Hi,

We throw three dice: A(1-1-1-2-2-3), B(1-2-3-4-5-6) and C(4-4-4-4-5-6), what are all different possibilities, ans what are the possibilities to have the number 1 exactly twice???

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- Apr 29th 2010, 11:41 AMbhitroofen01Dice
Hi,

We throw three dice: A(1-1-1-2-2-3), B(1-2-3-4-5-6) and C(4-4-4-4-5-6), what are all different possibilities, ans what are the possibilities to have the number 1 exactly twice??? - Apr 29th 2010, 12:14 PMshenanigans87
Just to clarify, only dice B is a standard 1-6 dice? The rest have 3 or 4 sides with the same number?

Under that assumption, the answer to the second half is

(1/2) * (1/6) = (1/12) = 8.33% chance that you will have two 1's (the third die is irrelevant in this case)

As for the first question, it's impossible to answer since you haven't clarified what counts as a possibility. For example, if the 3 dice end up being {1} {2} {4}, is that the same as {2} {1} {4}?

Assuming that 1 2 3 = 3 2 1 etc., there are 32 unique combinations total.

http://img580.imageshack.us/img580/126/probab0101.png - Apr 29th 2010, 08:10 PMSoroban
Hello, bhitroofen01!

Quote:

We throw three dice: A(1-1-1-2-2-3), B(1-2-3-4-5-6) and C(4-4-4-4-5-6).

What are all different possibilities?

. . $\displaystyle \begin{array}{cccccc}

(1,1,4) & (1,2,4) & (1,3,4) & (1,4,4) & (1,5,4) & (1,6,4) \\

(1,1,5) & (1,2,5) & (1,3,5) & (1,4,5) & (1,5,5) & (1,6,5) \\

(1,1,6) & (1,2,6) & (1,3,6) & (1,4,6) & (1,5,6) & (1,6,6)\end{array}$

. . $\displaystyle \begin{array}{cccccc}

(2,1,4) & (2,2,4) & (2,3,4) & (2,4,4) & (2,5,4) & (2,6,4) \\

(2,1,5) & (2,2,5) & (2,3,5) & (2,4,5) & (2,5,5) & (2,6,5) \\

(2,1,6) & (2,2,6) & (2,3,6) & (2,4,6) & (2,5,6) & (2,6,6) \end{array}$

. . $\displaystyle \begin{array}{cccccc}

(3,1,4) & (3,2,4) & (3,3,4) & (3,4,4) & (3,5,4) & (3,6,4) \\

(3,1,5) & (3,2,5) & (3,3,5) & (3,4,5) & (3,5,5) & (3,6,5) \\

(3,1,6) & (3,2,6) & (3,3,6) & (3,4,6) & (3,5,6) & (3,6,6) \end{array}$

Quote:

What are the possibilities to have the number 1 exactly twice?

, then shenanigans87 is absolutely correct!*probability*

There is basically one scenario:

. . $\displaystyle \begin{array}{ccccc}P(\text{1 on Die A}) &=& \frac{3}{6} &=& \frac{1}{2}\\ \\[-3mm]

P(\text{1 on Die B}) &=& \frac{1}{6} \\ \\[-3mm]

P(\text{any on Die C}) &=& \frac{6}{6} &=& 1 \end{array}$

Therefore: .$\displaystyle P(\text{two 1's}) \;=\;\frac{1}{2}\cdot\frac{1}{6}\cdot 1 \;=\;\frac{1}{12} \;=\;8\tfrac{1}{3}\%$