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I've been trying to calculate the probability of drawing exactly two aces and two kings, or exactly one ace and three kings when the player already has an ace in hand. The player draws 24 more cards, and the deck is full aside from the ace in hand.

I've calculated the probability of getting exactly two aces and two kings like so:

$\dfrac{{3\choose2}.{4\choose2}.{44\choose20}}{{51\choose24}} \approx 13.81\%$

Which seems a little high to me. However, moving on with the same equation for drawing exactly one ace and three kings:

$\dfrac{{3\choose1}.{4\choose3}.{44\choose20}}{{51\choose24}} \approx 9.20\%$

And so, the probability of getting one or the other is $13.81\% + 9.20\% = 23.01\%$.

Can someone tell me where I'm going wrong? Because I have trouble believing there's a $23.01\%$ chance of the described scenario arising.

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2 Answers 2

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For exactly two more aces and exactly two kings in 24 draws from a deck of 51 cards (missing an ace), I also get

$$ \frac{{3 \choose 2}{4 \choose 2}{44 \choose 20}}{{51 \choose 24}} = 0.138,$$

(to three places), computed in R as:

 18*choose(44, 20)/choose(51,24)
 ## 0.1380654

Here is a simulation of a million such draws with probabilities correct to 2 or 3 places.

 m = 10^6;  nr.ace = nr.kng = numeric(m)
 deck = 2:52 # aces are 1,2,3,4; kings 5,6.7,8
 for (i in 1:m) {
  draw = sample(deck, 24)
  nr.ace[i] = sum(match(2:4, draw, nomatch=0)>0)
  nr.kng[i] = sum(match(5:8, draw, nomatch=0)>0) }

 mean(nr.ace == 2)
 ## 0.357932
 mean(nr.kng == 2)
 ## 0.387775
 mean(nr.ace==2 & nr.kng==2)
 ## 0.137598  # approx 0.138 as in exact combinatorial result
 mean(nr.ace==1 & nr.kng==3)
 ## 0.092077

 AK = as.data.frame(cbind(nr.ace,nr.kng))
 table(AK)/m
           nr.kng
 ## nr.ace        0        1        2        3        4
 ##      0 0.007532 0.035196 0.055101 0.035380 0.007596
 ##      1 0.026295 0.109556 0.157987 0.092077 0.018387
 ##      2 0.027684 0.105594 0.137598 0.073657 0.013399
 ##      3 0.008706 0.030502 0.037089 0.017775 0.002889

The approximate probability of $P(A = 2, K = 2)$ is found in cell $(2,2)$ of the table. The approximate probability $P(A = 1, K = 3)$ is in cell $(1, 3).$

Related probabilities can also be approximated from the table. For example, the total probability $P(A = 2) \approx 0.358$ is found separately in the printout above and as the total of row 2 of the table.

 sum(c(0.027684, 0.105594, 0.137598, 0.073657, 0.013399))
 ## 0.357932

Its exact probability (to three places) is

$$ \frac{{3 \choose 2}{48 \choose 23}}{{51 \choose 24}} = 0.358,$$

 3*choose(48, 22)/choose(51, 24)
 ## 0.3578391

However, to get the probability of either 'two aces and two kings' OR 'one ace and three kings', you should add only two entries in the table $(2,2)$ and $(1,3).$

Addendum: 'Expanded' R code, demonstrating method of counting aces (cards 2 through 4) in 24 draws:

 draw = sample(deck, 24);  draw  # list of 24 cards drawn w/o replacement
 ## 17 40 46 51 44 30 50 24  4  7 52 31 13 28 25 18 22 42  3  5 43 48 19  8
 ace.posn = match(2:4, draw, nomatch=0);  ace.posn
 ## 0 19  9  # card 2 not drawn, 19th was card 3, 9th was card 4
 tf.aces = (ace.posn > 0);  tf.aces
 ## FALSE  TRUE  TRUE  # TRUE for each ace found
 nr.aces = sum(tf.aces);  nr.aces  # counts TRUE's
 ## 2  # count of aces in 'draw' is stored in vector 'nr.ace'

The count of the number of aces in a 'draw' does not depend on which aces or their order. This is done a million times.

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  • $\begingroup$ I'm not sure what all that test data means, but are you saying $23.01\%$ is correct, or incorrect? $\endgroup$ Commented Mar 26, 2016 at 21:25
  • $\begingroup$ Trying again to add my reply, browser lost the 1st attempt. Yes, 0.23 is correct. Add two cells from my table. (They agree with the first two combinatorial results in your question.) I don't see any double-counting in that. See the last sentence of my answer, added recently. – BruceET $\endgroup$
    – BruceET
    Commented Mar 26, 2016 at 21:43
  • $\begingroup$ I see. So it must be right, but it seems weird that there are such high odds for an "exact" scenario, doesn't it? I'm actually adding the odds for exactly one more ace and 2 of any other card (and I'm doing the same for exactly 2 more aces). The problem I'm running into is the massive amount of double-counting. Can you think of a more suitable algorithm for this? $\endgroup$ Commented Mar 26, 2016 at 21:46
  • $\begingroup$ I think that'll be messy. I think surprisingly high probabilities in your original problem are because you are drawing 24 cards. Intuition from 5-card poker and 13-card bridge does not carry over. $\endgroup$
    – BruceET
    Commented Mar 26, 2016 at 21:58
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    $\begingroup$ I rethought the issue, and find that even if we do not assume that a particular ace is missing, the result will be the same. I have revised my answer accordingly. (+1) $\endgroup$ Commented Mar 27, 2016 at 7:19
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Revised answer

Assuming you are respectively drawing $2$ or $1$ more aces, your answers are correct.

My suspicion that the first ace being unspecified would affect the result was unwarranted.

Here is a longer way for the first case (just to confirm the correctness).

The ace you have could be any of $4$, so numerator $= 4\times{3 \choose 2}{4 \choose 2}{44 \choose 20}$

But then you need to similarly structure the denominator as $4\times\binom{51}{24}$,

and the result will remain the same as what you got.

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  • 1
    $\begingroup$ Hmm.. Would you mind elaborating? I believe you that I've overcounted, but I'm having trouble figuring out why I need to divide.. $\endgroup$ Commented Mar 26, 2016 at 19:48
  • $\begingroup$ I am easily confused by combinatorial problems, but I too am puzzled by this analysis of overcounting. $\endgroup$
    – BruceET
    Commented Mar 26, 2016 at 20:48
  • $\begingroup$ Denominator count is unordered. In numerator, C(3,2) = 3. $\endgroup$
    – BruceET
    Commented Mar 27, 2016 at 4:21
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    $\begingroup$ I have reconsidered the queries raised, and concluded that there is no overcounting. $\endgroup$ Commented Mar 27, 2016 at 7:23

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