Playing Cards: Formula for match probability with card draw rules


I have resolved my confusion and developed the universal formula to solve this problem, so I am closing this question


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Ian F
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Ian F

"Mathematics is an exact fact; figures don’t lie, but liars sometimes figure" Col. L. F. Copeland, 1888 "Probability: a way to replace the words "maybe, can, should" with a number" "If you can't put a number to it, it is just an opinion and not science"

Updated on August 01, 2022


  • Ian F
    Ian F over 1 year

    This is a simplification of a larger problem I have been working on, calculating probabilities based upon a characteristic, with rules for selection based upon a second characteristic. I believe if I can, with your help, determine the correct calculation method for this problem I can then expand it to my larger problem. Thanks in advance!


    With a deck of 8 cards: ( Ace Clubs Black [ACb], Ace Diamonds Red [ADr], Ace Hearts Red [AHr], Ace Spades Black [ASb], King Clubs Black [KCb], King Diamonds Red [KDr], King Hearts Red [KHr], and King Spades Black [KSb].

    Therefore: the deck is (ACb, ADr, AHr, ASb, KCr, KDr KHr, KSb) arranged in fair random order.

    Draw Rules:

    Draw, without replacement, up to 4 cards. Stop drawing as soon as you have drawn 2 Aces or have drawn the 4th card.


    What is the probability P(A) of drawing exactly 2 black cards and the KHr card? Extra card, if drawn, must be red.

    Expected result (derived from counting of the exhaustive list of possibilities): $$P(A) = {296\over 1680} = {37\over 210}\approx 0.17619$$

    I need help developing generalized approach, as the underlying problem I am trying to solve has $_{416}P_6 \approx 5.0E6$ permutations, so full list enumeration and counting is not practical.

    Here's how I have been calculating it, but I have gotten stuck.

    [EDIT:] Upon reflection it seems each of the 5 hand options could each be treated like a pick from bags problem:

    E.g. You have 2 bags (A, K).

    Bag A contains 2 black beads and 2 red beads. Bag K contains 2 black beads, 1 red beads, and one green bead.

    For hand "AAK", draw 2 from bag A and 1 from bag K. What is the probability P(G2) that what you draw is B, B, G? $$P(G_2) = \left(\frac{2}{4}\right)\left ( \frac{1}{4} \right ) = 0.125$$ $$P(B_2) = \left(\frac{240}{1680}\right)$$ Then multiply this probability with the P(B2) to get the final hand probability. $$P(H_2) = P(B_2) * P(G_2) = \left(\frac{240}{1680}\right)*\left(\frac{1} {8}\right) = \left(\frac{240}{13440}\right) \approx 1.785714E-2$$ and then sum all together $$P(A) = \sum_{i=0}^5 P(G_i)P(B_i)$$

    Does that seem correct?


    Known permutation facts:

    Total Permutations of 4 cards: $T_p = _8P_4 = 1680$

    Possible hands drawn, sorted by rank, and their permutations (items in brackets are assumption that the undrawn cards can be any of the remaining):

    1. AA: $B_1 = _4P_2 * [_6P_2] * 1 = 360$ Hands: AAxx
    2. AAK: $B_2 = _4P_2 * _4P_1 * [_5P_1] * 2 = 240$ Hands: AKAx, KAAx
    3. AAKK: $B_3 = _4P_2 * _4P_2 * 3 = 432$ Hands: AKAK, AKKA, KKAA
    4. AKKK: $B_4 = _4P_1 * _4P_3 * 4 = 384$ Hands: AKKK, KAKK, KKAK, KKKA
    5. KKKK: $B_5 = _4P_4 * 1 = 24$ Hands: KKKK

    Probability of the hand n: $P(B_n) = {B_n\over T_P}$

    Permutations of desired outcome H (of 4 black pick 2, of 1 KHr pick 1, of 3 red pick 1, all permutations): $H_P = \binom{4}{2}\binom{1}{1}\binom{3}{1}*(_4P_4) = 432$

    And this is where I am having difficulty:

    Permutations based upon drawn hand Bn, where for nCr $\binom{n}{r}$ when r=n nCr = 1, when r>n, nCr = 0.

    1. For B1 [AA]: $H_1 = \binom{2}{2}\binom{2}{2}\binom{0}{1}\binom{1}{1}\binom{5}{2} * (_4P_4) = 0$
      Of the 2 Aces, Pick 2 and of those, pick from the 2 black. Of the 0 Kings, pick 1... Impossible
    2. For B2 [AAK]: $H_2 = \binom{2}{2}\binom{1}{1}\binom{5}{1} * (_4P_4) = 120$ All Permutations 4P4
    3. For B3 [AAKK]: $H_3 = \left(\binom{2}{2}\binom{1}{1}\binom{1}{1} + \binom{2}{1}\binom{2}{1}\binom{2}{1}\binom{1}{1} \right) * (_4P_4) = 168$
      Pick 2 black from the aces, 1 KHr, and one Kr. OR Pick 1 black from the aces, 1 red from the aces, 1 black from the kings, and 1 KHr. All permutations 4P4.
    4. For B4 [AKKK]: $H_4 = \left(\binom{2}{1}\binom{2}{1}\binom{1}{1}\binom{1}{1} + \binom{2}{1}\binom{2}{2}\binom{1}{1}\binom{1}{1}\right) * (_4P_4) = 144$
      Pick 1 black from the 2 for Aces, pick 1 black for the 2 for kings, pick 1 from KHr, and 1 from the remaining Kr OR pick 1 from the 2 red Aces, 2 from the 2 black kings, 1 from KHr and 1 Kr. All permutation 4P4
    5. For B5 [KKKK]: $H_5 = \binom{2}{2}\binom{1}{1}\binom{1}{1} * (_4P_4) = 24$
      Of the kings, pick 2 of the 2 blacks, of the remaining kings, pick the 1 KHr, of the remaining 1, pick 1. All permutations 4P4.

    The confusion I have is how to go from the Hn value combined with P(Bn) to derive the expected value for P(A). If it is $P(B_n|H_n)$ then I don't know the divisor for $B_n$

    As a check: The list enumerated quantities of Hn|Bn are:

    1. $H_1 = 0$
    2. $H_2 = 20$
    3. $H_3 = 108$
    4. $H_4 = 144$
    5. $H_5 = 24$

    Thank you for assistance. This is a re-ask of this problem, as my last attempt at the question was down voted twice without comment, so I have completely re-written it. If you do feel this is worthy of a down vote, can I please ask for the favor of a comment of advice on how to clarify the scenario to make it more acceptable. Thanks.

    • Andrew Woods
      Andrew Woods over 8 years
      Can I ask what the larger problem is?
    • Brian Tung
      Brian Tung over 8 years
      Is computer enumeration and evaluation of the larger problem impractical for some reason?