| medium | probability |
A & B are alternately picking balls from a bag without replacement. The bag has k black balls and 1 red ball. Winner is the one who picks the red ball. Who is more likely to win, the on who starts first, or second?
Look at k=0,1,2,3... Write down probability of starter being a winner.
Begin!
A & B keep on picking the balls without looking at their color. After all balls have been picked, the one who starts the game will have more balls (if k=even, total balls=odd) and hence higher probability of winning.
(prob of A winning in first chance = 1/(k+1)) + (prob of A winning in 3rd chance = (1-(1/k+1)) * (1-1/k) * 1/(k-1) = 1/(k+1)) + ... + (prob of A winning in 2r+1-th chance = 1/(k+1)) + ... . When k=2n+1, there are n+1 such terms, giving the prob as (n+1) * 1/(k+1)=1/2. When k=2n, there are n+1 such terms, giving prob as (n+1) * 1/(k+1)=(n+1)/(2n+1)>1/2. Hence, doesnt matter who starts first when k is odd. The first player has higher chance of winning when k is even