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I am trying to calculate some probabilities for a card game. Players have to draw 3 cards each time and the cards must add up to a certain value for them to win - the value changes depending on the stage of the game.

Is there a formula I can use to calculate the probabilities of drawing ≥ certain values - ≥12, ≥18 and so on? Players are drawing without replacement from a standard 52-card deck, no jokers. Aces, kings, queens, and jacks do not have a numerical value, so players are effectively drawing cards with values 2-10.

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    $\begingroup$ Doable, but exceedingly messy. I would suggest doing a large simulation, and use the resulting frequencies as a table of values. $\endgroup$
    – André Nicolas
    Commented Jul 4, 2015 at 5:33
  • $\begingroup$ Drawing with or without replacement? Do Aces and picture cards count $0$ in the sum or do you draw again? With a computer it would be easy enough to consider every equal probable draw. You could alternatively use a generating function approach though I doubt it would make the calculations easy. $\endgroup$
    – Henry
    Commented Jul 4, 2015 at 12:00
  • $\begingroup$ See chapter 3 of this book. $\endgroup$
    – user173262
    Commented Jul 4, 2015 at 13:20

1 Answer 1

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There is a generating function approach, but does not seem to have a closed form.

\begin{align*} G(x,y) &= \left((1+x)^4\prod_{i=2}^{10} (1+x\, y^i)\right)^4 \end{align*}

If the sum needs to be $\ge 12$, we need to find sum of the coefficients $[x^3 y^j]G(x,y)$ where $30 \ge j \ge 12$ and divide by $\binom{52}{3}$ to get the probability.

$$\mathbb{P}(sum \ge 12) = \frac{12236}{\binom{52}{3}} \approx 0.55366515837$$

$$\mathbb{P}(sum \ge 18) = \frac{4580}{\binom{52}{3}} \approx 0.207239819005$$

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