Basic question: Calculating expected value

Question

I'm sorry for posting such a basic question, but me and my friend were discussing a problem and we just cant find a common ground. The problem is:

A guy is playing a game. On n^th day the game ends with the probability of 1/(100-n). So on the first day probability that the game ends is 1/99, on the second, if it hasn't ended on the first, probability of ending would be 1/98, and so on until 99^th day when the game ends with probability 1 if it hasn't so far.

What is the expected value of duration of the game?

Answer 1

Probability game ends on day n is $$\frac{98}{99} \frac{97}{98} \ldots \frac{100-n}{100-n+1} \frac{1}{100-n} = \frac{1}{99}$$. So expected value of what day the game ends is $$ 1\frac{1}{99}+2\frac{1}{99}+...+99\frac{1}{99} = \frac{4950}{99} = 50. $$

Intuitively, as each day is equally likely to be finished on, the expected day is in the dead middle of the possible outcomes 1 through 99 (think about rolling a 6-sided die, you expect to obtain the average of the values which is 3.5).

Answer 2

The probabilities are (starting from $1$),

$$\frac1{99},\frac{98}{99}\frac1{98}=\frac1{99},\frac{98}{99}\frac{97}{98}\frac1{97}=\frac1{99},\cdots$$ and so on until day $99$. As all outcomes are equal, the expectation is the midpoint of $1$ and $99$, i.e. $50$.