Statistics - Expected win/loss for lottery question

Apr 2008
1
0
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
Mar 2008
5,618
2,802
P(I'm here)=1/3, P(I'm there)=t+1/3
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
Dec 2007
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Zeitgeist
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.
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.

If you're not allowed to include numbers like 001, 021 etc, then there are (9)(10)(10) = 900 different three digit numbers. The above logic stays the same .....