Rolling TWO SIX-SIDED DICE . Add the face values together. A failed roll (rollin

Question

Rolling TWO SIX-SIDED DICE. Add the face values together. A failed roll (rolling a 1) and A successful roll (not rolling a 1).

1. Find failed roll, (rounded to four decimal places)?

2. Find successful roll, (rounded to four decimal places)?

3. What is the mean value of a SUCCESSFUL roll? (Note, the lowest successful roll is a 4 (2+2). The largest successful roll is a 12 (6+6). The mean should be somewhere between these values.)

4.Let's assume the number of points earned for a successful roll is the mean calculated in the previous problem.The number of points earned for a failed roll would be 0, since your score is reduced to zero when you roll a 1. Expected Value = P(success) * value(success) + P(failure) * value(failure) P(success) is your answer from #2 value(success) is your answer from #3 value(failure) is zero, so P(failure) * value(failure) is always going to be zero. What is the expected value if you were to only roll once?(Round to four decimal places.)

5.What is the expected value if rolling successfully two times in a row? (Round to four decimal places.)

6.If your strategy was to roll two times, what would be your probabilty of failure? (Round to four decimal places.)

7.If you were to roll two times successfully, assume that your points earned would be two times the mean for a single roll. If you were to roll two times unsuccessfully, you would get 0 points. Find the expected value if you decided to roll twice? (Round to four decimal places.)

8.Using the logic discussed above, find the expected value for a strategy in which you roll three times? (Round to four decimal places.)

9.Using the logic discussed above, find the expected value for a strategy in which you roll four times? (Round your answer to four decimal places.)

10.Using the logic discussed above, find the expected value for a strategy in which you roll five times? (Round your answer to four decimal places.)

11.(A)Which strategy would you use? (B)Why?

Please when answering these questions number them like how I did

Explanation / Answer

1. Failed rolls are (1,1), (1,2) ,(2,1)

Pr(Failed rolls) = Pr(sum = 2) + Pr( sum =3) = 1/36 + 2/36 = 1/12

2. Pr (successul rolls) = 1 - 1/12 = 11/12

3. Mean value of successful rolls = ?

let say X is the sample space of successful rolls which can have value

X = 4, Pr( X= 4) = 3/36

X= 5, Pr(X = 5) = 4/36

X = 6; Pr( X = 6) = 5/36

X = 7; Pr(X = 7) = 6/36

and so on

so E(X) = (12/11) [4 * 3/36 + 5 * 4/36 + 6 * 5/36 + 7 * 6/36 + 8 * 5/36 + 9 * 4/36 + 10 * 3/36 + 11 * 2/36 + 12 * 1/36]

E(X) = 7.3939

4. expected value if you were to only roll * 1/12 + 7.3939 * 11/12 = 6.7778

5. Expected value if rolling successfully two times in a row

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