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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.

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  • $\begingroup$ 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
    Commented Apr 25, 2021 at 13:23

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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 $

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  • $\begingroup$ 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
    Commented 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
    Commented Apr 26, 2021 at 7:00
  • $\begingroup$ Since this question popped up again, I'll point out that "cards drawn are shuffled back into the deck" only occurs "If they go over 10", so the goal must be met while drawing without replacement. $\endgroup$
    – Daniel Mathias
    Commented Apr 2, 2023 at 0:49

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