1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ May 4, 2016 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, 2016 at 2:03
  • 1
    $\begingroup$ Check this see if it fits. $\endgroup$ May 4, 2016 at 2:21

2 Answers 2

2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

Hint: The expected value operator is a linear operator, which means that:

$$\mathbb{E}(X_1 + \cdots + X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n).$$

Is there any way you could write the number of winning tickets as a sum of other quantities whose expected values are easy to compute?

$\endgroup$
4
  • $\begingroup$ +1 I like it, but I give credit to @matteo for stating essentially the same thing. $\endgroup$
    – whuber
    Feb 12 at 21:32
  • $\begingroup$ @whuber: I think matteo is being quite a bit more generous in his hint than me! ; ) $\endgroup$
    – Ben
    Feb 12 at 21:37
  • 1
    $\begingroup$ After waiting more than four years to answer (as did Matteo), I think few people would object to going beyond just hinting :-). $\endgroup$
    – whuber
    Feb 12 at 21:38
  • 1
    $\begingroup$ @whuber: Ha ha, I didn't notice that part! (Just assumed this was a new question. Didn't realise it got bumped!) $\endgroup$
    – Ben
    Feb 12 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.