Question

In a lottery game, a player picks six numbers from 1 to 30. This means that there are 593775 ways to play the game. If the player gets the 1 combination that matches all six numbers, they win 50,000 dollars. The other 593774 combinations result in losing $1. What is the expected value of this game?

Answer 1

Given:

Numbers picked from 1 to 30 = 6 numbers

Number of ways = 593775

1 combination = $50,000

Given that the other 593774 combinations result in losing $1, let's find the expected value of losing the game.

Let's first find the number of ways of picking 6 numbers out of 30 numbers.

Apply the combination formula:

`\begin{gathered} ^(30)C_6=(30!)/(6!(30-6)!) \\ \\ =(30!)/(6!(24)!) \\ \\ =(30*29*28*27*26*25*24!)/(6*5*4*3*2*1*24!) \\ \\ =(427518000)/(720) \\ \\ =593775\text{ ways} \end{gathered}`

The probability of winning will be:

`P(win\text{ 50000\rparen = }(1)/(593775)`

The probability of losing is:

`P(lose\text{ 1\rparen = }(593774)/(593775)`

Hence, the expected value will be:

`\begin{gathered} E=50000*(1)/(593775)+((-1)*(593774)/(593775)) \\ \\ E=50000*(1)/(593775)-(593774)/(593775) \\ \\ E=(50000)/(593775)-(593774)/(593775) \end{gathered}`

Solving further, we have:

`\begin{gathered} E=(50000-593774)/(593775) \\ \\ E=-(543774)/(593775) \\ \\ E=-0.916 \end{gathered}`

Therefore, the expected value of this game is -0.916

ANSWER:

-0.916