Question 1030185: 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?
I'm not sure where to even start with this one.
Found 2 solutions by mathmate, stanbon:
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
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 average loss is $0.8514 per play.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 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?
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Random "winning" numbers::....... 20,000.........-1
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Corresponding probabilities:: ....1/24C6........1 - 1/(24C6)
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Expected value = 20,000*(1/24C6) - 1/[1-(1/24C6)]
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= 20000/134596 - 1/(1 - 7.43*10-6)
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= 0.1486-0.9999
= -$0.85 = -85 cents
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Cheers,
Stan H.
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