# Statistics - Expected win/loss for lottery question

#### 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. Please help! What theory/formula do I use here? Thank you. #### Moo MHF Hall of Honor 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 $$\displaystyle \frac{899}{900}$$ ----> $$\displaystyle E(X)=-1 \times \frac{899}{900}+750 \times \frac{1}{900} \approx -0.1656$$ #### mr fantastic MHF Hall of Fame 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.