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I have been scratching my head over this and I can't see to figure out the exact numbers to use for this problem.

The question is as stated:

"The numbers 1, 2, . . . , 20 are assigned at random to 10 people so that each one gets two numbers. A person gets happy if their two numbers add up to 20. On average, how many of them get happy?"

I figured I would use the formula for Expected Value for discrete distributions.

Where EX = Sum of (the probabilities of each i x i).

X is the random variable = the total number of people who are happy and i is the number of people who are happy, i = 1,2,...,9

As 2 of the numbers out of 1-20 can't make a pair that add to 20, namely 10 and 20, then for the first person I calculated that he/she had the probability of picking a successful pair to be:

(18/20)*(1/9) = 9/190

However, from here it is where I get stuck.

Some guidance will be greatly appreciated.

Thanks

Mike Campbell
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  • Are you sure it's a sum of 20, and not 21, which makes a person happy? It just seems to me that it would be a more natural thing to be happy about. – Arthur Oct 28 '18 at 20:41
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    @coffeemath I would assume the numbers are distributed without replacement. – Arthur Oct 28 '18 at 20:43
  • I assumed they were distributed without replacement. That's why I have been struggling with it, as it makes it a lot harder to calculate. The question as posted in quotation marks is all I've been given. But now you mention it, it doesn't specifically state whether numbers can be repeated or not. I just thought because they are assigned randomly to each person, then they couldn't be reassigned to someone else as well – Mike Campbell Oct 28 '18 at 21:04

3 Answers3

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Alternatively one may have a hypergeometric view, see https://en.wikipedia.org/wiki/Hypergeometric_distribution with $k=9$, $n=10$ and $N=\binom{20}{2}$.

Michael Hoppe
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Let $X_i$ for $1\leq i\leq10$ be the random variable which is $1$ of person $i$ is happy, and $0$ otherwise. Then what we are after is $$ E(X_1+X_2+\cdots+X_{10}) $$ Expectation is linear, meaning that this is equal to $$ E(X_1)+E(X_2)+\cdots+E(X_{10}) $$ Since the $X_i$ are identically distributed (although not independent), all the expectations in the expression above are equal, so this is just $$ 10E(X_1) $$ which is quite a bit more manageable.

Arthur
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First I would determine how many combinations of 2 can you make from the integers 1 through 20. ( 20 choose 2 ). Now how many of those combinations add up to 20? $(1, 19); (2, 18); ...; (9, 11)$

So there are 9 of these combinations.

( 20 choose 2 ) = 190

So on average, 9 out of 190 people would be happy. That is less than 5%, I would say on average, nobody is happy :-(

So here I would say that 9/190 is your expected value that somebody that gets a number would be happy. If you plug that into the expression from Arthur's answer, if order to get the expected number of happy people, it would be the number of people ( 10 ) multiplied times the expectation that somebody with a number would be happy ( 9/190 ).

So, $$10 * ( 9/190 ) = 90/190$$, which is less than half a person.

jeffery_the_wind
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  • Thank you for the answers. Greatly appreciated. If understood correctly, then each person has the same expectation of getting a pair i.e. being happy, whereas I was trying to work it out based on the fact that the probability of the 2nd person getting a pair would be lower if the first person had received a pair etc – Mike Campbell Oct 28 '18 at 22:15
  • @MikeCampbell I think my answer works if the numbers are given out with replacement, and I think that question is trying to imply that the numbers are handed out without replacement. So you were right in your intuition. There is math theory about this situation. Check out Michael Hoppe's answer, and I will try to update my answer. – jeffery_the_wind Oct 29 '18 at 13:09