Suppose someone gives you 8 to 2 odds that you can not roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 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 200 times? Explain.
Probability of rolling two even numbers
= (1/2)(1/2)=1/4
Expected value of the game, E(x) is the sum of the following:
outcome X, probability P(X), X*P(X)
2 evens=+$8, 1/4, +$2
at least one odd=-$2, 3/4, -$3/2
The sum of X*P(X) = $0.50
The expected value is thus $0.50 for each game.
Since odds of losing is 3 to 1, it would be reasonable to expect to lose the first game.
On the other hand, if we play 200 times, the law of large numbers would take over and there should be a winning of around $200*0.5=$100.
Well, well, well! Let's calculate the expected value and see if this game is a laughing matter or not.
To calculate the expected value, we multiply each possible outcome by its probability, and then sum them up. In this game, we have two possible outcomes: rolling two even numbers or not.
First, let's figure out the probability of rolling two even numbers. Out of the 36 possible outcomes when rolling two fair dice, there are 18 outcomes where both numbers are even. So, the probability of success is 18/36, which simplifies to 1/2.
Now, let's take a look at the possible winnings. If we win, we get $8, and if we lose, we lose $2. So, the winnings can be represented as +8 or -2, respectively.
To calculate the expected value, we multiply each outcome by its probability and sum them up:
(1/2 * 8) + (1/2 * -2) = 4 - 1 = 3
The expected value of this game is $3. This means that, on average, you can expect to win $3 per game in the long run.
But does that guarantee a win in the first game? Nope! Luck can be a fickle friend, and one game doesn't determine the whole circus. In any single game, you might win $8 or lose $2.
So, what can you expect if you play 200 times? Well, on average, for every game you play, you should expect to win $3. So, over 200 games, you can expect to win approximately 200 * $3 = $600.
Remember, though, this is just an expectation. The actual outcomes can still vary from this estimation. So, brace yourself for the rollercoaster of wins and losses, and may the dice be forever in your favor!
To calculate the expected value, we multiply the value of each outcome by its probability and sum them up. In this case, if you can roll two even numbers, you win $8, and if you can't, you lose $2.
First, let's determine the probability of rolling two even numbers.
There are 36 possible outcomes when rolling two fair dice (each die has 6 sides). Out of these, there are 18 outcomes where both numbers rolled are even (2, 4, or 6 on both dice). Therefore, the probability of rolling two even numbers is 18/36, or 1/2.
Next, we calculate the expected value:
Expected value = (Probability of winning * Value of winning) + (Probability of losing * Value of losing)
Expected value = (1/2 * $8) + (1/2 * -$2)
Expected value = $4 - $1
Expected value = $3
Therefore, the expected value of this game to you is $3.
In the first game, you can expect to neither win nor lose the expected value ($3) exactly since it is a probabilistic measure. However, over multiple games, you can expect to win or lose close to the expected value.
If you play 200 times, you can expect to win approximately 200 * $3 = $600 in total. However, in any individual game, you may win more or less than $3. The law of large numbers suggests that as the number of trials increases, the average outcome will approach the expected value more closely.
To find the expected value of a game, we multiply the probability of each possible outcome by the corresponding value and then sum them up.
In this case, there are 36 possible outcomes when rolling two fair dice, and we want to calculate the probability of rolling two even numbers. So let's start by determining the number of favorable outcomes.
When rolling a fair six-sided die, there are three even numbers (2, 4, and 6) and three odd numbers (1, 3, and 5). Since we need to roll two even numbers, the probability of rolling an even number on one die is 3/6, or 1/2. As the two rolls are independent events, we can multiply these probabilities together: (1/2) * (1/2) = 1/4.
The number of favorable outcomes is therefore one (rolling two even numbers) out of the total 36 possible outcomes. So the probability of rolling two even numbers is 1/36.
Now, let's calculate the expected value. We know that if we win, we gain $8, and if we lose, we lose $2. So the expected value is calculated as follows:
Expected Value = (Probability of winning * Value of winning) + (Probability of losing * Value of losing)
Expected Value = (1/36 * $8) + (35/36 * -$2)
Expected Value = ($8/36) + (-$70/36)
Expected Value = -$62/36 ≈ -$1.72
The expected value of this game is approximately -$1.72. This means, on average, you should expect to lose about $1.72 per game.
Now let's consider playing 200 times. To find the expected outcome for 200 games, we can multiply the expected value by the number of games:
Expected Outcome in 200 games = Expected Value * Number of Games
Expected Outcome in 200 games = -$1.72 * 200
Expected Outcome in 200 games = -$344
Therefore, if you play this game 200 times, you should expect to lose approximately $344 on average.
It's important to note that, even though the expected value is negative and you should expect to lose money over the long run, in any individual game, there is still a chance to win or lose. The expected value represents the average outcome over many repetitions of the game.