Question

In a gambling game, I roll two dice and win $1 if the sum of my two rolls is larger than 7. I lose $1 if the sum of my two rolls is smaller than 7. In any other cases, I don’t lose or win anything. I am going to repeat this game 72 times.

Write your answers as simplified fractions: What is the chance I win $1 on a single play?

Answer 1

The chance of winning $1 on a single play in the described dice game is 5/12, calculated by dividing the number of winning combinations (15) by the total possible outcomes (36).

Step-by-step explanation:

The question asks what the chance of winning $1 on a single play is in a gambling game where you roll two dice and the sum is compared to 7. To win $1, the total of the dice has to be greater than 7. To find the probability of this happening, we need to consider the possible outcomes when rolling two dice.

There are 6 faces on each die, making a total of 36 possible outcomes (6 × 6). The sums greater than 7 can be obtained with the following combinations: (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), and (6,6). There are 15 combinations that result in a sum greater than 7.

Therefore, the probability of winning $1 on a single roll is the number of winning combinations divided by the total number of combinations, which is 15/36. This fraction can be simplified to 5/12. So, the chance of winning $1 on a single play is 5/12.