Consider a dice game with the following rules.....

Question

Consider a dice game with the following rules:

• All players start with 10 points

• A dice is rolled, and if the number on the dice is 1, 2, 3, 4 or 5, this number is added

to each player’s score. If the number is 6, the score of all players is reduced to zero

• Before the dice is rolled each player is given the choice of continuing or taking their

current score as their final score.

• This process is repeated until a 6 is rolled and the game ends. All players still

playing at this point receive zero as their final score.

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Now suppose that a 6 was not rolled and you choose to continue playing

(v) On average, do you do worse or better than someone who chooses to stop playing

after 1 dice roll?

(vi) Is there an optimal number of dice rolls you should continue playing for? Explain

your reasoning.

(vii) If the rules are changed such that when the number is 2, 3, 4, 5, or 6, this number is

added to your score. While if the number is 1, your score is reduced to zero. Discuss

how this affects the optimal number of dice rolls you should continue playing for.

Answer 1

If, on the first roll, there was not a 6 rolled, as the average of 1-5 is 3, your expected value is 10+3= 13

Then, if you stop before the next roll, your expected value is, of course, 13 (assuming that everyone didn't already lose).

Your expected value if you roll is 5/6 * (13 + 3) = 5/6 * 16 = 80/6 = 40/3 = 13 1/3; you multiply by 5/6 because 1/6 of the time you get a 6 and get 0. Thus, you will do better than someone who stops after one roll, as 13 1/3 > 13

vi Your expected gain from the n'th roll is 5/6*3 - 1/6(10 + (n-1)3), as 10 + (n-1)3 is the amount that you expect to have after n - 1 rolls.

For n= 2, this is 5/6*3 - 1/6(10 + (2-1)3) = 5/2 - 13/6 = 1/3, the same answer we had before, as 13 1/3 - 13 = 1/3

We solve for n. 5/6*3 - 1/6(10 + (n-1)3) = 0

5/2 - 7/6 - n/2 = 0

n/2 = 8/6 = 4/3

n = 8/3

You should stop after two rolls, as 3 > 8/3 > 2

If you lose on a roll, the expected value of each roll if you win becomes 4.

Thus, we have 5/6*4 - 1/6(10 + (n-1)4)

20/6 - 1 - 2/3n = 0

2/3n = 14/6 = 7/3

n = 7/2

4 > 7/2 > 3

You should stop after 3 rolls.

Note that you could have determined the solution in a different way; rather than looking at things marginally, you could see the expected value after n rolls and take the biggest value. Your decision will be the same, and the math is not as pretty.

The expected value after n rolls is (5/6)^n*(10+3n), and this is maximized at n = 2.

For the other rule, the expected value after n rolls is (5/6)^n*(10+4n), and this is maximized at n = 3.