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.

Clearly, the winning prob. of A and B are equal out of symmetry considerations , so we need to find the prob. Pc , of C being the first to gather 2 wins.

This is very simple, because in order to achieve this he MUST win both the second game ( which is the first he participates in), as well as the next play ( the 3rd one). Reason: -

If he looses the 2nd play, then clearly the winner of the first play is the first winner of 2 games !

And if he looses the 3rd play ( after having won the second), then the situation is that each player will have gained one round, so that the fourth game, in which C does not participate, will yield a double win (of A or of B).

Calculating the probability of C winning the 2nd and 3rd play :

    Pc= PC2 * PC3/C2

 where :  PC2 is the prob. of C winning the 2nd game,

 which equals clearly PC2=0.5

         PC3/C2 denotes the prob. of C winning the 3rd game,

         given that he has already won the 2nd one.This 

         prob. is clearly again PC3/C2 = 0.5 .

   Therefore  Pc=0.5*0.5 = 0.25

   and consequently   Pa=Pb = (1-Pc)/2 = 3/8

                                   Dan Rosen

 

 

Posted by Dan Rosen on 2010-05-20 12:07:56