document.write( "Question 473709: If we roll 4 dice once, what is the chance the sum will be 12? \n" ); document.write( "
Algebra.Com's Answer #324952 by Edwin McCravy(20086)\"\" \"About 
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If we roll 4 dice once, what is the chance the sum will be 12?
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document.write( "The other tutor didn't even start to solve your problem.  Counting\r\n" );
document.write( "the number of ways to have sum 12 is NO EASY TASK!  And the denominator\r\n" );
document.write( "is not 24 at all.  It's 64.\r\n" );
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document.write( "There are these 11 sets of ways disregrading order that the 4\r\n" );
document.write( "dice can have sum 12.  However the dice are different so we must \r\n" );
document.write( "rearrange each of these in all possible distinguishable arrangements:\r\n" );
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document.write( "1.  1 1 4 6   <-- 4!/2! = 12 distinguishable arrangements     \r\n" );
document.write( "2.  1 1 5 5   <-- 4!/(2!2!) = 6 distinguishable arrangements\r\n" );
document.write( "3.  1 2 3 6   <-- 4! = 24 distinguishable arrangements\r\n" );
document.write( "4.  1 2 4 5   <-- 4! = 24 distinguishable arrangements\r\n" );
document.write( "5.  1 3 3 5   <-- 4!/2! = 12 distinguishable arrangements\r\n" );
document.write( "6.  1 3 4 4   <-- 4!/2! = 12 distinguishable arrangements\r\n" );
document.write( "7.  2 2 2 6   <-- 4!/3! = 4 distinguishable arrangements\r\n" );
document.write( "8.  2 2 3 5   <-- 4!/2! = 12 distinguishable arrangements\r\n" );
document.write( "9.  2 2 4 4   <-- 4!/(2!2!) = 6 distinguishable arrangements\r\n" );
document.write( "10.  2 3 3 4   <-- 4!/2! = 12 distinguishable arrangements\r\n" );
document.write( "11.  3 3 3 3   <-- 4!/4! = 1 distinguishable arrangement\r\n" );
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document.write( "The numerator of the desired probability is\r\n" );
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document.write( "12+6+24+24+12+12+4+12+6+12+1 = 125\r\n" );
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document.write( "The denominator is 64 since each die can land 6 ways:\r\n" );
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document.write( "So the answer is \"125%2F6%5E4\" = \"125%2F1296\"\r\n" );
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document.write( "Edwin
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