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Card Draw Percentages

  1. Feb 11, 2015 #1
    Hey all,

    I'm trying to figure out the percentage chance to draw a card based on the information below.

    The deck has 30 cards. 2 of them are Aces and I just need to get 1 of them in my hand.

    The draw goes like this, I draw 4 cards and if I don't have an ace I can put those 4 cards on the bottom (so no chance of being re-drawn yet) and draw 4 more.

    After this the deck is shuffled and I get to draw 1 more card.

    With these interactions in mind what is the percentage chance I draw an Ace with this sequence?

    I believe the chance are as follows but i may be forgetting something

    Draw 1 = 2/30
    Draw 2 = 2/29
    Draw 3 = 2/28
    Draw 4 = 2/27
    Draw 5 = 2/26
    Draw 6 = 2/25
    Draw 7 = 2/24
    Draw 8 = 2/23

    Draw 9 = 2/26

    I for the life of me can't figure it out and would love to know the method to figure it out.

    Thanks in advance.
  2. jcsd
  3. Feb 11, 2015 #2


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    For your second draw, it should be 2/29*28/30 + 1/29*2/30. --You can draw a tree diagram to evaluate the probability of each branch.
    Is the question asking the probability of at least one ace or do you stop once you have one?
  4. Feb 11, 2015 #3
    Yes it is the probability of having one Ace at the end of the drawing and you also can stop if you get an Ace.
  5. Feb 12, 2015 #4


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    Okay, then just use the first term in my post #2.
    For each draw the probability of drawing an ace is P(ace)*P(no ace in prior draws).
    So, as you said, p(Draw 1) = 2/30. Draw 2 = 2/29*28/30 or 2/29*[1-p(Draw 1)]. etc.
    It looks like the base probability you have in post #1 is good. Add the weights and the total probability will be the sum of all the cases.
  6. Feb 12, 2015 #5


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    When I did it, I got a total probability of just over 50%. Summing over the base probabilities without weights you had in post #1 gives almost 70%.
  7. Feb 13, 2015 #6
    Thanks heaps RUber, I also came to the conclusion of ~51%, 50.98% to be exact.

    Forgive me as I'm not the best with terminology (or math for that matter), but the 70% figure you propose 'without weights' is there a way to explain that to me?

    Thanks again this has helped immensely.
  8. Feb 13, 2015 #7


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    No Problem, Maccadin. When I said without weights, I meant the raw probability of drawing an ace from the deck like you had listed in your initial post.
    I was considering the probability of reaching a certain draw the "weight", so that you could use the information you already had and just add an additional factor.
  9. Feb 20, 2015 #8
    May I offer my solution (I am actually checking my understanding of this type of problems)

    The way I would tackle this problem is:
    getting no ace in first four cards: q1 = C(28,4)/C(30,4). p1=1-q1
    getting no ace on second four cards: q2 = C(24,4)/C(26,4). p2=1-q2
    and the last draw:
    p3= 1/30, as the whole deck has been reshuffled?

    Hence, the p(At least 1 Ace) = p1 + q1*p2 + q1*q2*p3 = 0.4867

    Please tell me where did I go wrong...
    Last edited: Feb 20, 2015
  10. Feb 20, 2015 #9


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    p3 should be 2/30.
  11. Feb 20, 2015 #10
    ahh, yes of course 2/30.
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