By standard deck of cards, I mean a 52-card deck used for games like Blackjack and Poker. Also, let's assume Ace can only be 11. So {1,9} is not a possible choice.
Here is my logic:
The total amount of possible 2-card hands I can draw is $C(52,2)$ which is my denominator.
Now, counting all possible hands that sum to 10. I have {2,8}, {3,7}, {4,6}, and {5,5}. I think it makes sense for {2,8}, {3,7}, {4,6} to be handled separately from {5,5} as demonstrated below.
Case {2,8}, {3,7}, {4,6}:
For my first card, I must pick a card out of any of these ranks {2,3,4,6,7,8}. And I have 4 cards in each rank, and I must pick 1 out of my available options. => $C(24,1)$.
For the second card, I must pick a card out of the rank that completes the sum (i.e. if 8 was 1st, then I must pick 2). There should only be 1 rank, which has 4 cards => $C(4,1)$.
Case {5,5}:
For both cards, they must be of the same rank, 5, and there are only 4 available in the deck => $C(4,2)$.
Using my above logic I get this equation:
$( C(24,1)*C(4,1) + C(4,2) ) / C(52,2) )$
However, this answer is incorrect. Where did I go wrong in my logic?
Best Answer
In my opinion, first let us choose any one from your 3 groups {2,8}, {3,7}, {4,6}, and then choosing one card from each set. So, we will get C(3,1)C(4,1)C(4,1)+C(4,2) for {5,5}. Thus, the required probability=$\frac{54}{C(52,2)}$