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An Ace has a value of 11 in this problem, and a face card has a value of 10. Thus, my idea for solving this problem was to separate the problem into 7 different cases (drawing a 2 first, 3 first, 4 first, etc.), because drawing anything greater than an 8 will result in a sum greater than 11 no matter what.

So for the case of drawing a 2 first, the number of ways was 4 x ((6x4) + 3), because 6 possible numbers could be drawn plus the three 2's that were left to add up to less than 11. For the case of 3, it would be 4 x ((5x4) + 3). I think I'm on the right track, but I'm unsure of whether or not I would multiply each case by 2! to account for the order of cards drawn not mattering?

Any help would be appreciated, or even a different solution to this problem :)

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$\newcommand{\cb}{\color{brown}}$Any pair that includes a $10$, a face card, or an ace totals at least $11$, so you can focus on pairs drawn from the $32$ cards whose denominations are between $2$ and $9$ inclusive. Make a chart showing the possible pairs of values and the resulting totals:

$$\begin{array}{r|cc} &2&3&4&5&6&7&8&9\\ \hline 2&4&5&6&7&8&9&10&\cb{11}\\ 3&5&6&7&8&9&10&\cb{11}&\cb{12}\\ 4&6&7&8&9&10&\cb{11}&\cb{12}&\cb{13}\\ 5&7&8&9&10&\cb{11}&\cb{12}&\cb{13}&\cb{14}\\ 6&8&9&10&\cb{11}&\cb{12}&\cb{13}&\cb{14}&\cb{15}\\ 7&9&10&\cb{11}&\cb{12}&\cb{13}&\cb{14}&\cb{15}&\cb{16}\\ 8&10&\cb{11}&\cb{12}&\cb{13}&\cb{14}&\cb{15}&\cb{16}&\cb{17}\\ 9&\cb{11}&\cb{12}&\cb{13}&\cb{14}&\cb{15}&\cb{16}&\cb{17}&\cb{18} \end{array}$$

The combinations that you want are the black ones, and they’re not hard to count. If you’re drawing with replacement there are $\sum_{k=1}^7k=28$ different black combinations $\langle k,\ell\rangle$, and each can be drawn in $4^2=16$ ways. If, as appears to be the case, you’re drawing without replacement, it’s a little messier: you’ll have to distinguish the $\langle k,k\rangle$ cases (for $k=2,3,4,5$ from the others. There are $4^2=16$ possible $\langle 3,4\rangle$ cases, for instance, but only $\binom42=6$ possible $\langle 3,3\rangle$ cases. But it’s still not hard to count the successful pairs.

Then count the total number of different ways to draw a pair of cards, with or without replacement as appropriate, and you’ll have your probability.

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  • $\begingroup$ Thank you so much for your answer, I really appreciate it! I would like to clarify one thing though, if you don't mind. For the case of (k, k) I was wondering why you used a combination instead of doing a permutation in which the possible cases would be 4 x 3 = 12, which accounts for the dependency? Thanks again! $\endgroup$ – Brian Nov 16 '13 at 0:03
  • $\begingroup$ @Brian: Because I’m counting only the combinations of two cards that you can draw. There are $16$ different pairs consisting of a two and a three, but only $6$ consisting of two threes. If I were counting permutations, it would be $32$ and $12$. You’re welcome! $\endgroup$ – Brian M. Scott Nov 16 '13 at 0:07
  • $\begingroup$ Oh okay perfect, and last question just to make sure: The (3,4) case and the (4,3) case are two separate possibilities correct? So there are 16 ways to draw a 3 then a 4, and another 16 ways to draw a 4 and a 3. $\endgroup$ – Brian Nov 16 '13 at 1:58
  • $\begingroup$ @Brian: Yes: that’s why it would be $32$ if I were counting permutations. Here that’s an unnecessary complication: it’s easier to count combinations, and the numbers are smaller, too. $\endgroup$ – Brian M. Scott Nov 16 '13 at 2:04

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