At the beginning of each round of a game of chance the player pays 12.5. The player then rolls one die with outcome N. The player then rolls N dice and wins an amount equal to the total of the numbers showing on the N dice. All dice have 6 sides and are fair.

Calculate the probability that a player starting with 15,000 will have at least 15,000 after 1000 rounds using the normal approximation.

• 0.01
• 0.04
• 0.06
• 0.09
• 0.12

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: E

In each round,

N = result of first roll, to see how many dice you will roll
X = result of for one of the N dice you roll
S = sum of X for the N dice

[[math]] \operatorname{E}( X ) = \operatorname{E}( N ) = 3.5 [[/math]]

[[math]] \operatorname{E}( X ) = \operatorname{E}( N ) = 2.9167 [[/math]]

[[math]] \operatorname{E}( S ) = \operatorname{E}( N ) \operatorname{E}( X ) = 12.25 [[/math]]

[[math]] \operatorname{E}( S ) = \operatorname{E}( N )\operatorname{E}( X ) + \operatorname{E}( N ) \operatorname{E}( X )^2 = 3.5(2.9167) + 2.9167(3.5) 2 = 45.938 [[/math]]

Let [math]S_{1000}[/math] the sum of the winnings after 1000 rounds

[[math]] \operatorname{E}(S_{1000}) = 1000(12.25) - 12,250 [[/math]]

[[math]] SD(S_{1000}) = \sqrt{1000(45.938)} = 214.33 [[/math]]

After 1000 rounds, you have your initial 15,000, less payments of 12,500, plus winnings for a total of 2,500 + [math]S_{1000}[/math]. Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater than 15000-0.5. In this problem, that continuity correction has negligible impact.

[[math]] \begin{aligned} \operatorname{Pr}(2,500 + S_{1000} \gt 14,999.5 ) &= \operatorname{Pr}(S_{1000} \gt 12,499.5 ) \\ &\approx \operatorname{Pr}( Z \gt \frac{12, 499.5 − 12, 250}{214.33} = 1.17)\\ &= 0.12 \end{aligned} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.