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For the following questions A,B and C, find the expected values of the game given that the game you are playing has the following initial rules, you pick a card from a standard deck of cards and bet a dollar.

A) If your card comes up you get your dollar back and win a dollar B) If your card comes up you win 48 dollars. C) Based on the expected values you found in (A) and (B) in last two problems, is either game worth playing?

Please explain all A,B and C answers in detail.

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Wikipedia tells us that the expected value is:

$E[x] = x_1p_1 + x_2p_2 + \ ...\ + x_kp_k$, where the $x_i$ are the values that $x$ can take, and $p_i$ are the corresponding probabilities that $x$ will take on that value.

There are 52 cards in the deck.

For game (A), the value of drawing your selected card is 1 dollar, and the value of drawing any other card is 0. The respective probabilities are 1/52 and 51/52. So we have:

$E = 1 \cdot \dfrac{1}{52} + 0 \cdot \dfrac{51}{52} = 1.9$ cents

Can you do game (B)?

(Added) You might want to consider how to generalize this, just in case you are accosted in a dark alley and forced to wager your life on a discrete random variable.

Let's say that you choose $1$ out of $n$ cards/roulette slots/numbers/etc., and you will receive $R$ if you are right, and $0$ if you are wrong. Then the expectation is

$E = R \cdot \dfrac{1}{n} + 0 \cdot \dfrac{n-1}{n} = \dfrac{R}{n}$

If you compare this with what it cost you to play the game, you can see what the payoff would need to be for you to "break even".

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And in regard to your request that I/we "explain all answers... in detail", I would first ask for you to show some work. – The Chaz 2.0 Dec 7 '11 at 18:43

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