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.

  • $\begingroup$ Do the players draw the cards alternatively? Does the game end if A or B first reach 10? Does the game end (without a winner) if they both go over 10? $\endgroup$
    – user
    Apr 25, 2021 at 13:23

1 Answer 1


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 $

  • $\begingroup$ Thanks for your help. Though as the game is involving a deck of cards it's not possible to have five aces. $\endgroup$
    – C. de Haer
    Apr 26, 2021 at 4:42
  • $\begingroup$ you are right but I am going by the question that "cards drawn are shuffled back into the deck" so I can draw $5$ Aces in $5$ draws with probability $(\frac{4}{34})^5$. $\endgroup$
    – Math Lover
    Apr 26, 2021 at 7:00

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