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Suppose someone gives you 5 to 1 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win ​$5 if you succeed and you lose ​$1 if you fail. What is the expected value of this game to​ you? Should you expect to win or lose the expected value in the first​ game? What can you expect if you play 100 ​times? Explain.

User Greg R
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2 Answers

5 votes

Final answer:

The expected value of the game where you are given 5 to 1 odds on rolling two even numbers with two dice is $0.50. You would average a win of $0.50 per game over an infinite number of plays, but actual outcomes can vary in the short term.

Step-by-step explanation:

To calculate the expected value of the game where you get 5 to 1 odds on rolling two even numbers with two dice, we first need to determine the probability of rolling two even numbers. Since a die has three even numbers (2, 4, 6) and three odd numbers (1, 3, 5), the probability of rolling an even number on one die is 3/6 or 1/2. To get two even numbers, you must roll an even number on both dice, which means the probabilities are independent and must be multiplied together.

The probability of success (rolling two even numbers) is (1/2) × (1/2) = 1/4. The probability of failure (not rolling two even numbers) is 1 - 1/4 = 3/4.

Now we can calculate the expected value:
(Expected value) = (Probability of success) × (Winnings) + (Probability of failure) × (Losses)
(Expected value) = (1/4) × (+$5) + (3/4) × (-$1)
(Expected value) = ($5/4) - ($3/4)
(Expected value) = $5/4 - $3/4
(Expected value) = $2/4 or $0.50

The expected value of this game is $0.50. This means if you were to play the game an infinite number of times, you would average a win of $0.50 per game. However, in the first game, you cannot expect to win or lose exactly $0.50, because the outcomes are discrete - you either win $5 or lose $1. Over the long term, such as playing the game 100 times, you would tend to approach the average expected value of $0.50 per game. However, variance can cause actual results to deviate from the expected average in the short term.

User Ksogor
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2 votes

Answer:

The expected value of this game to me is 5 odd x $5 = $25.

There is 1/2 chances of winning and losing.

The expectations after playing 100 times is 100x5 oddx$5=$2500

Step-by-step explanation:

Odd is the probability or likelihood of an event occurring.

User Mike Menko
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