Card Draw Percentages

  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
RUber
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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?
 
  • #3
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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.
 
  • #4
RUber
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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.
 
  • #5
RUber
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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%.
 
  • #6
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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.
 
  • #7
RUber
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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.
 
  • #8
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1
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:
  • #9
RUber
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p3 should be 2/30.
 
  • #10
20
1
ahh, yes of course 2/30.
thanks.
 

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