I am stuck at the following question.
What is the probability that two throws with three dice each will show the same configuration if (a) the dice are distinguishable (b) they are not.
Solution.
(a) $P(\text{Two throws of three dice s.t. each show the same configuration})=6^{3}/6^{6}=1/216$.
(b)Firstly from the second set of throws, the $3$ dies with the same face value as the first ones can be chosen in $3!=6$ ways.
Place $r=3$ dies in $n=6$ cells, such that order does not matter with repetition, can be done in ${{3+6-1}\choose{3}}={8\choose 3}=56$
${\displaystyle P(\text{Two throws of three indistinguishable dice show the same configuration})=\frac{56\times6}{6^{6}}}$
However, my answer to the second part of the problem is incorrect. Could someone help me think correctly about the problem.
Best Answer
(b)
Three distinct faces :
$$P(\text{getting the same configuration}) = \frac{(6\cdot5\cdot4)\times(3\cdot2\cdot1)}{6^6}$$
One repetition:
$$P(\text{getting the same configuration}) = \frac{{3\choose2}(6\cdot5)\times{3\choose2}}{6^6}$$
Two repetitions:
$$P(\text{getting the same configuration}) = \frac{6}{6^6}$$
Thus, the required probability is,
$$P(\text{getting the same configuration}) = \frac{996}{6^6}$$