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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 :)

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    $\begingroup$ Your method $2$ produces the probability that the first four cards are aces...thus the "odd" card must be in the last slot. As there are five slots in which to place the odd card, the true answer must be $5$ times this value. $\endgroup$ – lulu Mar 20 '16 at 13:41
  • $\begingroup$ Method 1 is correct. By method 2 you found the probability of drawing $AAAAN$ in order where $N$ stands for non ace. There more possibilities ($5$ acutally). Eg. $ANAAA$ is one of them. $\endgroup$ – drhab Mar 20 '16 at 13:41
  • $\begingroup$ @drhab I understand what you say. But my confusion lies here , if four aces and a non ace is a group of five cards then why does the order matter? Any specific reason behind it? $\endgroup$ – ViX28 Mar 20 '16 at 13:53
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    $\begingroup$ Actually the order does not matter in this case. However in method 2 you calculate in a way such that the order does matter. Try it simpler. There is one blue ball and one red ball. If I draw $2$ of the $2$ balls, then what is the probability to end up with a blue and a red ball? Calculating this with method 2 (where e.g. blue ball takes the place of ace) gives you the wrong probability $\frac12\times1=\frac12$. $\endgroup$ – drhab Mar 20 '16 at 14:01
  • $\begingroup$ See also (possible duplicate): math.stackexchange.com/questions/804058/… $\endgroup$ – wolfies Mar 20 '16 at 14:03
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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

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The difference between your two methods is about "how" you select your cards. The first example using combinations is an example of selecting 5 cards at once. The equation you provided is correct in the sense that it tells us how many ways we can select 4 ace's out of 5 cards that are selected at once out of the total possible 5 card hands.

The second example using permutations is different because it is an example of selecting 5 cards one at a time. You are assuming that choosing one card will affect the next one because the total cards left is one less. However that is not the case in your question because for it to affect the next card, you have to reshuffle the deck and then select. Otherwise you are just selecting 5 cards at once from one deck and hence must use the combinations method.

Another way to think about this is thinking about what the question is asking. It never specifies that the cards are chosen one at a time from a randomly shuffled deck. Thus this leads me to believe (I might be wrong if I didn't understand the question) that I can't use actual probability to solve this question and will have to find a way to count the possibilities of 4 aces in a hand of 5 cards out of all possibilities of 5 card hands.

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