## Question

###### Tve been playing a lot of King of Toyko lately: There are six dice (at least at first), with sides labeled 1, 2, 3 and On each turn you get three dice rolls; each time; you keep the ones you like and can reroll any of the others_ Now , am currently in Tokyo; so does me nO good (you have to leave Tokyo to heal). Im aiming for many as possible: If I end up with at least three of one number, though will get some victory points (YOu get the value of the die if you have three; and one more for each

Tve been playing a lot of King of Toyko lately: There are six dice (at least at first), with sides labeled 1, 2, 3 and On each turn you get three dice rolls; each time; you keep the ones you like and can reroll any of the others_ Now , am currently in Tokyo; so does me nO good (you have to leave Tokyo to heal). Im aiming for many as possible: If I end up with at least three of one number, though will get some victory points (YOu get the value of the die if you have three; and one more for each additional occurance of that number) So here's my strategy: on each roll _ keep all the get , and keep all the 3 s get, but [ reroll everything else. What is the probability that end up with three and three victory points (meaning the other three dice are all 3s)? The following parts break down the computation Fix 0 < jk < 3 and find the conditional probability that end up with three and three 3*s, given that start my third roll with j k3s, and 6 - j - k other things. Now fix p and q with 0 < p < j and 0 < q < k and find the conditional probability that I start my third roll with k3*s, and 6 - j - k other things, given that I start my second roll with p q 3*s, and 6 q other things. Use the previous three parts to calculate the desired probability: Please explain you reasoning and simplify YOur answer to the extent possible: your final answer should involve only one multinomial coefficient BONUS: actually get a numerical aswer