Show that the Mathematical expectation of what the man earns is $\frac {S}{n (2n-1)}$

expected valueprobability

A bag contains $2n$ counters of which half are marked with odd numbers and half with even numbers, the sum of all the numbers being $S$. A man is to draw two counters. If the sum of the numbers drawn is odd, he is to receive that number if dollars; if even, he has to pay that amount of dollars. Show that his expectation is $\dfrac {S}{n(2n-1)}$

Attempt: Let the odd numbers be $\{a_1,\cdots,a_n \}$ and the even numbers be $\{b_1,\cdots,b_n \}$

Let $X_{i,1}, X_{i,2}$ represent the $i_{th}$ counters. Let $X_i$ be an indicator variable which takes the value $1$ if $X_{i,1}+ X_{i,2} $ is odd and the value $-1$ if $X_{i,1}, X_{i,2}$ is even.

Then $P(X_i=1) = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot 2= \dfrac{1}{2};~P(X_i=-1)=\dfrac {n}{2n} \cdot \dfrac {n-1}{2n} \cdot 2=\dfrac {n-1}{2n}$

Thus, $E[X_i]= \dfrac{1}{2}~[ ~\sum_{k=1}^n a_k – \dfrac {n-1}{n} \sum_{k=1}^n b_k ~]~=\dfrac{1}{2} [S – \dfrac {2n-1}{n}\cdot \sum_{k=1}^n b_k ]$

$\sum_1^n E[X_i]=\dfrac{n}{2} [S – \dfrac {2n-1}{n}\cdot \sum_{k=1}^n b_k ]$

I feel stuck here. Could someone please show me a direction from here.

Thanks a lot!

Best Answer

Note that it is more likely that the sum of his numbers is odd than that it is even:

  • There is a total of $(2n)\cdot(2n-1)$ possible choices.
  • If the sum of the two chosen numbers should be odd, then that means that he chose $(a_i, b_{\tilde i})$ or $(b_i,a_{\tilde i})$ for any $i,\tilde i=1\dots n$. So there are $2\cdot (n\cdot n)$ choices such that the sum is odd.
  • However, for even numbers, we must have either $(a_i, a_{\tilde i})$ or $(b_i,b_{\tilde i})$ for some distinct $i,\tilde i$. So there are $2\cdot (n\cdot (n-1))$ choices.

Hence, $\mathbb P(\text{odd sum})=\frac{n}{2n-1}$ and $\mathbb P(\text{even sum})=\frac{n-1}{2n-1}$.

Let $W$ denote the win. Then $$\mathbb E(W)=\frac{n}{2n-1}\cdot\mathbb E(W\mid \text{odd sum})+\frac{n-1}{2n-1}\cdot\mathbb E(W\mid \text{even sum}).$$

Now (tiny exercise), note that $-\mathbb E(W\mid \text{even sum})=\mathbb E(W\mid \text{odd sum})=\frac{S}{n}$. Hence, $$\mathbb E(W)=\frac{S}{n\cdot(2n-1)}.$$

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