Poker dice is played by simultaneously rolling 5 dice. The probability of two pairs is approximately 0.2315. This probability can be derived by the following method: $$P\{two\ pairs\} = \frac{{6\choose2}{5\choose2}{3\choose2}{4\choose1}}{6^5}$$
However, I am trying out another approach, that is:
$$\frac{6\cdot1\cdot5\cdot1\cdot4 }{6^5}$$
My thought is the 6's represents the die could lie on any numbers. The following 1's represents that it must be the same number as the previous one. The 5's represents it could be any number except the previous number and the following 1's means it must be the same number as the last occurring one, producing two pairs. Finally, the 4's represents it could be any number other than the first two pairs.
The above calculation only considers one specific sequence. Hence, to include all possible arrangement, I multiple it with $$Total\ arrangement = \frac{5!}{2!\cdot 2!}$$ This idea comes from the problem of letter arrangement. For example, given the question how many different letter arrangements can be made from the letters {A,A,B,B,C}. I presume the solution is the same as the above calculation. Nevertheless, I got it wrong. In fact, I get: $$\frac{6\cdot1\cdot5\cdot1\cdot4 }{6^5}\cdot \frac{5!}{2!\cdot 2!} = 2\cdot \frac{{6\choose2}{5\choose2}{3\choose2}{4\choose1}}{6^5}$$ Can someone please explain my misconception?
Probability – Calculating Two Pairs in Poker Dice and Letter Arrangements
combinatoricsprobability
Best Answer
The error you made is in double-counting the arrangements. Once you say (in the first part) that the first two dice are some particular number, and the second two the other paired number, then you don't want to also say that any possible arrangement of the five numbers works.
Frr example, say the first number in you "6" in part 1 is 4, and the second is 2.
Then you want to count up the arrangements of 4422x whre x is neither 4 nor 2.
If you start counting all arrangements of 5 dice, you will count 4422x and also 2244x. But by your counting in part one, that should appear only once in part 2.
So your calculation is high by a factor of two.