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In the problem statement of project Euler problem 121, the following information is given:

A pouch contains one black chip and one white chip. In a simple game, one player takes a chip at random and makes a note of the colour. After every turn the chip is returned to the pouch and an extra black chip is added, and another chip is taken at random.

A player has to pay Rupee 1 to play and if he is considered to have won if they have taken more white chips than black chips at the end of the game. If the game is played for four turns, the probability of a player winning is exactly 11/120.

How did they arrive at 11/120 for the example?

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up vote 4 down vote accepted

In the first round the chance is 1/2 to get a white chip, in the second round, it is 1/3 and so on. Now you list all possibilities to have at least three white chips, and sum over the products of the corresponding probabilities.

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is the probability of winning 1,2,3 and loosing 4th round $\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}$ ? – nariknahom Oct 16 '11 at 17:11
@narik No, this is $\frac 12 \cdot \frac 13 \cdot \frac 14 \cdot \frac 45$. – Phira Oct 16 '11 at 17:14
thanks a lot! now i get it. – nariknahom Oct 16 '11 at 17:20

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