Originally Posted by

**Volga** iv. Finally, the fun part.

The probability that the person who starts the game wins the game.

I consider the sequence

$\displaystyle \frac{1}{3},...\frac{2^{2m}}{3^{2m+1}}, m=0,1,2,...$ which represents probabilities of winning the game (picking the red disc) by the person who makes the first move.

Total probability is the sum

$\displaystyle \Sigma_{m=0}^{\infty}\frac{2^{2m}}{3^{2m+1}}=\Sigm a_{m=0}^{\infty}(\frac{2^2}{3^2})^m\frac{1}{3}=\fr ac{1}{3}\Sigma_{m=0}^{\infty}(\frac{4}{9})^m=\frac {1}{3}\frac{1}{1-4/9}=\frac{3}{5}$

Then the probability of the other person winning is 1-3/5=2/5

Since 3/5>2/5, I would choose to start. (makes sense since I can pick up the red disc on the first move already, without giving the other person a chance to make a single move)