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?

Hint

Look at k=0,1,2,3... Write down probability of starter being a winner.

Answer

Begin!

Solution

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.

Solution by Palak Bhushan:

(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

Solution by Palak Bhushan:

(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

Source: 40 Puzzles

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