I am trying to calculate the probability of taking from a regular deck of cards (52) 5 cards with 4 different suits, my go:
first we think about picking 'diamond' (just to make it more simple) it is: $\frac{13}{52}$
second we think about picking 'heart': $\frac{13}{51}$
same thing for the two other suits: $\frac{13}{50}$ and $\frac{13}{49}$ respectively.
The fifth card needs to be the same one of the suits, so: $\frac{12}{48}$ because we took each card from each suit, so we have only $12$ remaining in that same suit.
But we have to consider the order that we draw the cards, so we multiply by $5!$
And we have my final answer of: $5! \cdot \frac{13}{52} \cdot \frac{13}{51} \cdot \frac{13}{50} \cdot \frac{13}{49} \cdot \frac{12}{48}$
Sadly, this answer is only half the real answer: $\approx 0.26375$ mine gives: $\approx 0.13187$
Where does the $ \cdot 2$ need to come from? I don't see another orientation other than the suit we pick the second card to be of, so we don't need to multiply by $2$ but by $4C2 = 6$ and that gives me an answer that times $3$ larger than the real one, so in that case, where does the $\frac{1}{3}$ need to come from?
Thank you!
Best Answer
Obviously, the "correct" hand should contain two cards of a suit and one card of each of the other 3 suits. There are $\binom41$ ways to choose the suit with the double, and $\binom{13}2$ ways to choose the double out of the suit. Therefore the overall number of "correct" hands is: $$ \binom41\binom{13}2\binom{13}1^3. $$ Dividing this by the overall number of hands: $\binom{52}5$ you obtain the correct probability.
You way fails because of two reasons: first you double-count the hands and second your miss the factor 4 because the "last" card may be of any of 4 suits.