Question

In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 40,000 dollars. Otherwise, they lose $1. Hint: There are 296010 possible ways of choosing six numbers, and you want just the one winning combination.

What is the expected value of this game?

Answer 1

Answer:

We conclude that  expected value of this game is -0.865$.

Step-by-step explanation:

We know that in a lottery game, a player picks six numbers from 1 to 27.

We know that

$C_6^{27}=296010$

As there is only one advantageous combination, we conclude that the number of non-winning combinations is 296009.

He can win 40,000 dollars.

We calculate:

$E(X)=\frac{1}{296010}\cdot 40000\ - \frac{296009}{296010}\cdot 1\ \\\\E(X)=\frac{40000-296009}{296010}\, \ \\\\E(X)=-0.865\,$

We conclude that  expected value of this game is -0.865$.