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May 14th, 2014, 08:07 AM   #1
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2 person die roll 2 times

If A and B both rolled two standard dice two times then what is the probability that B gets the same number as A but not necessarily in the same order?

I don't understand the question. Can you please solve it?

Thanks in advance and Regards
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May 14th, 2014, 01:25 PM   #2
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Hello, Shen!

Quote:
If A and B both rolled two standard dice two times
then what is the probability that B gets the same number as A
but not necessarily in the same order?

I don't understand the question. $\displaystyle \;\;$Neither do I!

Because of the phrase "but not necessarily in the same order",
I believe it asked for the probability that B gets the same two numbers as A.

I listed the outcomes and counted.

$\displaystyle \quad \begin{array}{cccccc} 11 & 12 & 13 & 14 & 15 & 16 \\
21 &22 & 23 & 24 & 25 & 26 \\
31&32&33&34&35&36\\
41&42&43&44&45&46 \\
51&52&53&54&55&56\\
61&62&63&64&65&66 \end{array}$


Suppose A rolls a "double".
This happens $\displaystyle \tfrac{6}{36} = \tfrac{1}{6}$ of the time.
Then B must match it; probability: $\displaystyle \tfrac{1}{36}$
$\displaystyle P(\text{Double and match}) \:=\:\tfrac{1}{6}\cdot\tfrac{1}{36} \:=\:\tfrac{1}{216}$

Suppose A rolls some "Other".
This happens $\displaystyle \tfrac{30}{36} = \tfrac{5}{6}$ of the time.
Then B must match it; probability: $\displaystyle \tfrac{2}{36}$
$\displaystyle P(\text{Other and match}) \:=\:\tfrac{5}{6}\cdot\tfrac{2}{36} \:=\:\tfrac{10}{216}$

Therefore: $\displaystyle \:P(\text{Match}) \;=\;\tfrac{1}{216} + \tfrac{10}{216} \;=\;\tfrac{11}{216}$

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May 14th, 2014, 02:09 PM   #3
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Hi soroban

Thanks. This is the answer. I was stuck as i did not divide the double rolls and single rolls. Now i understand.

Thanks again
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