Case A) The dice are distinguishable.
Case B) The dice are not distinguishable.
This exercise is solven on my text book but I dont understand why the B) case is solved like it is. It use permutations and variations to get the favorable cases and then they divide by te variation of 6 elements taken 6 by 6.
$$
p2 = \frac{V_{6,3}\cdot P_{3} + V_{6,2}\cdot RP_{3}^{2,1,0}\cdot RP_{3}^{1,2,0} + V_{6,1}\cdot RP_{3}^{3,0,0}}{RV_{6,6}}
$$
That is the answer they give, and the solution is 83/3888. Can someone explain me why should I solve this this way and what is the real diference between the case A and case B?
Best Answer
I don't understand the notation, but I can tell you how to solve the problem.
There are $6^3=216$ possible rolls. In $6$ rolls, all three dice show the same number. There are $6\cdot5\cdot3=90$ rolls where two of the dice are the same and the third different and $6\cdot5\cdot4=120$ rolls where all three numbers are different.
The probability that the second person rolls the same as the first is $$\frac6{216}\frac1{216}+\frac{90}{216}\frac3{216}+\frac{120}{216}\frac{6}{216}=\frac{996}{216^2}=\frac{83}{3888}$$
In each term, the first factor is the probability that the first player rolls a result of a particular type, and the second factor is the probability that the second player matches the roll.
Does this make sense? Feel free to ask if it doesn't.