Algebra homework help probability and odds

Question

Problem : Jason gets two chances to roll a sum of 8 with the two dice. To further clarify, he can win by either rolling an 8 on his first roll, and if he fails to roll an 8 on his first roll, he can win by rolling an 8 on his second roll. The wager for each game is $2. 

The rules are as follows: If an 8 is rolled on either the first roll or the second roll, Jason will win a profit of $5. If neither roll is an 8, Jason will lose his $2 wager. 

1. Should Jason expect to win money or lose money playing this game? 

1. What would Jason’s profit for winning need to be for him to have the same likelihood of winning and losing this game? 

Answer 1

As this is an Algebra question, I'm assuming we're not working with binomial probability. Let me know if that's not the case.

Let's start by figuring out how many combinations of the dice will result in a sum of 8.

6 + 2 = 8

5 + 3 = 8

4 + 4 = 8

3 + 5 = 8

2 + 6 = 8

We have a total of (6 * 6 = 36) options, so there is a 5/36 chance that an 8 will be rolled.

That's the probability of case 1: an 8 is rolled the first time– 5/36. We need to calculate the probability of case 2: a non-8 is rolled, then and 8 is rolled.

The probability of a non-8 being rolled is (36 - 5)/36, or 31/36. To calculate the probability of two conditions both happening, we need to multiply the probabilities. In this case, we'll multiply the probability of a non-8 being rolled (step 1) and the probability of an 8 being rolled (step 2). (31/36) * (5/36). This is roughly equivalent to 0.11960.

Now, to calculate the probability that case 1 OR case 2 will occur, we need to add the probabilities. The probability of case 2 is a rather nasty fraction (155/1296), so we'll use decimals to calculate, rounded to 5 places for calculations, then 4 for our answer. 5/36 = 0.13889.

Now, let's do the actual adding. 0.11960 + 0.13889 = 0.25849. Rounded to four places for our final answer, that will be 0.2585.

Ideally, always answer your questions in context: The probability that James will win the game is 0.2585.

It's up to you to interpret that response–would you expect to win a game that has roughly 1-in-4 odds? I certainly wouldn't.

The profit question appears to be worded incorrectly–the chance of the game is constant, independent of the profit from winning. I'll interpret it in an alternative manner: What would Jason's profit for winning need to be for him to walk away with no money gained or lost in the long run?

In this problem, there is a 25.85% chance that he will win, but we'll round it to 25% for simplicity's sake. This means that for every four games he plays, he should win 1 and lose 3. In order to have 0 net sum, the profit needs to be 3 times the wager, so with a $2 wager, the profit would need to be $6. Winning $6 would be negated by the three $2 losses (6 - 2 - 2 - 2 = 0).

Answer 2

First, we need his chances of rolling an 8 with two dice: 2(2-6) + 2(3-5) + 4-4 or 5 ways out of 36 possibilities.

His chances of winning are 5/36 on first roll, and then he can win 5/36 * 31/36 on the second roll if he misses on the first.

P of win is 5/36 + 5*31/36² = 0.258 His P of losing is (1-P) = 0.742

a) Expected value of game = $5*P - $2(1-P) = -$0.194 or -19 cents

b) What does the payoff need to be for him to break even?

Make his EV zero and let x be the payoff:

0 = x(.258) - 2(.742) x = $ 5.75

Take care.