The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:

An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all the cards are gone. If after a certain number of tries, the athlete has done a total of 12 pushups, then the probability of him/her picking a 7 as the very next card is $\frac{a}{b}$. What is a + b?

I know that I have to count the possibilities of getting a 12:


*

*2 cards
The possibilities would be 2 and 10, 3 and 9, 4 and 8, 5 and 7, which would make 4 possibilities.


*

*3 cards
1, 2, 9 | 1, 3, 8 | 1, 4, 7 | 1, 5, 6 | 2, 3, 7 | 2, 4, 6 | 3, 4, 5 | which would be 7 possibilities.


*

*4 cards
1, 2, 3, 6 | 1, 2, 4, 5 | which is 2 possibilities.
There cannot be more than 4 cards: 1, 2, 3, 4, 5 would make 15.
The total number of possibilities for getting a 12 is 17, and out of those, there are 3 of them that will pick the card 7, which means 14 out of the 17 possibilities will be valid for the problem.
I don't know how to continue from here. Should I solve for the probability each of the three cases listed above separately, then add them, or is there a better method?

$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$


*

*The number of decks for which the first $2$ cards sum to $12$ is $(4)(2!)(8!)$, where

*

*The factor $4$ counts the ordered pairs $(2,10),\;(3,9),\;(4,8),\;(5,7)$.

*The factor $2!$ allows each pair to be permuted.

*The factor $8!$ counts the permutations of the remaining cards.


*The number of decks for which the first $2$ cards sum to $12$ and the next card is $7$ is $(3)(2!)(7!)$, where

*

*The factor $3$ counts the ordered pairs $(2,10),\;(3,9),\;(4,8)$.

*The factor $2!$ allows each pair to be permuted.

*The factor $7!$ counts the permutations of the cards remaining after the $7$.



$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$


*

*The number of decks for which the first $3$ cards sum to $12$ is
is $(7)(3!)(7!)$, where

*

*The factor $7$ counts the ordered triples $(1,2,9),\;(1,3,8),\;(1,4,7)\;(1,5,6),\;(2,3,7),\;(2,4,6),\;(3,4,5)$.

*The factor $3!$ allows each triple to be permuted.

*The factor $7!$ counts the permutations of the remaining cards.


*The number of decks for which the first $3$ cards sum to $12$  and the next card is $7$ is $(5)(3!)(6!)$, where

*

*The factor $5$ counts the ordered triples $(1,2,9),\;(1,3,8),\;(1,5,6),\;(2,4,6),\;(3,4,5)$.

*The factor $3!$ allows each triple to be permuted.

*The factor $6!$ counts the permutations of the cards remaining after the $7$.



$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$


*

*The number of decks for which the first $4$ cards sum to $12$ is
is $(2)(4!)(6!)$, where

*

*The factor $2$ counts the ordered quadruples $(1,2,3,6),\;(1,2,4,5)$.

*The factor $4!$ allows each quadruple to be permuted.

*The factor $6!$ counts the permutations of the remaining cards.


*The number of decks for which the first $4$ cards sum to $12$ and the next card is $7$ is $(2)(4!)(5!)$, where

*

*The factor $2$ counts the ordered quadruples $(1,2,3,6),\;(1,2,4,5)$.

*The factor $4!$ allows each quadruple to be permuted.

*The factor $5!$ counts the permutations of the cards remaining after the $7$.



$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
Hence, the probability that the next card is $7$, given that the previous cards sum to $12$ is
$$
\frac
{\;\,(3)(2!)(7!)+(5)(3!)(6!)+(2)(4!)(5!)}
{\;\,(4)(2!)(8!)+(7)(3!)(7!)+(2)(4!)(6!)}
=
\frac{8}{79}
$$
