SOLUTION: 4 non standard 6 sided dice 1) 3,3,3,3,3,3 2) 4,4,4,4,0,0 3) 5,5,5,1,1,1 4) 6,6,2,2,2,2. 3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s.for the four pairs of dice listed above, analyze eac

Algebra ->  Probability-and-statistics -> SOLUTION: 4 non standard 6 sided dice 1) 3,3,3,3,3,3 2) 4,4,4,4,0,0 3) 5,5,5,1,1,1 4) 6,6,2,2,2,2. 3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s.for the four pairs of dice listed above, analyze eac      Log On


   



Question 1208306: 4 non standard 6 sided dice 1) 3,3,3,3,3,3 2) 4,4,4,4,0,0 3) 5,5,5,1,1,1 4) 6,6,2,2,2,2. 3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s.for the four pairs of dice listed above, analyze each case using the knowledge of Probability. Make a theoretical analysis about which one of the four dice has the highest probability of winning (just determine which one of the pair is more likely to win, i.e., p>0.5).
Found 2 solutions by mccravyedwin, ikleyn:
Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
Are you saying you have 
a)  two dice each with all 6 sides marked 3.
b)  two dice each with 4 sides marked 4 and 2 sides marked 0
c)  two dice each with 3 sides marked 5 and 3 sides marked 1
d)  two dice each with 2 sides marked 6 and 4 sides marked 2

How do you win?  By getting the highest sum?

3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s

Who or what are the "3s", "4s", "5s" and "6s"?

Sorry, I'm confused by the "vs." statements.  

I can't tell who is playing whom and with what pair of dice?

For the pair of dice 3,3,3,3,3,3, you can only get 6
P(sum=6) = 1

For the pair of dice 4,4,4,4,0,0 you can only get 8 or 4 or 0
P(sum=8)= P(1st die=4)*P(2nd die=4) = (4/6)(4/6) = (2/3)(2/3) = 4/9
P(sum=0)= P(1st die=0)*P(2nd die=0) = (2/6)(2/6) = (1/3)(1/3) = 1/9
P(sum=4)= 1 - 4/9 - 1/9 = 4/9

For the pair of dice 5,5,5,1,1,1 you can only get 10,6, or 2
P(sum=10)= P(1st die=5)*P(2nd die=5) = (3/6)(3/6) = (1/2)(1/2) = 1/4
P(sum=2)= P(1st die=1)*P(2nd die=1) = (3/6)(3/6) = (1/2)(1/2) = 1/4
P(sum=6)= 1 - 1/4 - 1/4 = 1/2

For the pair of dice 6,6,2,2,2,2 you can only get 12 or 8 or 4
P(sum=12)= P(1st die=6)*P(2nd die=6) = (2/6)(2/6) = (1/3)(1/3) = 1/9
P(sum=8)= P(1st die=2)*P(2nd die=2) = (4/6)(4/6) = (2/3)(2/3) = 4/9
P(sum=4)= 1 - 1/9 - 4/9 = 4/9
 
Sorry I can't help more.  I don't understand 

"3s vs 4s, 4s vs 5s, 5s vs 6s, 6s vs 3s".  

Edwin

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

This post asks about winning,  but does not explain what the game is and what is the winning formula.

Without this info, this post is simply soup of words.


This my statement is as clear as  2 x 2 = 4  to anyone starting from kindergarten,
and does not require any additional explanations.


As submitted,  this post does not deserves any serious consideration.

On the contrary,  it deserves serious censure.