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Given a 52-card deck, if we pick 10 cards, what's the probability of having all four aces among the 10 cards we picked?

My attempt was defining $\Omega = \{(1,2,...,52)^{10}\}$. Now $|\Omega|=\binom{52}{10}$

Let $A$ be the event that we're looking for. $A=\{\{1,2,3,4\}\cup(5,6,...,52)^{6}\}$. $|A|=\binom{48}{6}$

Now $Pr(A)=\dfrac{|A|}{|\Omega|}=\dfrac{\binom{48}{6}}{\binom{52}{10}}$

I don't have the answer but I noticed I got a really small number ($0.07\%$), so I don't think I solved it right.

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    $\begingroup$ I dont see a probem with your solution. Getting 4 specific cards in a 10 card draw is going to be a low probability event. $\endgroup$ – user625 Jun 19 '18 at 1:18
  • $\begingroup$ I got an even lower answer: $\frac{6}{7735}$ $\endgroup$ – Joseph Eck Jun 19 '18 at 1:27
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    $\begingroup$ @JosephEck, what do you mean by "lower"? $6/7735=0.00077569489\approx0.0776\%$. $\endgroup$ – Barry Cipra Jun 19 '18 at 1:47
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    $\begingroup$ @BarryCipra Lol, brain fart, forgot to convert my decimal to a percent $\endgroup$ – Joseph Eck Jun 19 '18 at 1:52
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For any 4 cards from 10 there are $^{10}C_4 = 210$ combinations.

For any 4 cards from 52 there are $^{52}C_4 = 270725$ combinations.

$P(4A) = \frac{^{10}C_4}{^{52}C_4} = \frac{210}{270725} = .0007757$

In other words, your $10$ cards only comprises of $210$ four card combinations from a possible $270725.$

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  • $\begingroup$ Ah ha! It took me a moment to realise what you were counting here. Rather than "selecting 4 cards from 10", I would say "selecting 4 places from 10 (when shuffling the deck)". $\endgroup$ – Graham Kemp Jun 19 '18 at 2:27
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Comment: You have several correct answers already--including your own! I just wanted to show the connection with the hypergeometric distribution.

The number $X$ of Aces among ten cards chosen at random without replacement from a 52 card deck has a hypergeometric distribution. There are four favorable cards (Aces) and 48 unfavorable cards (non-Aces).

In R statistical software, one can compute the probabilities $P(X = k)$ for $k = 0, 1, 2, 3, 4.$

cbind(k, pdf)
     k          pdf
     0 0.4134453782
     1 0.4240465417
     2 0.1431157078
     3 0.0186166774
     4 0.0007756949

You seek $$P(X=4) = \frac{{4 \choose 4}{48 \choose 6}}{{52 \choose 10}} = 0.0007756949.$$

You are correct that getting all four Aces, even among ten cards, has a very small probability. The figure below shows a bar chart of the hypergeometric distribution of $X.$ Your probability corresponds to the very short bar at the far right.

enter image description here

Notes: (1) As an experiment, I tried doing ten million draws of 10 cards and counting the Aces each time. The expected number of aces is $10^7 \times 0.0007756949 \approx 7757.$ In my simulation I got 7591, which is fewer than 7757, but within the margin of simulation error.

(2) In many versions of poker each player's hand consists of five supposedly randomly chosen cards. A poker player would be delighted to find all four Aces in his/her hand. But the probability of that is even smaller than your probability: $1.847 \times 10^{-5}.$

dhyper(4, 4, 48, 5)
## 1.846893e-05

(3) If you sample the ten cards with replacement, then the number $Y$ of Aces in ten independent draws has $Y \sim \mathsf{Binom}(10, 1/13).$ Then $P(Y = 4) = 0.004548553.$ This is larger than your probability because the Aces don't get "used up" as they get chosen. (Each Ace could possibly be chosen more than once.)

dbinom(4, 10, 1/13)
## 0.004548553
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  • $\begingroup$ So the chance of getting at least 1 ace is higher than having no aces at all? $\endgroup$ – steenbergh Jun 19 '18 at 7:47
  • $\begingroup$ Seems so...by a smidge. That's when 10 cards are drawn. But that's not true for a 5-card poker hand: dhyper(0:1, 4, 48, 5) returns 0.6588420 for no Ace and 0.2994736 for one Ace. $\endgroup$ – BruceET Jun 19 '18 at 7:58
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Note that we pick the ten cards without "putting them back", so the space $\Omega$ is the set of all subsets with ten elements of $\Omega_0=\{1,2,\dots,52\}$. This is not the cartesian product $\Omega_0^{\times 10}$. (Which allows repetitions, and also knows the order the ten cards came to the hand.)

Our convention is to always sort the hand. (First w.r.t values, than w.r.t. colors.) And the aces are last after sorting.

The number of all "possible cases" is thus $$ |\Omega|=\binom{52}{10}\ . $$ Now let us consider the "favorable cases". Yes, the aces are at the end, for the remained $6$ places we have the remained $48$ cards, there are $$ \binom{48}6 $$ good cases. This is the posted solution. The quotient of the above two binomial coefficients is the searched probability.


We may think alternatively as follows. The cards are extracted one by one, so the new modelling space is $\Omega'$ the space of all ordered tuples with different elements from $\Omega_0$. There are $$ |\Omega'|=(52-0)(52-1)\cdots(52-9)=\frac {52!}{(52-10)!} $$ possible cases. Out of them, come cases are favorable. Let us count them. We have $\binom{10}4$ choices for the positions of the aces among the ten places of a "good ordered (tuple) hand". We have $4!$ possibilities to insert into these places the four aces. The remained positions can be filled with the remained cards in $$(48-0)(48-1)\cdots(48-5)=\frac {48!}{(48-6)!}$$ different ways. So there are totally $$ \binom{10}4\cdot 4!\cdot \frac {48!}{(48-6)!}= \frac{10!}{6!}\cdot \frac {48!}{(48-6)!} $$ "good cases". The probability, computed in this model is: $$ \frac {\frac{10!}{6!}\cdot \frac {48!}{(48-6)!}} {\frac{52!}{42!}} = \frac {\frac{48!}{42!6!}} {\frac{52!}{42!10!}} = \frac{\binom{48}6}{\binom{52}{10}}\ , $$ same as in the first modelling.


The probability is approximatively:

sage: binomial(48,6) / binomial(52,10)
6/7735
sage: _.n()
0.000775694893341952

Let us simulate some cases to convince ourselves, sage.

N = 10**5    # trials
C = Combinations( [1..52], 10 )

hitcounter = 0    # so far
for trial in xrange(N):
    hand = C.random_element()    # it is already sorted
    if hand[:4] == [1,2,3,4]:
        hitcounter += 1
print ( "N=%s trials, %s hits, quote = %s = %f" 
        % (N, hitcounter, hitcounter/N, hitcounter/float(N)) )

This gives this time:

N=100000 trials, 84 hits, quote = 21/25000 = 0.000840

Here is the list of probabilities to get four aces (4A) if we extract $k$ cards,

sage: for k in [4..52]:
....:     print ( "4A in a hand with %2s cards come with probability ~ %f"
....:             % (k, binomial(48,k-4)/binomial(52,k)) )
....:     
4A in a hand with  4 cards come with probability ~ 0.000004
4A in a hand with  5 cards come with probability ~ 0.000018
4A in a hand with  6 cards come with probability ~ 0.000055
4A in a hand with  7 cards come with probability ~ 0.000129
4A in a hand with  8 cards come with probability ~ 0.000259
4A in a hand with  9 cards come with probability ~ 0.000465
4A in a hand with 10 cards come with probability ~ 0.000776
4A in a hand with 11 cards come with probability ~ 0.001219
4A in a hand with 12 cards come with probability ~ 0.001828
4A in a hand with 13 cards come with probability ~ 0.002641
4A in a hand with 14 cards come with probability ~ 0.003697
4A in a hand with 15 cards come with probability ~ 0.005042
4A in a hand with 16 cards come with probability ~ 0.006723
4A in a hand with 17 cards come with probability ~ 0.008791
4A in a hand with 18 cards come with probability ~ 0.011303
4A in a hand with 19 cards come with probability ~ 0.014317
4A in a hand with 20 cards come with probability ~ 0.017896
4A in a hand with 21 cards come with probability ~ 0.022107
4A in a hand with 22 cards come with probability ~ 0.027020
4A in a hand with 23 cards come with probability ~ 0.032708
4A in a hand with 24 cards come with probability ~ 0.039250
4A in a hand with 25 cards come with probability ~ 0.046726
4A in a hand with 26 cards come with probability ~ 0.055222
4A in a hand with 27 cards come with probability ~ 0.064826
4A in a hand with 28 cards come with probability ~ 0.075630
4A in a hand with 29 cards come with probability ~ 0.087731
4A in a hand with 30 cards come with probability ~ 0.101228
4A in a hand with 31 cards come with probability ~ 0.116225
4A in a hand with 32 cards come with probability ~ 0.132829
4A in a hand with 33 cards come with probability ~ 0.151150
4A in a hand with 34 cards come with probability ~ 0.171303
4A in a hand with 35 cards come with probability ~ 0.193407
4A in a hand with 36 cards come with probability ~ 0.217582
4A in a hand with 37 cards come with probability ~ 0.243956
4A in a hand with 38 cards come with probability ~ 0.272657
4A in a hand with 39 cards come with probability ~ 0.303818
4A in a hand with 40 cards come with probability ~ 0.337575
4A in a hand with 41 cards come with probability ~ 0.374070
4A in a hand with 42 cards come with probability ~ 0.413445
4A in a hand with 43 cards come with probability ~ 0.455850
4A in a hand with 44 cards come with probability ~ 0.501435
4A in a hand with 45 cards come with probability ~ 0.550356
4A in a hand with 46 cards come with probability ~ 0.602770
4A in a hand with 47 cards come with probability ~ 0.658842
4A in a hand with 48 cards come with probability ~ 0.718737
4A in a hand with 49 cards come with probability ~ 0.782624
4A in a hand with 50 cards come with probability ~ 0.850679
4A in a hand with 51 cards come with probability ~ 0.923077
4A in a hand with 52 cards come with probability ~ 1.000000

and note that even if we have a hand with $51$ cards, there is still a $4/52=1/13$ probability to miss an $A$...

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Yes, the probability for selecting four specific cards (and six others) when selecting ten cards from a standard 52 card deck (without replacement) is $${\left.{\dbinom 44}\dbinom{48}6\middle/\dbinom{52}{10}\right.}$$

The probability that those four cards shall be placed among the top ten places in the deck is $${\left.\dbinom{10}4\middle/\dbinom{52}{4}\right.}$$

Which is the same small value $6/7735$.

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The chance that A♣️ is in your hand is $10/52$. Given that, the chance the A♦️ also is is then $9/51$, etc, so the answer is $\dfrac{10\cdot9\cdot8\cdot7}{52\cdot51\cdot50\cdot49}$.

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