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We have a fair die and have up to 4 die rolls at our disposal. The objective of the game is to maximize the expected value of the *last* roll. For example, we roll once and get a 6; this cannot be improved upon, so we should stop, i.e., it is not optimal to roll again. Conversely, if the first roll gives us 1, then naively we should roll again.
1. What is the optimal strategy?
A. Stop if first roll gives 4 or 2, or second roll gives 3 or 6, or third roll gives 2 or higher.
B. Stop if first roll gives 3 or 6, or second roll gives 3 or 6, or third roll gives 2 or higher.
C. Stop if first roll gives 5 or 6, or second roll gives 5 or 6, or third roll gives 4 or higher.
D. Stop if first roll gives 1 or 5, or second roll gives 1 or 4, or third roll gives 3 or higher.
E. Stop if first roll gives 4 or 6, or second roll gives 4 or 5, or third roll gives 4 or higher.
2. What is the expected value of a game played using the optimal strategy?
A. Expected value of game played optimally is 3.9
B. Expected value of game played optimally is 4.9
C. Expected value of game played optimally is 5.5
D. Expected value of game played optimally is 3.5
E. Expected value of game played optimally is 4.5
My solution: We calculate the expected value of each of the 5 given optimal strategy as follows:
For choice A, Expected value(=6*1/6+9*1/6*2/3+(2+3+4+5+6)*1/6*2/3*2/3=67/27)
choice B, Expected value(=9*1/6+9*1/6*2/3+(2+3+4+5+6)*1/6*2/3*2/3=3.9 )(approximately)
choice C, Expected value(=11*1/6+11*1/6*2/3+(4+5+6)*1/6*2/3*2/3=4.16)
choice D, Expected value(=1+5/9+(3+4+5+6)*1/6*2/3*2/3=2.89)
choice E, Expected value(=10/6+9*1/6*2/3+(4+5+6)*1/6*2/3*2/3=3.77)
Based on these results, C should be the correct answer. However, the expected value of a game using this strategy based on my calculation doesn't match any of the 5 given answer choices of problem 2. I think my argument is wrong, but I can't figure out where. Can someone help me with this problem?
1. What is the optimal strategy?
A. Stop if first roll gives 4 or 2, or second roll gives 3 or 6, or third roll gives 2 or higher.
B. Stop if first roll gives 3 or 6, or second roll gives 3 or 6, or third roll gives 2 or higher.
C. Stop if first roll gives 5 or 6, or second roll gives 5 or 6, or third roll gives 4 or higher.
D. Stop if first roll gives 1 or 5, or second roll gives 1 or 4, or third roll gives 3 or higher.
E. Stop if first roll gives 4 or 6, or second roll gives 4 or 5, or third roll gives 4 or higher.
2. What is the expected value of a game played using the optimal strategy?
A. Expected value of game played optimally is 3.9
B. Expected value of game played optimally is 4.9
C. Expected value of game played optimally is 5.5
D. Expected value of game played optimally is 3.5
E. Expected value of game played optimally is 4.5
My solution: We calculate the expected value of each of the 5 given optimal strategy as follows:
For choice A, Expected value(=6*1/6+9*1/6*2/3+(2+3+4+5+6)*1/6*2/3*2/3=67/27)
choice B, Expected value(=9*1/6+9*1/6*2/3+(2+3+4+5+6)*1/6*2/3*2/3=3.9 )(approximately)
choice C, Expected value(=11*1/6+11*1/6*2/3+(4+5+6)*1/6*2/3*2/3=4.16)
choice D, Expected value(=1+5/9+(3+4+5+6)*1/6*2/3*2/3=2.89)
choice E, Expected value(=10/6+9*1/6*2/3+(4+5+6)*1/6*2/3*2/3=3.77)
Based on these results, C should be the correct answer. However, the expected value of a game using this strategy based on my calculation doesn't match any of the 5 given answer choices of problem 2. I think my argument is wrong, but I can't figure out where. Can someone help me with this problem?
