What is the probability of drawing 2 aces and 3 jacks from a deck of 52 cards?

Question

Here is the calculation I used, but it is incorrect. I am not sure why. The equation makes sense to me.

My calculation:

Original problem with actual solution:

I don't understand why my calculation is logically different from the actual solution.

Answer 1

Draw Aces First $\cdot$ Draw Jacks Second $\cdot$ Number of permutations of aces and jack draws =

$ (\frac{4}{52} \cdot \frac{3}{51}) \cdot (\frac{4}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}) \cdot \frac{5!}{2!3!}$ = 0.000009234 = 0.0009234%

You may convince yourself that no matter if you draw aces first, jacks first, or a mixture, the probability will be the same, that is why we only have to compute one of them, and multiply by the permutations.

Answer 2

We extract five cards only. The modeling probability space is $\Omega$, it has as elements tuples of the shape $(c_1,c_2,c_3,c_4,c_5)$, the components being different cards. The order in the tuple counts. So there are $$(52-0)(52-1)(52-2)(52-3)(52-4)$$ elements in the space. These are "all events". How many of them are "good events", i.e. such that we have -up to reordering- a mixed JJJAA distribution. For this, we count first the possibilities to place the JJJ among the five places, there are $\binom 53=10$ such possibilities. For each such possibility, we isolate separately, in the same order, the JJJ places and the AA places. There are $(4-0)(4-1)(4-2)$ possibilities to fill in the JJJ places, and $(4-0)(4-1)$ possibilities for the $AA$. So the probability is $$ \frac{\binom 53\cdot 4\cdot3\cdot 2\cdot 4\cdot 3}{52\cdot 51\cdot 50\cdot 49\cdot 48} = 1/108290 \approx 0.00000923446301597562\dots\ . $$

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Computer simulation, here sage:

sage: A = Arrangements( [1..52], 5 )
sage: A.random_element()
[43, 47, 9, 41, 12]
sage: success_counter = 0
sage: for trial in xrange(10**6):
....:     c = A.random_element()
....:     c.sort()
....:     if len( Set(c).intersection( Set([1,2,3,4]) ) ) != 3:    continue 
....:     if len( Set(c).intersection( Set([5,6,7,8]) ) ) != 2:    continue
....:     print "JJJAA at trial %s" % trial
....:     success_counter += 1
....:     
JJJAA at trial 17976
JJJAA at trial 220707
JJJAA at trial 421770
JJJAA at trial 465984
JJJAA at trial 468250
JJJAA at trial 549154
JJJAA at trial 589489
JJJAA at trial 917291
sage: success_counter / 10**6
1/125000
sage: binomial(5,3)*4*3*2*4*3 / ( 52*51*50*49*48)
1/108290

Answer 3

Here is the calculation I used, but it is incorrect. I am not sure why. The equation makes sense to me.

$$\dbinom{52}{2,3,47}\dfrac{4^24^347^{47}}{52^{52}}$$

That is the probability for drawing exactly two aces and three jacks when drawing with replacement (and shuffling) from a standard deck fifty two times.

The problem specified that a poker hand of 5 cards was to be drawn.   That is drawing without replacement just five times.

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There are $\binom {52}{5}$ distinct (and equally probable) ways to select 5 from 52 cards, and among these are $\binom{4}2\binom 43$ ways to select 2 from 4 aces and 3 from 4 jacks.

You can also evaluate the probability for drawing some arrangement of 2 aces and 3 jacks. $$\begin{align}&\binom 52\left(\dfrac{~~4\cdot ~~3\cdot ~~4\cdot ~~3\cdot~~ 2}{52\cdot 51\cdot 50\cdot 49\cdot 48}\right)\\=&\frac{5!}{3!~2!}\frac{4!~4!~47!}{2!~1!~52!}\\=&\frac{4!}{2!~2!}\frac{4!}{3!~1!}\frac{5!~47!}{52!}\\=&\left.\binom 42\binom 43\middle/\binom{52}{5}\right.\end{align}$$