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March 5th, 2018, 03:15 PM   #1
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Question 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.
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March 5th, 2018, 05:53 PM   #2
jks
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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.
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March 5th, 2018, 06:09 PM   #3
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Quote:
Originally Posted by sita View Post
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}
I think there is a slightly more elegant solution.

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 $216-96 = 120$ 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

$P[R< B< G] = \dfrac{20}{216}= \dfrac{5}{54}$
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March 15th, 2018, 04:49 PM   #4
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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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