SOLUTION: Trey has a deck of 10 cards numbered 1 through 10. He is playing a game of chance. This game is this: Trey chooses one card from the deck at random. He wins an amount of money equ

Question 1201591: Trey has a deck of 10 cards numbered 1 through 10. He is playing a game of chance.
This game is this: Trey chooses one card from the deck at random. He wins an amount of money equal to the value of the card if an even numbered card is drawn. He loses $7 if an odd numbered card is drawn.
(a). Find the expected value of playing the game.
(b). What can Trey expect in the long run, after playing the game many times?
(He replaces the card in the deck each time.)
Found 2 solutions by Glaviolette, math_tutor2020:
Answer by Glaviolette(140) (Show Source): You can put this solution on YOUR website!

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Part (a)

C = card value = 1,2,3,...,9,10
W = winnings in dollars

If C is odd, then W = -7 to indicate he lost $7.
If C is even, then W = C (eg: C = 8 means Trey won $8)
Each P(W) value is 1/10 = 0.1 assuming each card is likely to be chosen.

Here is one way to write out the probability distribution.

C W P(W)
1 -7 0.1
2 2 0.1
3 -7 0.1
4 4 0.1
5 -7 0.1
6 6 0.1
7 -7 0.1
8 8 0.1
9 -7 0.1
10 10 0.1

Feel free to replace each 0.1 with 1/10 if you prefer.

Or we could condense things like so

C W P(W)
2 2 0.1
4 4 0.1
6 6 0.1
8 8 0.1
10 10 0.1
odd number -7 0.5

Multiply each W and P(W) value to form a new column.

C W P(W) W*P(W)
2 2 0.1 0.2
4 4 0.1 0.4
6 6 0.1 0.6
8 8 0.1 0.8
10 10 0.1 1
odd number -7 0.5 -3.5

Then add up the results of that new column.
0.2+0.4+0.6+0.8+1+(-3.5) = -0.5

This is the expected value.
The average amount Trey expects to lose on any given game is $0.50 aka 50 cents.

Answer: -0.50

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Part (b)

Let's say he plays this game 1000 times.
This would mean he expects to lose 0.50*1000 = 500 dollars over the course of those 1000 games.
He might get lucky to win money once in a while, but the long run odds are stacked against him.

In general, if he plays n games, then he'll expect to lose 0.50n dollars on average.

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C	W	P(W)
1	-7	0.1
2	2	0.1
3	-7	0.1
4	4	0.1
5	-7	0.1
6	6	0.1
7	-7	0.1
8	8	0.1
9	-7	0.1
10	10	0.1

C	W	P(W)	W*P(W)
2	2	0.1	0.2
4	4	0.1	0.4
6	6	0.1	0.6
8	8	0.1	0.8
10	10	0.1	1
odd number	-7	0.5	-3.5