If the game costs $5 to play, what is the standard deviation?

Question

In a new card game, you start with a well- shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $55. If you draw 3 black cards, you win $25. For any other draws, you win nothing.

If the game costs $5 to play, what is the standard deviation?

Answer 1

There are 26 black cards in a deck of cards. Thus there are 26C3 = 26*25*24/6 = 26*25*4 = 2600 different combinations of 3-card-hands in which all the cards are black that you can form from a 52-card deck. Note that because these are combinations rather than permutations, getting {Jack of Spades, Jack of Clubs, Queen of Clubs} would be counted as the same hand as {Queen of Clubs, Jack of Spades, Jack of Clubs} but a different from {Queen of Spades, Jack of Spades, Jack of Clubs} because the order of the cards does not matter when calculating combinations, only the cards themselves.

Similarly, there are 52C3 = 52*51*50/6 = 52*17*25 = 22,100 different total combinations of 3-card-hands from a 52-card deck.

Since any combination of 3 cards in a hand is equally likely as any other combination, it follows that the probability of getting 3 black cards is simply (# of combinations of 3 cards from a 52-card deck in which all cards are black)/(# of total combinations of 3 cards from a 52-card deck) which equals 2600/22100 = 26/221 = 2/17

Using the same logic since there are 13 hearts in a 52-card deck there are 13C3 = 13*12*11/6 = 13*2*11 = 286. Thus the probability of getting 3 heart cads is simply (# of combinations of 3 cards from a 52-card deck in which all cards are hearts)/(# of total combinations of 3 cards from a 52-card deck) which equals 286/22100 = 22/1700 = 11/850.

Thus, P(3 hearts) = 11/850 and P(3 black cards) = 2/17. Thus the probability of not winning = P(getting neither 3 hearts nor 3 black cards). Since getting 3 hearts and getting 3 black cards are mutually exclusive outcomes, this probability = 1 - 11/850 - 2/17 = 1 - 11/850 - 100/850 = 739/850.

The game costs 5 dollars to play. Our three possible events (or outcomes) in our card game are {Neither 3 hearts nor 3 black cards, 3 hearts, 3 black cards}

Thus the following are our payoffs for each possible event.

Neither 3 hearts nor 3 black cards: -$5 (with probability 739/850)

3 hearts: -$5 + $55 = $50 (with probability 11/850)

3 black cards: -$5 + $25 = $20 (with probability 2/17)

The variance of our winnings in this card game is E[X^2] - E[X] if we define X = total winnings in dollars

E[X^2] = (-5)^2 * 739/850 + (50)^2 * 11/850 + (20)^2 * 2/17 = 101.147059

E[X] = -5*739/850 + 50*11/850 + 20*2/17 = -1.347

Var[X] = E[X^2] - (E[X])^2 = 101.47059 - (-1.347)^2 = 99.656.

Our standard deviation is simply sqrt(Var[X]) = sqrt(99.656) which is approximately $9.98