Three backgammon players of equal game skills compete for a prize.

The prize will be awarded to the winner of two games in a row.

A and B, following a drawing, play the first game, then the winner will
face C.

Next game, if needed, will be by C and the player who lost the 1st
game and so on.

Determine a priori chances of winning for each of the 3 players, assuming
Lady Luck treats them without discrimination.

Rem: There are no draws in backgammon.

Let p = probability that A ultimately wins = probability that B ultimately wins

Then 1-2p = probability that C ultimately wins.

Let q = probability of ultimately winning if you have just won a match and need one more to win

Let r = probability of ultimately winning if you have just lost a match and the winner still needs one more to win

Then, at the start, from A's point of view

(1) p = .5*q + .5*r

After match 1, from the winner's point of view,

(2) q = .5*1 + .5*r

After match 1, from C's point of view

(3) (1-2p) = .5*q

Subtracting (1) - (2) gives:

(4) p = 1.5q - .5

Substituting (3) into (4)

(5) p = 3*(1-2p) - .5

7p = 2.5

p = 5/14, just as Daniel and Charlie have found

So at the start, the chances of (A,B,C) ultimately winning are (5/14,5/14,4/14) respectively.

After the first match, the winner's chance of ultimately winning improves to q = 2 - 4p = 8/14,

and the loser's chances of ultimately winning drops to 2/14