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March 5th, 2018, 03:15 PM  #1 
Newbie Joined: Oct 2017 From: Rumba Posts: 20 Thanks: 0  Probability question
Three fair dice are rolled: one red die, one blue die and one green die. Let R, B and G be the numbers respectively shown by the red, blue and green die. Compute the probability P {R < B < G} Last edited by sita; March 5th, 2018 at 03:31 PM. 
March 5th, 2018, 05:53 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications 
Hi Sita, I think that this problem can be done by brute force, especially if each die has six sides, as is typical. I worked the problem by starting with the green die. Obviously, for G=1,2 there are no possibilities. For G=3, then only G=3, B=2, R=1 (denoted as 321) gives a true outcome. For G=4, then 421, 431, and 432 give a true outcome. For G=5, then 521, 531, 532, 541, 542, and 543 give a true outcome. Can you do likewise for G=6, and can you calculate the total number of possibilities? Adding the number of true outcomes for G=3,4,5,6 and dividing by the total number of possibilities will give the answer. 
March 5th, 2018, 06:09 PM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 2,098 Thanks: 1093  Quote:
First let's determine how many groupings include equality. There are 6 ways all three dice are equal. There are $(3)(6)(5) = 90$ ways that only 2 of the dice are equal. $3 = \dbinom{3}{2}$ There are 6 numbers that the two dice that are equal could be. There are 5 numbers that the third die could be and not equal the other two. So there are at total of 96 rolls that involve equality. There are $6^3 = 216$ total possible rolls. Thus there are $21696 = 120$ rolls where strict inequality is observed. Now... the problem is symmetric in the colors and thus each ordering will get 1/6 of these rolls, as there are 6 ways to arrange RGB. Thus the particular ordering of $R < B < G$ will have $\dfrac{120}{6}=20$ associated rolls and thus $P[R< B< G] = \dfrac{20}{216}= \dfrac{5}{54}$  
March 15th, 2018, 04:49 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 
What's quicker than what romsek said is that the answer is 6C3 = 20. In combinations, the order matters, and an item can only be chosen once, which is what we want.


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