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A lottery consists of choosing 11 different numbered balls (the winning numbers) at random from a group of 100 numbers. A player chooses 11 numbers on a ticket. What's the expected number of winning numbers chosen on a ticket?

I tried this method: $$\begin{array}{rcl} && 0\cdot\mathbb{P}(0\text{ winning numbers}) + ....+ 11\cdot \mathbb{P}(11\text{ winning numbers}) \\ & = &0 \cdot\left(\binom{11}{0}/\binom{100}{11}\right) + ....+11\cdot\left(\binom{11}{11}/\binom{100}{11}\right) \end{array}$$

This should be correct but it's a pain to calculate by hand (as I would have a few minutes to do this without a calculator on an exam).

Question: Can anyone think of a faster way?

I'm trying to find a family that represents this process, but can't think of any. Perhaps we can use a new random variable, which is a sum of 1s if the $i$th number matches and 0s if the $i$th number does not match.

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    $\begingroup$ You should probably add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung - Reinstate Monica May 4 '16 at 1:51
  • $\begingroup$ Thanks for the suggestion - I'll add the self-study tag! I feel like I've gone as far as my knowledge will allow, made a good faith attempt, and provided a working solution already. I'm wondering if anyone can think of a new way to approach this. $\endgroup$ – Mico May 4 '16 at 2:03
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    $\begingroup$ Check this see if it fits. $\endgroup$ – Antoni Parellada May 4 '16 at 2:21
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Perhaps you could try factoring as follows: $$ \frac{1}{\binom{100}{11}}\left[\sum\limits_{n=1}^{11} \binom{11}{n} + \sum\limits_{n=2}^{11}\binom{11}{n} + \cdots + \sum\limits_{n=10}^{11}\binom{11}{n} + \binom{11}{11} \right] $$ $$ = \frac{1}{\binom{100}{11}}\left[ \sum\limits_{m=1}^{11}\sum\limits_{n=m}^{11} \binom{11}{n} \right] \tag{1} $$ Using a special case of the binomial theorem, this would reduce to: $$\frac{1}{\binom{100}{11}}\left[ \left(2^{11} - \binom{11}{0}\right) + \left(2^{11}- \sum\limits_{k=0}^1\binom{11}{k}\right) + \cdots + \left(2^{11} - \sum\limits_{k=0}^9 \binom{11}{k} \right)+\left(2^{11} - \sum\limits_{k=0}^{10}\binom{11}{k} \right) \right]\\ $$ $$ = \frac{1}{\binom{100}{11}} \left[2^{11} - \sum_{j=0}^{10} \sum_{k=0}^j \binom{11}{k} \right] \tag{2}$$ Keep in mind that further reduction/simplification along these lines is not possible, since: $$\sum_{j=0}^{10} \sum_{k=0}^j \binom{11}{k} = 2^{11} - \sum\limits_{m=1}^{11}\sum\limits_{n=m}^{11} \binom{11}{n} $$ In other words, attempting to iterate the argument just leads one to cycle between the results (1) and (2). Needless to say, it is at best arguable or questionable if a double sum of binomial coefficients actually could make one's life easier, but I thought I would at least write it out.

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It seems much simpler than what you are attempting, but maybe I am missing something.

You have $N=100$ possible numbers, out of which $K=11$ are winning ones.

When a player chooses their 11 symbols, each of these has a probability of $\frac{K}{N}=\frac{11}{100}=0.11$ of being one of the winning numbers.

Assuming the player choose numbers for their ticket without replacement, i.e. the same number never appear twice in the same ticket, the expected number of winning numbers per ticket is given by the "number of numbers" in each ticket times the probability of each being the winning one, that is $11 \times \frac{K}{N}= 1.21$

As @Antoni Parellada suggested in the comments, this correspond to the expected value of an hypergeometric distribution

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