A box contains 12 balls of which 4 are white and 8 are red. Three players A, B, and C draw a ball in succession replacing each ball after it is drawn. The first player to draw a white ball wins the game. Find that probability that:
A wins the game
B wins the game
C wins the game
Best Answer
As an alternative approach, we can model this as an infinite series.
Starting with player A:
$$ {P(A\ wins)\ =\ P(A\ wins\ 1^{st}\ turn)\ +\ P(A\ wins\ 2^{nd}\ turn)\ +\ P(A\ wins\ 3^{rd}\ turn)\ +\ ...} $$
First looking at the individual cases for player A:
$$ P(A\ wins\ 1^{st}\ turn)\ =\ (A\ chooses\ white\ ball)\ $$ $$ =\ \frac{4}{12}\ =\ \frac{1}{3} $$ $$ P(A\ wins\ 2^{nd}\ turn)\ =\ (A,B,C\ all\ choose\ nonwhite\ balls)(A\ chooses\ white\ ball)\ $$ $$ =\ \bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\frac{1}{3}\ =\ \bigg(\frac{2}{3}\bigg)^3\frac{1}{3} $$ $$ P(A\ wins\ 3^{rd}\ turn)\ =\ (A,B,C\ all\ choose\ nonwhite\ balls)(A,B,C\ all\ choose\ nonwhite\ balls)(A\ chooses\ white\ ball)\ $$ $$ =\ \bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\frac{1}{3}\ =\ \bigg(\frac{2}{3}\bigg)^6\frac{1}{3} $$
Combining these cases back into the total probability: $$ P(A\ wins)\ =\ \frac{1}{3}\ +\ \bigg(\frac{2}{3}\bigg)^3\frac{1}{3}\ +\ \bigg(\frac{2}{3}\bigg)^6\frac{1}{3}\ +\ ... $$
This series can be modeled as an infinite sum:
$$ P(A\ wins)\ = \frac{1}{3} \sum\limits_{n=0}^{\infty}\bigg[\bigg(\frac{2}{3}\bigg)^3\bigg]^n $$
This is a infinite sum is a geometric series and can be represented as: $$ P(A\ wins)\ = \frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{9}{19} $$
The same geometric series holds for players B and C and we only need to multiply by appropriate probabilities for their cases:
$$ P(B\ wins)\ = \frac{2}{3}\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{6}{19} $$ $$ P(C\ wins)\ = \bigg(\frac{2}{3}\bigg)^2\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{4}{19} $$