Here is a solution to yesterday’s probability question

**Question-** We play a game where I roll a dice up to a maximum of four times. After each roll, you can take the dollar amount of the number showing or you may ask me to roll again. (Once you accept the dollar amount of any dice roll, the game is over. If you pass three times, you must accept the dollar amount of the fourth roll.) What is the optimal strategy to follow and what is the expected value of your winnings?

**Answer: –** The best way to solve this is by working backward.

1) Assume only one roll of the dice is left. What is the expected value of that one roll? This is pretty easy-

Expected Value ( 1 die) = (1/6) (6) + (1/6)(5)+(1/6)(4)+(1/6)(3)+(1/6)(2)+(1/6)(1)= $3.50

2) Now if I offer you two dice rolls, you can either take the first roll’s result or turn it down and take the second roll. So the best strategy to follow (e.g. the stopping strategy) is to take the first roll (or stop) if the first dice roll is higher than the expected value of the second roll, or else continue. So you should take the first roll (or stop) if it is 4 or greater, else continue. So the expected value of 2 dice rolls is:

Expected Value ( 2 dice) = (1/6) (6) + (1/6)(5)+(1/6)(4)+(1/2)(3.5)= $4.25

3) Next, we repeat the same calculation for three dice. Now we only stop after the first roll if it is a 5 or a 6.

Expected Value( 3 dice) = (1/6)(6) + (1/6)(5) + (2/3)(4.25)= 4 2/3 or $4.67 rounded to the nearest penny.

4) Finally we can compute the expected value for four dice! Again we only stop after the first roll if it is 5 or a 6.

Expected Value (4 dice)= (1/6)(6) + (1/6)(5) + (2/3)(4 2/3)= 4 17/18= $4.94 rounded to nearest penny.

Stopping Strategy- First roll:Stop on 5,6 Second roll:Stop on 5,6 Third Roll: Stop on 4,5,6

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