Probability Distribution

The game follows geometric distribution with p=2/6=1/3. Let X be the no. of draws until a red ball is drawn. Then X~Geo(1/3) and E(X)=1/p=3. This means that after a log run, John will make three draws in each game averagely. So, the expected gain of John until a red ball is drawn is 100*2=200 dollars.

P(ended after drawing 0 times)=(1/3)
P(ended after drawing 1 times)=(1/3)(2/3)
P(ended after drawing 2 times)=(1/3)(2/3)^2
.
.
.
P(ended after drawing n times)=(1/3)(2/3)^n
Expected number of drawing,
x=(0)(1/3)+(1)(1/3)(2/3)+(2)(1/3)(2/3)^2+...+(n)(1/3)(2/3)^n
So
3x=(1)(2/3)+(2)(2/3)^2+(3)(2/3)^3+...
2x=............(1)(2/3)^2+(2)(2/3)^3+...
So x=(2/3)+(2/3)^2+(2/3)^3+...
By sum to infinity geometric sequence
x=a/(1-r)=(2/3)/(1-2/3)=2
Expected number of drawing=2

Expected gain=100x2=$200


2010-11-27 23:21:21 補充：
Sorry, I have made a mistake.
All 'drawing' in my answer should be replaced by 'winning'.

That is,
P(ended after winning 0 times)=(1/3)
P(ended after winning 1 times)=(1/3)(2/3)
.
.
P(ended after winning n times)=(1/3)(2/3)^n
..
Expected number of winning=2
Expected gain=100x2=$200