Create a playful illustration which symbolizes chance and expectation. In the center, there are two rolling dice, each on a separate trajectory. On one side of the image, visualize the winning numbers 2, 3, 4, 10, 11, and 12 surrounded by symbols of prosperity such as coins and money envelopes. On the other side, represent the losing numbers 5, 6, 7, 8, 9 with visual symbols of loss or subtraction, for instance a fallen stack of coins or an empty wallet. The image should be fashioned in an engaging and sportsmanlike manner. No text should be included.

A dice game involves rolling 2 dice. If you roll a 2, 3, 4, 10, 11, or a 12 you win $5. If you roll a 5, 6, 7, 8, or 9 you lose $5. Find the expected value you win (or lose) per game.

Prob(2) = 1/36

prob(3) = 2/36
prob(4) = 3/36
prob(12) = 1/36
prob(11) = 2/36
prob(10) = 3/36
total of above , your cases of winning = 12/36

so the prob of remaining cases = 24/36

expected value of game
= (12/36)(5) + (24/36)(-5)
= (1/3)(5) - (2/3)(5
= -5/3

You would be expected to lose $1.67

Prob(2) = 1/36

prob(3) = 2/36
prob(4) = 3/36
prob(12) = 1/36
prob(11) = 2/36
prob(10) = 3/36
total of above , your cases of winning = 12/36

so the prob of remaining cases = 24/36

expected value of game
= (12/36)(5) + (24/36)(-5)
= (1/3)(5) - (2/3)(5
= -5/3

You would be expected to lose $1.67

$1.67

Ah, the thrilling world of dice games! Let's calculate the expected value, or as I like to call it, the "anticipated amusement quotient."

To find the expected value, we'll look at the probabilities of winning or losing with each possible outcome.

There are 36 equally likely outcomes with two dice, so let's break it down:

Outcomes that win $5:
2 (1+1), 3 (1+2 or 2+1), 4 (1+3, 2+2, or 3+1),
10 (4+6 or 6+4), 11 (5+6 or 6+5), and 12 (6+6).
There are a total of 6 winning outcomes.

Outcomes that lose $5:
5 (1+4, 2+3, 3+2, or 4+1), 6 (1+5, 5+1, 2+4, or 4+2),
7 (1+6, 6+1, 2+5, 5+2, 3+4, or 4+3), 8 (2+6, 6+2, 3+5, or 5+3),
and 9 (3+6, 6+3, 4+5, or 5+4).
There are a total of 15 losing outcomes.

Now, let's calculate the expected value:

Probability of winning = 6/36 = 1/6
Probability of losing = 15/36 = 5/12

Expected value = (Probability of winning * Amount won) - (Probability of losing * Amount lost)
= (1/6 * $5) - (5/12 * $5)
= $5/6 - $25/12
= ($25 - $25)/12
= $0/12
= $0

So, after all the calculations, the expected value per game is a whopping $0. That means you can expect to neither win nor lose anything. It's like playing a game with the universe, where the entertainment is priceless!

Remember, though, the numbers are just probabilities, and clowning around with dice is all about having fun.

To find the expected value, we need to calculate the probability of each possible outcome and multiply it by the corresponding amount of money won or lost.

There are 6 possible outcomes for each dice: 1, 2, 3, 4, 5, and 6. So, the total number of possible outcomes for two dice is 6 * 6 = 36.

Let's calculate the probability of winning $5. There are 6 possible outcomes that result in winning: (1,1), (1,2), (2,1), (1,3), (3,1), and (2,2). So, the probability of winning is 6/36 = 1/6.

Now, let's calculate the probability of losing $5. There are 10 possible outcomes that result in losing: (1,4), (4,1), (1,5), (5,1), (1,6), (6,1), (2,3), (3,2), (2,4), and (4,2). So, the probability of losing is 10/36 = 5/18.

To calculate the expected value, we subtract the probability of losing from the probability of winning and multiply it by the corresponding amount:

Expected value = (Probability of winning * Amount won) - (Probability of losing * Amount lost)
= (1/6 * $5) - (5/18 * $5)
= $5/6 - $25/18

Now, let's calculate the expected value:

Expected value = $5/6 - $25/18
= ((5 * 3)/(6 * 3)) - (25/18)
= $15/18 - $25/18
= -$10/18
= -$5/9

Therefore, the expected value per game is -$5/9, which means you can expect to lose approximately $0.556 per game.