# Probability of the sum of two drawn cards being less than 11?

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 :)

$\newcommand{\cb}{\color{brown}}$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.
• @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! – Brian M. Scott Nov 16 '13 at 0:07
• @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. – Brian M. Scott Nov 16 '13 at 2:04