Expected minimum lottery number

I'll solve this problem in general, where $b$ distinct numbers are chosen from $1$ to $a$ . Let $E(a,b)$ represent the expected value.

In order to find the expected value, we need to find the probability of $k$ being the smallest number, where $1 \le k \le a$ . If $k$ is the smallest number, that leaves $a-k$ possibilities for the other $b-1$ numbers. Hence the number of combinations where $k$ is the smallest chosen number is given by: $\displaystyle\binom{a-k}{b-1}$ . And the total number of possible combinations is $\displaystyle\binom{a}{b}$ .

Hence, the expected value is given by: $E(a,b) = \dfrac{1}{\binom{a}{b}} \sum_{k=1}^{a} k \binom{a-k}{b-1}$ Now I say that $\displaystyle\sum_{k=1}^{a} k \displaystyle\binom{a-k}{b-1} = \displaystyle\binom{a+1}{b+1}$ . Why? This can be proved algebraically, or combinatorially. I'll give you the latter. Clearly $\displaystyle\binom{a+1}{b+1}$ is the number of ways to choose $b+1$ objects out of $a+1$ objects. To count it another way, let $k+1$ represent the $2^{nd}$ smallest number chosen. Then there are $k$ choices for the smallest number, and the number of ways to choose the rest of the objects is $\displaystyle\binom{(a+1)-(k+1)}{b-1} = \displaystyle\binom{a-k}{b-1}$ . Hence the number of ways to choose $b+1$ objects out of $a+1$ objects can also be calculated as $\displaystyle\sum_{k=1}^{a} k \displaystyle\binom{a-k}{b-1}$ . $\square$

Therefore $E(a,b) = \dfrac{\displaystyle\binom{a+1}{b+1}}{\displaystyle\binom{a}{b}}$ . Using the factorial expansions, this can be simplified to $E(a,b) = \dfrac{a+1}{b+1}$ .

So, for this particular problem, we have $E(49, 7) = \dfrac{50}{8} = \boxed{6.25}$ .