TITLE: Translating mathematical problem statement into natural language
QUESTION [7 upvotes]: Let us consider following problem :
One thousand tickets are sold at $\$1$ each for a color television valued at $\$350$. What is the expected value of the gain if you purchase one ticket?
I would like to describe my basic problem related to this task. English is not my native language therefore I did not understand  logic of statement. What I know is that  we had $1000$ tickets that are sold at $\$1$ each. This means that  the total revenue from $1000$ tickets is $\$1000$. Is this correct? We have a color television that costs $\$350$. 
My question is: 
What is the expected value of the gain if you purchase one ticket, really I don't understand the connection of purchase=buy. If I bought one ticket that cost  $\$1$, and if I bought a television which costs $\$350$, then my  gain will be $\$349$ correct? If I lost, then my gain will be  $\$-1$ because I  paid this $\$1$. What about their probability? If the probability that out of $\$1000$ I will  win  is $0.001$, then the probability that I will loose is $0.999$, so expected value will be
$$349\times 0.001+(-1)\times 0.999=-0.65$$
But still I do not understand logically this statement, could you  describe please in a  simple manner this  problem?

REPLY [2 votes]: I can understand the results of such calculations if I can imagine that the experiment described can be repeated many times under the same circumstances and independently. Say, somebody organizes the television experiment $1\ 000\ 000$ times and you take part always and there are always $999$ other gamblers present. The question is your average gain. 
For sure, you lose $1$ dollar every time but you win the television set with a chance of $1$ to $1\ 000$. That is, you will win (about) $1\ 000$ times out of the $1 \ 000\ 000$ cases. So your total loss is $ \$1 \ 000 \ 000$ and your total gain is $\$350\ 000$. The bottom line is  $\$350\ 000-\$1\ 000\ 000=-\$650\ 000.$
As far as the average gain: You simply divide your total gain by $1\ 000\ 000$, the number of experiments.
$$\frac{-\$650\ 000}{1\ 000\ 000}=-\$0.65.$$
Probability theory does not work such a naïve way. We say that the probability of winning is $\frac1{1\ 000}=$ and the probability of not wining is $1-\frac1{1\ 000}=0.999.$ The expected gain is the sum of the products of the gains and the probabilities:
$$(-1)\cdot0.999+349\cdot0.001=-0.999+0.349=-0.65.$$