Drawing Cards Without Replacement

Discussion in 'Physics & Math' started by chuk15, Feb 17, 2010.

  1. chuk15 Registered Senior Member

    Messages:
    30
    Lets say you have a deck of 52 normal everyday cards.

    You draw cards without replacement (aka do not put the card you draw back into the deck).

    You draw cards until you get to the ace of spades.

    How do you calculate the total odds of drawing the Ace of Spades after X amount of turns?

    My hypothesis is that the odds of drawing the Ace of Spades increases as you draw cards as the card pool steadily decreases as you draw. (1/52 becomes 1/51, and 1/51 becomes 1/50 the next draw and so on.)

    I've attempted to calculate the probability by hand as follows:

    Odds of drawing Ace of Spades with 52 cards in deck * Odds of not drawing Ace of Spades with 52 cards in deck
    +
    Odds of drawing Ace of Spades with 51 cards in deck * Odds of not drawing Ace of Spades with 51 cards in deck...

    It looks like this mathematically. (I think)

    (1/52)(51/52) + (1/51)(50/51) . . .

    Using variables,

    (52-X)/(53-X)^2

    Using definite integrals I found the area underneath this equation from x=0 to x=33 and found the number to be greater than 1 which doesn't make sense. The probability of drawing an Ace of Spades after 33 turns can't be greater than 1 because what if the Ace of Spades was the bottom card in the deck O_O

    Help!
     
    Last edited: Feb 18, 2010
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  3. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Almost...

    Odds of drawing Ace of Spades with 52 cards in deck
    +
    Odds of drawing Ace of Spades with 51 cards in deck * Odds of not drawing Ace of Spades with 52 cards in deck
    +
    Odds of drawing Ace of Spades with 50 cards in deck * Odds of not drawing Ace of Spades with 51 cards in deck
    +
    ...
     
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  5. Asguard Kiss my dark side Valued Senior Member

    Messages:
    23,049
    depends, are you trying to work out the odds independently or on a specific turn?

    what i mean is "you throw a dice 99 times and it comes up 6 what is the probability that the 100th will be a 6" = 1 in 6 (assuming its not a loaded dice).

    If your saying that what is the probability of drawing it on turn x and not on any other turn then you have to go through the proccess pete is saying, if however your saying "after 6 turns of not drawing the ace whats the chances of doing it this time" then its 1 in 46
     
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  7. chuk15 Registered Senior Member

    Messages:
    30
    I'm asking that but with one important difference.

    Let's say you're rolling the dice till you get a 6. Any other number that comes up before that is crossed out and you will not count against you if you roll one of the marked numbers. So if you roll the die and a 3 comes up, you mark it off, and every time you roll a three after that it doesn't count. So on the first turn the probability of rolling a 6 is 1/6. The second turn it would be 1/5 if u rolled a three because you mark off the three. 1/4 if u mark off another number, etc.

    My question is how would you calculate the odds if you keep in mind that you're marking off each side.

    I'm not sure if it would be 1/46 because these events aren't independent. You may have a 1/46 chance when there are 46 cards left, but what about the 1/47 chance right before that, or the 1/48 chance before that?
     
  8. chuk15 Registered Senior Member

    Messages:
    30
    In mathematical terms would it be:

    (1/52) + (1/51)(51/52) + (1/50)(50/51) + . . .?
     
  9. Asguard Kiss my dark side Valued Senior Member

    Messages:
    23,049
    thats what im asking, are you taking the chance as "im going to do this till i get x and what is the chance that x will happen on turn n" OR are you saying "ok i have thrown my 99 6s in a row but i wonder what the chance is that i will throw it this time"

    If its the first then its

    1/52 + 1/51 + 1/50 + 1/49....
    If the second then its simply 1/how ever many you have left
     
  10. chuk15 Registered Senior Member

    Messages:
    30
    I believe it's something like this, yes. I'm drawing cards till I get the Ace of Spades, and I want to know the total probability of drawing it after an X amount of turns. (so for 10 draws I'd add up all nine previous probabilities and the 10th one).


    I've tried that but when you add 1/52 + 1/51 + 1/50 + 1/49... at around the 33rd term the odds of getting the Ace of Spades increases past one, which doesn't make sense.

    I did this by evaluating the area under the curve of 1/(52-X) from zero to 33 and I got a probability of 1.0068 (over 100% chance).

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    Even worse, at the 45th term (45th draw) I got a probability of 2.0053. (200%)
     
  11. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Wait, I made a mistake... try again:

    Odds of drawing Ace of Spades with 52 cards in deck
    +
    Odds of drawing Ace of Spades with 51 cards in deck *
    Odds of not drawing Ace of Spades with 52 cards in deck​
    +
    Odds of drawing Ace of Spades with 50 cards in deck *
    Odds of not drawing Ace of Spades with 51 cards in deck *
    Odds of not drawing Ace of Spades with 52 cards in deck​
    +
    Odds of drawing Ace of Spades with 49 cards in deck *
    Odds of not drawing Ace of Spades with 50 cards in deck *
    Odds of not drawing Ace of Spades with 51 cards in deck *
    Odds of not drawing Ace of Spades with 52 cards in deck​
    +
    ...

    So...
    (1/52) + (1/51)(51/52) + (1/50)(50/51)(51/52) + (1/49)(49/50)(50/51)(51/52) + . . .

    Which of course breaks down to...
    n/52

    Which in hindsight makes perfect sense!
    The question is the same as:

    Odds first card is Ace of Spades
    +
    Odds second card is Ace of Spades
    +
    Odds third card is Ace of Spades
    +
    ...
    +
    Odds nth card is Ace of Spades
     
  12. chuk15 Registered Senior Member

    Messages:
    30
    That looks to be the correct answer, but I'm not sure I understand why you have to multiply by the odds of not drawing the ace of spades with 52 cards in deck every term as well as previous terms.
     
  13. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Because the only way you can draw the Ace of Spades on the 4th turn is by...
    Not drawing it on the first, and
    Not drawing it on the second, and
    Not drawing it on the third, and
    Drawing it on the fourth.
     
  14. paulfr Registered Senior Member

    Messages:
    227
    Isn't Probability Fun ?

    Double checking the OPs question, he said ..
    "How to you calculate the total odds of drawing the Ace of Spades after X amount
    of turns?"

    So to Pete I say, yes your wording [formula] seems correct to me.
    BUT
    If it is then calculating we get ...........

    P[ drawing the AceSpades on the 4th card] =

    51/52 x 50/51 x 49/50 x 1/49 = 1/52

    and we can see this would be the SAME ANSWER for any N.

    [ Note we can multiply P's because the P[not drawing at position k+1]
    is independent of the P[not drawing at position k] ]

    How can this be ?
    How can the probability not be higher for higher N ?

    I think the answer is that this problem is equivalent to
    laying all 52 cards face down and asking what is the P
    of finding the AceSpades in the Nth position.
    It is clearer that the card could be in any position with equal P.
    And drawing cards up to that position does not change the physical situation.

    For comparison;
    This is different than the problem ....
    What is the Probability that the card is in the 4th position,
    GIVEN that it was not in the first three positions.
    That is a Conditional Probability.
    And that probability would be 1/49 which IS GREATER than 1/52

    Cheers
     
    Last edited: Feb 18, 2010
  15. chuk15 Registered Senior Member

    Messages:
    30
    Ah, thanks for the clarification.
     

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