A box contains 10 colored discs of which 2 are red A box contains 10 colored discs of which 2 are red. A man pays **10 cent** to play a game in which discs are pulled out on at a time, without replacement. If his first draw is a red disc, he will win **25 cents**. If his second draw is a red, then he will win **20 cents**. If his third draw is a red disc, then he will win **5 cent**. Calculate his expected profit or loss My answer: $$\left(\left(\frac{2}{10}\times0.25\right) + \left(\frac 19\times 0.2\right) + \left(\frac18\times 0.05\right)\right) - (0.1) \\\= -\frac{31}{1440} = -0.0215277777777778$$ the correct answer: 9.33

I would say that there are 3 'winning' cases:

* win on the first draw: $25 $ with probability $2/10$

* win on the second draw: $20$ with probability $8/10 \times 2/9$

* win on the third draw: $5 $ with probability $8/10 \times 7/9 \times 2 / 8$




The expected gain is then $9.33\dots$.

But you need to substract the cost of the game! this yields $9.33\dots - 10 = -2/3$ cent as the final result.