# Lottery probability in a year

I have a few exam review questions that I don't know how to solve. Maybe someone has a solution to it?

A lottery chooses a winning number x in the set S = {0, 1, 2, 3, ..., 999}

If you want to play, you pay $1 and choose a number y in S. If y=x, then you receive \$700. So your net dollars is \$699. Otherwise, you lose$1.

Assume that you play this game once a day for one year (365 days)

each day, the lottery chooses a new winning number

each day, you choose a random y uniformly at random from the set S, independently from previous choices.

Define the random variable X to be the total amount of dollars that John wins during one year. Determine the expected value E(x).

It also gives you a hint: use linearity of expectation.

Expected Value of one day is $(699*1/1000)+(-1*999/1000)=-300/1000=-0.3$

Expected Value of one day times days in a year is $-0.3*365=-109.5$, expected value of a year is -109.5

Standard Deviation of a year is $\sqrt{365}*standard DeviationOfOneDay$

Standard deviation is written as $\sqrt{((occurence*(Value1-mean)^2+ occurence* (Value2-mean)^2)/1000)}$

Standard Deviation of 1 day is $\sqrt{((1*999.3^2+ 999* 0.7^2)/1000)}\approx 31.6084$

So the standard deviation of the year is $\approx 603.877$

There are two random variables;

------$Y$ = John receives $700 ------$Z$= Jonh reeceives nothing$E(Y) = 699* 1/1000 = 699/1000 E(Z) = -1* 999/1000 =-999/1000 $By linearity of Expectation, -------$E(X)= E(Y+Z) = E(Y) +E(Z)= 699/1000-999/1000= -3/10 $That was for a day. Now for a year, you multiply that with$365\$

Hope this helps