Question 1030185
 
Question:
In a lottery game, a player picks six numbers from 1 to 24. If the player matches all six numbers, they win $ 20,000. Otherwise, they lose $1. Assume it costs nothing to play.
What is the expected value of this game?
 
Solution:
To start solving the problem, we need to understand the definitions of words, the most important of which is "expected value".
 
Wiki defines expected value as:
"The expected value is also known as the expectation, mathematical expectation, EV, average, mean value, mean, or first moment. More practically, the expected value of a discrete random variable is the probability-weighted average of all possible values."
 
To mathematically calculate the expected value, we will need to do the sum of the product
x*P(x)
for all possible (outcome) values of x, that is, all values of x for which the probability is non-zero.
 
Here the possible outcomes are x=-1 (lose) or x=20000 (win)
Total possible number of outcomes = C(24,6) [24 choose 6]
=24!/(6!18!)
= 134596
Out of which there is only one winning combination.
Therefore we conclude:
P(win 20000)=1/134596
P(lose 1)=134595/134596
and hence the expected value is:
20000*(1/134596)+(-1)*(134595/134596)
=-114595/134596
=-0.8514 (rounded to four places after decimal)

Interpretation:
In the long run, if the game is played many, many times, the <i>average</i> loss is $0.8514 per play.