圖片參考：http://imgcld.yimg.com/8/n/HA00041588/o/701102220034213873383200.jpg
In the 1st row, it's an example of "full house", a set of 3 cards of the same denomination, plus a set of 2 cards of the same denomination.
In the 2nd row, it's an example of "two pairs", two sets of pairs of the same card denomination plus a random card.
When 5 cards are randomly drawn from a pack of 52 playing cards, the ways of calculating the two combinations above are different.
For "full house":
P(f.h) = 13C1 x 4C3 x 12C1 x 4C2 over 52C5
For "two pairs":
P(t.p) = 13C2 x 4C3 x 4C2 x 44C1 over 52C5
If we replace the 13C2 in the calculation of P(t.p) by 13C1 x 12C1 the result will be different and wrong.
Explain why we can't replace the 13C2 in the calculation of P(t.p) by 13C1 x 12C1 but we use 13C1 x 12C1 in the calculation of P(f.h).
更新1:
There is a simple correction, P(t.p) = 13C2 x 4C3 x 4C2 x 44C1 over 52C5 should be P(t.p) = 13C2 x 4C2 x 4C2 x 44C1 over 52C5, where 4C3 is replaced by 4C2.
