Let E1, E2 be two events such that
E1 = the captain of team ‘A’ gets a six.
E2 = the captain of team ‘B’ gets a six.
Here P (E1) = 1/6 , P (E2) = 1/6
P(E1) = 1 - 1/6 = 5/6, P(E2) = 1 - 1/6 = 5/6
Now P (winning the match by team A)
= 1/6 + 5/6 x 5/6 x 1/6 + 5/6 x 5/6 x 5/6 x 5/6 x 1/6 + ....
= 1/6 + (5/6)2 x 1/6 + (5/6)4 x 1/6 + ...
= (1/6)/(1 - 25/36) = 6/11
∴ P (winning the match by team B) = 1 - 6/11 = 5/11
The decision of re free was not fair because the probability of winning match is more for that team who start to throw dice.