Cam and Kayla are playing the game SIXES The way it works is

Cam and Kayla are playing the game SIXES. The way it works is that each person starts with three fair die (remember die have the numbers 1, 2, 3, 4, 5 and 6) and each die is set to 1. Kayla rolls all of her die that are not a 6, and that counts as her turn; any die that land on a six are removed from the rest of the game (thus on her first turn she rolls all three die, but if one lands on a 6 then on her next turn she would only roll two die). Then Cam rolls all his die that are not a 6, and that counts as his turn; any die that lands on a six are removed from the rest of the game (thus on his first turn he rolls all three die, but if two land on a 6 then on his next turn he would only roll one die). They then continue taking turns rolling die, alternating and only rolling the die that arc not a 6. The game continues until each person has all their die on a 6; you win if and only if you get all 6\'s in fewer turns than the other. (a: 20 points) What is the probability Cam wins? If possible do not leave as a sum. (b: 20 points) What is the expected number of turns Kayla needs before all of her die are showing 6? (c: 40 points) Write a computer program to simulate a modification of the game, where now each player starts with 10 (ten) die. You are only allowed to run your code with 10 die, you may not change it to check your answers for (a) or (b). Play the game at least 1,000,000 times and use the results of your simulations to estimate the number of rolls it takes Kayla before all her die are showing 6.

Solution

a) This is an unbiased game (assuming all dice are unbiased and identical), hence the probability of any person winning would be 0.5 or 50%

b) Probability that any die shows 6 is 1/6, Probability that kayla\'s game ends at nth turn would be, probability that previous two sixes occured at any of n - 1 turns, and then last one occured at nth turn, breaking it down even further, last one occured at nth turn, 1st one occured at any turn from 1 to n, 2nd one occured at any turn from 1st one to n, so its like choosing any one from 1 to n and choosing 2nd one from any of 1st to n. which is nC1 * nCn/2 * (1/6) ^ 3