SOLUTION: Three six sided dice are thrown and produce a random set of numbers eg 1 4 6 = 11. Now a second set of three six sided dice are thrown. Before we worry about the sum total o

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Question 360489: Three six sided dice are thrown and produce a random set of numbers
eg 1 4 6 = 11.
Now a second set of three six sided dice are thrown.
Before we worry about the sum total of the second set here is the question.
Before any dice are thrown including the first set what is the probability that the sum total for both dice set A will be equal to the sum total of dice set B
before we know the result of A.
MIke Manttan



sets will be exactly equal. Not just 11 but any number
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

3 dice can fall in exactly 63 or 216 ways.  

3 dice can have sum 3 in exactly 1 way.
3 dice can have sum 4 in exactly 3 ways.
3 dice can have sum 5 in exactly 6 ways.
3 dice can have sum 6 in exactly 10 ways.
3 dice can have sum 7 in exactly 15 ways.
3 dice can have sum 8 in exactly 21 ways.
3 dice can have sum 9 in exactly 25 ways.
3 dice can have sum 10 in exactly 27 ways.
3 dice can have sum 11 in exactly 27 ways.
3 dice can have sum 12 in exactly 25 ways.
3 dice can have sum 13 in exactly 21 ways.
3 dice can have sum 14 in exactly 15 ways.
3 dice can have sum 15 in exactly 10 ways.
3 dice can have sum 16 in exactly 6 ways.
3 dice can have sum 17 in exactly 3 ways.
3 dice can have sum 18 in exactly 1 way.

Note that 1+3+6+10+15+21+25+27+27+25+21+15+10+5+3+1 = 216 = 63.

So there are 216×216 or 46656 ways both sets of dice can fall.
That will be the denominator of the desired probability.

The number of ways both sets of dice can have sum 3 is 1×1 or 1.
The number of ways both sets of dice can have sum 4 is 3×3 or 9.
The number of ways both sets of dice can have sum 5 is 6×6 or 36.
The number of ways both sets of dice can have sum 6 is 10×10 or 100.
The number of ways both sets of dice can have sum 7 is 15×15 or 225.
The number of ways both sets of dice can have sum 8 is 21×21 or 441.
The number of ways both sets of dice can have sum 9 is 25×25 or 625.
The number of ways both sets of dice can have sum 10 is 27×27 or 729.
The number of ways both sets of dice can have sum 11 is 27×27 or 729.
The number of ways both sets of dice can have sum 12 is 25×25 or 625.
The number of ways both sets of dice can have sum 13 is 21×21 or 441.
The number of ways both sets of dice can have sum 14 is 15×15 or 225.
The number of ways both sets of dice can have sum 15 is 10×10 or 100.
The number of ways both sets of dice can have sum 16 is 6×6 or 36.
The number of ways both sets of dice can have sum 17 is 3×3 or 9.
The number of ways both sets of dice can have sum 18 is 1×1 or 1.

1+9+36+100+225+441+625+729+729+625+441+225+100+36+9+1 = 4332

Therefore the desired probability is  = 

or about 0.0928497942

Edwin
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