Question

The odds in favor of A winning a game of chess against B are 3:2. If 3 games are to be played, what are the odds a. in favor of A winning at least 2 games out of the 3, b. against A losing the first 2 games to B?

Answer 1

a) Odds in favor of A winning =3:2 =3/2

Probability of A winning = O/(1+O) = (3/2)(1+(3/2)) = 3/5

a)

to winning at least 2 games is equivalent to winning 2 games or winning 3 games out of 3 games. We can use binomial to calculate the probability.

= C(3,2) * (3/5)^2 * (2/5)^1 + C(3,3) * (3/5)^3 * (2/5)^0

=81/125

Odds = P/(1-P) = (81/125)/(1-(81/125)) = 81/44 = 81:44

2)

to losing first 2 games is equivalent to (2 lose, 1win)  or (losing 3 games) out of 3 games. We can use binomial to calculate the probability.

Odds in favor of A losing =2:3 =2/3

Probability of A losing = O/(1+O) = (2/3)(1+(2/3)) = 2/5

P(2 loss and last win) = (2/5)*(2/5) *(3/5)= 12/125

P(2 loss and last loss) = (2/5)*(2/5) *(2/5)= 8/125

Probability of at 2 loss= P(2 loss and last win) +P(2 loss and last loss)

=(12/125) +(8/125) =20/125

Odds in favor of A losing = P/(1-P) = (20/125)/(1-(20/125)) = 20/105 = 4/21 = 4:21

Odds in against A losing = 21:4