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An ace is always worth 11, all face cards (Jack, Queen, King) are worth 10, and all number cards are worth the number they show. Given a shuffled deck of 52 cards what is the probability that you draw 2 cards and they sum $21$?

I have counted the number of possible hands $(52C2) = 1326$ and then the combination that adds to $21$ which is $10+11$, there are 16 cards of value 10: 4 10s, Js, Qs, Ks, and then 4 cards of value 11: 4 aces.

So the probability that the 2 drawn cards sum to $21$:

$(16*4)/1326$

However, that is an incorrect answer...

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    $\begingroup$ Who says it's incorrect? $\endgroup$
    – Arthur
    Commented Jul 30, 2019 at 11:40
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    $\begingroup$ In $(52C2)$, you don't consider the order in which cards are drawn. In $(16 \cdot 4)$, however, you do. $\endgroup$
    – Dirk
    Commented Jul 30, 2019 at 11:44
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    $\begingroup$ I can confirm your result. The probability drawing firstly a face card or a 10 and then a 11 is $\frac{16}{52}\cdot \frac{4}{51}$. Since we can draw the carde in reverse order we have to multiply it by $2$. $\endgroup$ Commented Jul 30, 2019 at 11:50
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    $\begingroup$ I see where I went wrong, I decided that 16*4 = 66, not 64... $\endgroup$ Commented Jul 30, 2019 at 11:54
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    $\begingroup$ @LiborZachoval Such flaws are not unusual. Here the user just omit a $0$ of a figure. $\endgroup$ Commented Jul 30, 2019 at 12:03

2 Answers 2

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You're right. The probability is $$ \frac{64}{1326} $$ However, depending on how this is graded, they might expect an answer like $$ \frac{32}{663},\quad4.8\% $$ or something similar. The other possibility is that whoever marks your answer is just wrong. It happens from time to time. Or, as it turns out in the comments, $16\cdot 4$ can quickly become $66$ if you're being a bit too quick.

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Considering ordered hands,

  • total hands: $52\cdot51$.

  • winning hands: $4\cdot16+16\cdot4.$

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  • $\begingroup$ Yes, and that gives the same answer. $\endgroup$
    – drhab
    Commented Jul 30, 2019 at 11:52

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