Question

In European roulette, the wheel is divided into 37 compartments numbered 1 through 36 and 0. (In American roulette there are 38 compartments numbered 1 through 36, 0, and 00.) One-half of the numbers 1 through 36 are red, the other half are black, and the number 0 is green. Find the expected value of the winnings on a $7 bet placed on black in European roulette. (Round your answer to three decimal places.)

Answer 1

Answer:

The expectation is -$0.189.

Step-by-step explanation:

Consider the provided information.

In European roulette, the wheel is divided into 37 compartments numbered 1 through 36 and 0.

One-half of the numbers 1 through 36 are red, the other half are black, and the number 0 is green.

We need to find the expected value of the winnings on a $7 bet placed on black in European roulette.

Here the half of 36 is 18.

That means 18 compartments are red and 18 are black.

The probability of getting black in European roulette is 18/37

The probability of not getting black in European roulette is 19/37. Because 18 are red and 1 is green.

If the ball lands on a black number, the player wins the amount of his bet.

The bet is ball will land on a black number.

The favorable outcomes are 18/37 and unfavorable are 19/37.

Let S be possible numerical outcomes of an experiment and P(S) be the probability.

The expectation can be calculated as:

E(x) = sum of S × P(S)

For$S_1 = 7$

$P(S_1) = \frac{18}{37}$

For $S_2 = -7$(negative sign represents the loss)

$P(S_2) = \frac{19}{37}$

Now, use the above formula.

$E(x) = 7\times \frac{18}{37}-7\times \frac{19}{37}\\E(x) = -0.189$

Hence, the expectation is -$0.189.