Two people are playing a card game. They are using a reduced deck of cards, consisting of A, 2, 3, …, 9 for each of the four suits (i.e. 36 cards). In this game an ace has a value of 1.
Player A deals a single card to themselves and Player B.
Player A has a 5 of diamonds.
Player B has a 7 of spades.
Player A then draws one card at a time trying to reach a sum of 10, with the starting card. If they go over 10, they lose, and the cards drawn are shuffled back into the deck.
What is the probability that A wins, and what is the probability that B wins. (Winning means they can form a sum of 10).
The answers given are 0.1536 for A, and 0.1468 for B.
I can get the answer for B as follows:
$\frac{nCr(4,1)}{nCr(34,1)}+\frac{nCr(4,1)}{nCr(34,1)}\times\frac{nCr(4,1)}{nCr(33,1)}\times2+\frac{nCr(4,3)}{nCr(34,3)}$
which is the probability of a 3 + probability of a 2 and an A + probability of three Aces.
However, I can't get the answer for A, even trying very similar techniques.
Best Answer
The answer for $A$ seems to have approximation error. Here is how I look at $A$ getting to sum of $10$.
In one draw - gets one of the remaining $3$ cards with face value $5$.
In two draws - $(4,1)$ or $(3,2)$
In three draws - $(1, 1, 3)$ or $(2, 2, 1)$
In four draws - $(1, 1, 1, 2)$
In five draws - $(1, 1, 1, 1, 1)$.
So desired probability $ = \displaystyle \small \frac{3}{34} + 2 \cdot 2! \big(\frac{4}{34}\big)^2 + 2 \cdot \frac{3!}{2!} \big(\frac{4}{34}\big)^3 + \frac{4!}{3!} \big(\frac{4}{34}\big)^4 + \big(\frac{4}{34}\big)^5$
$ = \displaystyle \small \frac{437763}{2839714} \approx 0.154 $