Question

You play a game in which you roll a number n of dice. You win if you roll at least one 6, but no 1s or 2s; and, otherwise, you lose. What is the optimal number of dice to roll in this situation? Choose the best answer.

Answer 1

Answer:

2

Step-by-step explanation:

probability of 6 = 1/6

probability of not coming 6 = 1 - 1/6 = 5/6

in n throws

probability of coming atleast 1 six

= 1 - probability of 0 six

= 1 - ⁿC₀(1/6)⁰(5/6)ⁿ

= 1 - (5/6)ⁿ

Probability of 1 or 2 = 2/6 = 1/3

Probability of not coming 1 or 2 = 1 - 1/3 = 2/3

Probability of not coming 1 or 2 even once

= probability of coming 1 or 2  , 0 times

= ⁿC₀(1/3)⁰(2/3)ⁿ

= (2/3)ⁿ

Total probability = (1 - (5/6)ⁿ)(2/3)ⁿ

P = (1 - (5/6)ⁿ)(2/3)ⁿ

n              P

0             0

1              1/9

2             11/81

3             91/729

4             671/6561

optimal number of times dice to roll in this situation = 2