# Thread: Statistics - Expected win/loss for lottery question

1. ## Statistics - Expected win/loss for lottery question

Hello! I need help on this question. I've been thinking about it for ages and I have absolutely no idea where to start.

The local lottery sets up a game wherein a player has a chance of collecting $750. The player must choose a 3-digit number and if it matches, the player receives$750.
It costs $1 to play the game. What is the expected profit (or loss) for each player? I'm not sure if there was a thread on this before. I googled but the link was not valid anymore. Please help! What theory/formula do I use here? Thank you. 2. Hello, Let X denote the earning for each game.$\displaystyle X=-1$if the number doesn't match.$\displaystyle X=750$if the number matches. The value you're looking for is$\displaystyle E(X)=\sum_x x P(X=x)$, with x taking values from$\displaystyle \{-1;750\}$Now let's calculate the probabilities. Choosing a 3 digit number is equivalent to choosing randomly three times a number between 0 and 9 (except for the first digit). YZT$\displaystyle Y \in \{1;2; \dots ; 9\}$-> probability to get the matching one :$\displaystyle \frac{1}{9}\displaystyle Z \in \{0;1;2; \dots ;9 \}$-> probability to get the matching one :$\displaystyle \frac{1}{10}\displaystyle T \in \{0;1;2; \dots ;9 \}$-> probability to get the matching one :$\displaystyle \frac{1}{10}$Another way to do it, more direct, is to know that there are$\displaystyle 999-100+1=900$number between 100 and 999. Hence, the probability of getting the right number is$\displaystyle \frac{1}{900}$The probability of getting a wrong number is$\displaystyle \frac{899}{900}$---->$\displaystyle E(X)=-1 \times \frac{899}{900}+750 \times \frac{1}{900} \approx -0.1656$3. Originally Posted by defygravityy Hello! I need help on this question. I've been thinking about it for ages and I have absolutely no idea where to start. The local lottery sets up a game wherein a player has a chance of collecting$750. The player must choose a 3-digit number and if it matches, the player receives $750. It costs$1 to play the game.
What is the expected profit (or loss) for each player?

I'm not sure if there was a thread on this before. I googled but the link was not valid anymore.

Excluding 0 but including numbers like 001, 021 etc, there are 999 different three digit numbers. So your probability of winning $750 is 1/999 and of losing$1 (that is, winning $-1) is 998/999. Expected profit ($) = (750)(1/999) + (-1)(998/999) = -248/999, which represents a loss.