If a card is drawn at random from a standard deck of 52, the probability of drawing one of the 26 black cards is ; the probability of drawing an ace is . Now, examine each of the three scenarios.

(1) If the ace of spades is replaced with the joker, this leaves 25 black cards and 3 aces out of a total of 52 cards. The probability of drawing one of the 25 black cards is ; the probability of drawing an ace is .

(2) If the joker is added, there are still 26 black cards and 4 aces, but the deck now has 53 cards. The probability of drawing one of the 25 black cards is ; the probability of drawing an ace is .

(3) If the ace of spades is removed, this leaves 25 black cards and 3 aces out of a total of 51 cards. The probability of drawing one of the 25 black cards is ; the probability of drawing an ace is .

In all three cases, both probabilities have changed.

If the odds in favor of an event are to , the probability of that event is . Setting  and , the probability of the event is

.

The odds in favor of an event are equal to the ratio of the probability of the event to that of the opposite event. Therefore, we want to determine :

,

or 17 to 8 in favor.

The fairness or unfairness of the game is a function of the expected value, which can be calculated by multiplying the probability of each outcome by its value, and adding the products.

We will examine the value of the game to Jack; a positive value indicates a gain to Jack, and a negative value indicates a gain to Jill.

Since only ranks matter in this game, there are thirteen equiprobable outcomes. Four are favorable to Jill, and the other nine are favorable to Jack. There are two events, a win for Jill and a win for Jack, which we will call  and , respectively. Their probabilities and their values to Jack are:

: A king, queen, jack, or ten is drawn, and Jill wins.

Probability: 

Value to Jack:

: A card of any of the other nine ranks is drawn, and Jack wins:

Probability: 

Value to Jack: 

The expected value of one play of the game to Jack is

in Jack's favor.

Two cards are drawn from the deck without regard to order, so the sample space  is the set of all combinations of two cards from a set of twelve. The size of this sample space is 

The event is the set of all combinations of two cards from the set of four kings. The size of this event space is

The probability of the event is

The fairness or unfairness of the game is a function of the expected value, which can be calculated by multiplying the probability of each outcome by its value, and adding the products.

We will examine the value of the game to Jack; a positive value indicates a gain to Jack, and a negative value indicates a gain to Jill.

Since only suits matter in this game, there are four equiprobable outcomes. One is favorable to Jill, and the other three are favorable to Jack. There are two events, a win for Jill and a win for Jack, which we will call  and , respectively. Their probabilities and their values to Jack are:

: A spade is drawn, and Jill wins.

Probability: 

Value to Jack: 

: A card of any of the other three suits is drawn, and Jack wins:

Probability: 

Value to Jack: 

The expected value of one play of the game to Jack is

 The game is fair.

We will consider the value of each event to Jill; positive values refer to Jack paying Jill, and negative values refer to Jill paying Jack.

There are 53 equiprobable outcomes. Along with their probabilities and their values to Jill, they are:

: The joker is drawn. 1 out of 53 outcomes is favorable to this event, so

The value to Jill, , is what is to be determned.

: A club is drawn. 13 out of 53 outcomes are favorable to this event, so

Since Jill wins $100 in this event, the value to her is

: A card other than a club or a joker is drawn. 39 out of 53 outcomes are favorable to this event, so

Since Jill loses $20, the value to her is 

.

The expected value of any game can be calculated by multiplying the probability of each event by its value, and adding the products. That is,

The game is to be fair, meaning that . Making this and the other substitutions, we get

Solve for

Jill must agree to pay Jack $520 in the event of the draw of the joker.

This is a conditional probability problem.

There are thirteen spades and thirteen clubs; since the first card is known to be black, the probability of the first card being a spade (or club) is

,

The probability of the second card being a spade depends on the suit of the first card. The probability of the second card being a spade, given that the first card was also a spade, is

The probability of the second card being a spade, given that the first card was a club, is

The overall probability that the second card will be a spade is

First, recall that each die is rolled independently.

Now, it is known that the first die is an odd number therefore the possible options are:

From here, find the number of combinations each one will have to make a sum of 8 with the other die.

By investigation, die number one cannot be one because it along with any value from die two will not equal eight.

Therefore, investigate when die one is three.

Next, if die one is five then the possible sum is,

Therefore there are a total of two possibilities of rolling a sum equaled to eight.

From here, identify the total number of options the sum could be when it does not equal eight.

Since each trial can have one of two outcomes, heads or tails, this is an example of a Bernoulli process.

The probability that trials will result in  successes in a Bernoulli process is

,

where is the probability of a success in any given trial and is the probability of a failure.

Since there are six tosses, set . We can call heads the success with probability , and tails the failure with probability . Accordingly, the probability that exactly heads will come up is

Since we are looking for the probability of heads coming up at least five times, evaluate as follows:

Of the five choices, 0.25 comes closest.