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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

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  • $\begingroup$ What did you try? $\endgroup$ – Zafer Cesur Mar 13 '14 at 17:25
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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:

$$ P(B\ wins)\ = \frac{2}{3}\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{6}{19} $$ $$ P(C\ wins)\ = \bigg(\frac{2}{3}\bigg)^2\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{4}{19} $$

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Let $a$ be the probability $A$ wins.

If A draws a white immediately, she wins. If she draws a red, and so do the other two, then her probability of winning after that is $a$. Thus $$a=\frac{1}{3}+\left(\frac{2}{3}\right)^3 a.$$ This gives a linear equation for $a$. Solve.

Another way: Let $a$ be the probability A wins. For B to win, A must fail on her first try, and B must ultimately win. Given that A has failed on her first trial, the probability B wins is $a$. Thus the probability B wins is $\frac{2}{3}a$. Similarly, the probability C wins is $\left(\frac{2}{3}\right)^2a$.

We have $$a+\frac{2}{3}a+\left(\frac{2}{3}\right)^2a=1,$$ and now we can compute the three probabilities.

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