D/E})$$

I would like to generalize the problem for n players . What is the function for n players case?

I haven't seen such probability formula during my search. Please let me know if you know a research about it.

Thanks
Best Regards