Question

You pick a card at random from an ordinary deck of 52 cards. If the card is a diamond, you get some points; if not, you lose 2 points. What value for the diamonds would make the game fair?

a) 12
b) 6
c) 24
d) 3

Answer 1

Answer:12

Step-by-step explanation:

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Answer 2

The value for the diamonds that would make the game fair is given by: Option B: 6

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

How to find the mean (expectation) of a random variable?

Supposing that the considered random variable is discrete, we get:

$Mean = E(X) = \sum_{\forall x_i} f(x_i)x_i$

where $x_i; \: \: i = 1,2, ... ,n$ is its n data values

and $f(x_i)$ is the probability of $X = x_i$

For the game to be fair, the expected number of points that one can loose should be equal to the expected number of points one can gain.

Thus, for game to be fair, we need:

Expected number of points gained from picking diamond = expected number of points lost for not picking diamond.

Let we get P points if we pick diamond.

Let X be the random variable tracking the score of the player for each single draw.

The probability of picking a diamond from deck of 52 cards is 13/52 = 1/4

It is because there are total 52 ways to choose a card but only 13 ways t to choose a diamond.

Thus, probability of getting P points = 1/4  = 0.25

The probability of not getting a diamond = 1 - probability of getting diamond = 1 - 0.25 = 0.75

The points we get when not getting diamond = -2 (negative since we actually loose it, so we can take it as negative gain).

Then we need E(X) = expectation of X = 0 (since the negative points of loosing would equate to positive point of winning, thus, making the expected overall score 0, thus, keeping it a fair game.)

Now we have:

• P(X = -2) = P(choosing non-diamond card) = 0.75
• P(X = p) = P(choosing diamond card) = 0.25

There are the only values X can have.

Thus, we get:

$E(X) = -2 \times P(X = -2) + p \times P(X = p)\\ E(X) = -2 \times 0.75 + p \times 0.25\\0 = -1.5 + 0.25p\\\\p = \dfrac{1.5}{0.25} = 6$

Thus, the value for the diamonds that would make the game fair is given by: Option B: 6

Learn more about expectation of a random variable here:

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