Matt Jake and Kim are playing a game with dice. They both are trying to roll a h
ID: 3226066 • Letter: M
Question
Matt Jake and Kim are playing a game with dice. They both are trying to roll a higher number then the other. But, Matt only has 2 dice whereas Jake has 3 dice, and Kim has 4 dice. How what is the probably that Matt beats both Jake and Kim? (Please use calculus II knowledge *all dice are six sided) Matt Jake and Kim are playing a game with dice. They both are trying to roll a higher number then the other. But, Matt only has 2 dice whereas Jake has 3 dice, and Kim has 4 dice. How what is the probably that Matt beats both Jake and Kim? (Please use calculus II knowledge *all dice are six sided)Explanation / Answer
No of favourable outcomes for Matt to win: Please note, he cannot win if he rolls (1,1), (1,2),(2,1) or (2,2) since Jakes Minimum value would be (1,1,1) = 3 and Kims minimum value would be (2,1,1) = 4
Mat should get a sum of 5,6,7,8,9,10,11 or 12.
No of ways is
5 = (1,4),(4,1)(2,3),(3,2) = 4 ways
6 = (1,5),(5,1),(2,4)(4,2)(3,3) = 5 ways
7 = (1,6)(6,1)(2,5)(5,2)(3,4)(4,3) = 6 ways
8 = (2,6)(6,2)(3,5)(5,3)(4,4) = 5 ways
9 = (3,6)(6,3)(4,5)(5,4) = 4 ways
10 = (4,6)(6,4)(5,5) = 3 ways
11 = (5,6)(6,5) = 2 ways
12 = (6,6) = 1 way
There are 26 ways to get a no more than Kim or Jake
Total no of nos possible = 6x6 = 36
Therefore the required probability = 30/36 = 5/6