Answer to Question #124501 in Statistics and Probability for Dan Degrossa

Question

In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 50,000 dollars. Otherwise, they lose $1.

What is the expected value of this game? $

Answer 1

probability of winning: six out of total 27 numbers:

=

$\frac{6}{27}\frac{5}{26}\frac{4}{25}\frac{3}{24}\frac{2}{23}\frac{1}{22}$ = $\frac{1}{296010}$

Probability of losing :

1- $\frac{1}{296010}$ =$\frac{296009}{296010}$

We have the following table for x: amount we win or lose and p(x) probability of win or lose

x: 50000 -1

P(x): $\frac{1}{296010}$ $\frac{296009}{296010}$

Expected value : $\Sigma x*P(x)$

= $\frac{1}{296010}$ *50000 +$\frac{296009}{296010}$ *(-1)

= -0.831

answer: expected value of the game = $-0.831