What is the probability that a five-card poker hand has four ACES?
When I was solving the above stated problem, I got confused while trying different methods :
Assume a normal $52$ deck of cards.
Method 1:
Selecting the $4$ aces from total $4$ aces can be done in $\mathsf C(4,4)$ ways and selecting any non ace element from rest $48$ cards can be done by $\mathsf C(48,1)$ ways. Any $5$ cards can be drawn from $52$ deck of card in $\mathsf C(52,5)$ ways.
So the probability is
$$\frac{\mathsf C(4,4)\times \mathsf C(48,1)}{\mathsf C(52,5)}$$
Method 2:
We have $4$ aces in total. so probability of selecting an ace from $52$ cards is
$4/524$ , then we are left with $51$ cards and selecting again another ace gives probability $3/51$. Similarly for next two aces probability will be $2/50$ and $1/49$. Now we are left with total $48$ cards and we can obviously choose any of these $48$ card which gives probability of $48/48$.
Multiplying the probabilities gives us
$$\frac{(4\cdot 3\cdot 2\cdot 1\cdot 48)}{(52\cdot 51\cdot 50\cdot 49\cdot 48)}.$$
Method $1$ produces result which is $5$ times more than method $2$. What am I doing wrong ?
Help appreciated 🙂
Best Answer
I would like to add something that may be of use to you in future:
In drawing w/o replacement,
when order doesn't matter, using combinations gives the answer directly:
$$\text{e,g, here,}\;\; \frac{\binom44\binom{48}1}{\binom{52}{5}}$$
whereas if order does matter, multiplying probabilities gives the answer directly,
$$\text{e.g. if the non-ace must be second},\;\; \frac4{52}\cdot\frac{48}{51}\cdot\frac3{50}\cdot\frac2{49}\cdot\frac1{48}$$
and if you chose instead to use the other process,
you must use a multiplication/division factor to compensate