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May 14th, 2014, 08:07 AM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  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 
May 14th, 2014, 01:25 PM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Shen! Quote:
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: $\displaystyle \:P(\text{Match}) \;=\;\tfrac{1}{216} + \tfrac{10}{216} \;=\;\tfrac{11}{216}$  
May 14th, 2014, 02:09 PM  #3 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
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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