Question 1086169: A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.
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If the card is a face card, and the coin lands on Heads, you win $8
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If the card is a face card, and the coin lands on Tails, you win $2
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If the card is not a face card, you lose $2, no matter what the coin shows.
Question: Find the expected value for this game (expected net gain or loss). (Round your answer to two decimal places.)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let's define four events:
F = event of drawing a face card
N = event of drawing a non-face card
H = event of the coin landing on heads
T = event of the coin landing on tails
The events F and N are complementary. Meaning that one event or the other, but not both, must happen. We either draw a face card (F) or we don't (N). This is why the probabilities add to 1
P(F) + P(N) = 1
Solve for P(N) to get
P(N) = 1 - P(F)
If we knew the probability of drawing a face card, P(F), then we could find the probability of not getting a face card P(N).
There are 4 suits with 3 face cards per suit (King, Queen, Jack). So 4*3 = 12 face cards out of 52 cards total.
Therefore,
P(F) = probability of drawing a face card
P(F) = (number of face cards)/(number of cards total)
P(F) = 12/52
P(F) = 3/13
And,
P(N) = probability of drawing a non-face card
P(N) = 1-P(F)
P(N) = 1-3/13
P(N) = 13/13-3/13
P(N) = 10/13
An alternative is to think "there are 40 non face cards (52-12 = 40) out of 52, so 40/52 = 10/13"
Those two probabilities, P(F) and P(N), will be used later. Let's move on to the next two probabilities
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Assuming we have a fair coin (either side is likely to be landed on), this means
P(H) = 1/2
P(T) = 1/2
Note how P(H) + P(T) = 1
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So far we have the four simple probabilities
P(F) = 3/13, P(N) = 10/13
P(H) = 1/2, P(T) = 1/2
Assuming the events of drawing a card and flipping a coin are independent, then we can form the compound probabilities
P(F & H) = P(F)*P(H) = (3/13)*(1/2) = 3/26
P(F & T) = P(F)*P(T) = (3/13)*(1/2) = 3/26
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Let's introduce some new notation. Similar to the probability P(X) notation, let's introduce the function V(X) where V is the net value and X is the general event. To be more specific, writing V(F) represents the net value of drawing a face card.
The three cases we're concerned with are:
V(F & H) = net value for getting face card and heads = 8
V(F & T) = net value for getting face card and tails = 2
V(N) = net value for getting non face card = -2
The negative value (-2) indicates a loss of 2 dollars.
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When we play the game out, there are three cases:
Case A = drawing a face card and the coin landing on heads
Case B = drawing a face card and the coin landing on tails
Case C = drawing a non-face card
What we do is multiply the probabilities for each case happening with the net values for each case
For case A, we have the probability P(F & H) = 3/26 and the net value V(F & H) = 8 so
P(F & H)*V(F & H) = (3/26)*8 = 24/26 = 12/13
Similarly for case B
P(F & T)*V(F & T) = (3/26)*2 = 6/26 = 3/13
and finally case C
P(N)*V(N) = (10/13)*(-2) = -20/13
Once we have those three results (12/13, 3/13, -20/13), we add them up
(12/13) + (3/13) + (-20/13) = (12+3+(-20))/13
(12/13) + (3/13) + (-20/13) = -5/13
The fraction -5/13 converts to the approximate decimal value -0.384615 which rounds to -0.38
What does this value mean? It is the expected value, ie the expected loss. We expect to lose about 38 cents for each game played. This is not a fair game (because expected value isn't 0). The game clearly favors the house instead of the player.
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Answer: -0.38 (represents an average loss of $0.38 or 38 cents per game)
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