consider this problem:

Draw cards repeatedly, without replacement, from a standard 52-card deck until we find the ace of spades. What is the probability that we draw between 20 and 30 cards?

The solution I came across: $$P(A)=\frac{11}{52}$$ where the numerator denotes the sum from the 20th to the 30th draw, and the denominator the sum of all possible draws from the 1st until the 52nd.

I've thought about this problem a lot, and I just do not understand the thinking put behind it, this problem clearly requires more attention, order does matter and the fact that the cards drawn can not be replaced made me doubt the above solution.

The solution that I believe is correct: $$P(A)=\frac{1}{33}+\frac{1}{32}+\frac{1}{31}+...+\frac{1}{23}$$ Why? Well the first 19 drawn cards must have resulted in other values than the ones required and considering the fact that cards drawn can not be replaced, the denominator or sample space always decreases by 1 until exactly the criterion of having to draw between 20 and 30 cards is met.

I would like to be assured if first of all the solution I gave is correct, or corrected if I am wrong, I would also appreciate if you could provide other helpful ideas, methods which I could to tackle these kind of problems.

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    $\begingroup$ Note: taking your proposed solution and logic, the probability that the ace of spades is within the last ten cards is $P(A)=\frac{1}{10}+\frac{1}{9}+\dots+\frac{1}{1}$ which is larger than one, a statistical impossibility. There is no such thing (under common definitions) of a probability of greater than one. $\endgroup$ – JMoravitz Jan 17 '16 at 22:18
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    $\begingroup$ The problem with your logic is that you are drawing with replacement. If you find it slot #20 it's not in #21 but you are not considering this. $\endgroup$ – Loren Pechtel Jan 18 '16 at 0:09
  • $\begingroup$ @LorenPechtel: no, it is not correctly with replacement, either. The denominators would have to be $52$ if it were. $\endgroup$ – Ross Millikan Jan 18 '16 at 0:15
  • $\begingroup$ @RossMillikan Ok, I didn't express it perfectly. Every draw is from a deck containing the ace, but the size of the deck drops as cards are removed. $\endgroup$ – Loren Pechtel Jan 18 '16 at 3:09
  • $\begingroup$ That would be drawing without replacement, but assuming you somehow know the ace of spades is not in the first $19$ cards. If that is what you mean, you need to say so. Even so, the chance for card $21$ would still be $\frac 1{33}$ because it might have been card $20$. Look at JMoravitz' example. $\endgroup$ – Ross Millikan Jan 18 '16 at 6:47

When you shuffle the deck, the ace of spades is as likely to be in any spot as any other. You win if it is in $11$ of the $52$ spots. You can do the calculation your way, but you have to start with the chance that the ace is not drawn in the first $19$ cards. The chance of that is $\frac {51}{52} \cdot \frac {50}{51} \cdot \dots \cdot \frac {33}{34}=\frac {33}{52}$ Then the chance that it is in the next $11$ cards is $\frac {11}{33}$. The product of these is $\frac {11}{52}$ as it should be. If you want to keep going your way, the chance it is the $20$th card given it has not been found is $\frac 1{33}$. The chance it is the $21$st card, given that it is not found by card $19$, is $\frac {32}{33} \cdot \frac 1{32}=\frac 1{33}$ You can keep going this way, but it is a much harder approach and will still come out $\frac {11}{52}$ in the end.


The probability that the $i-th$ card is the ace of spades is $\frac{1}{52}$for every $i=1,...,52$. So, the desired probability is $\frac{11}{52}$ because $[20,30]$ contains $11$ numbers.


This is a pretty straightforward application of the Negative Hypergeometric Distribution which is used to model the number of trials to achieve a success when sampling without replacement. If you have a population of N (52 cards) with m successes (1 Ace of Spades), and you wish to draw till you achieve j successes (draw the Ace of Spades once), then the probability that you need to draw k times is

$$ P(x=k) = \frac{{k-1 \choose j-1}{N-k \choose m-j}}{N \choose m} $$

Thus, what you are interested in is (as each event k=1, k=2, etc is mutually exclusive)

$$ \sum_{k=20}^{30} P(x=k) = \sum_{k=20}^{30} \frac{{k-1 \choose 0}{52-k \choose 0}}{52 \choose 1} = \sum_{k=20}^{30} \frac{1*1}{52} = \frac{11}{52} $$


You aren't actually doing your calculations right.

The probability of drawing on the 20th is the probability of one in the 33 cards remaining BUT you have to take into account that it also has to be the case that you never drew the card in the first 19 draws.

The probability of never having drawn the ace in the first 19 draws is the probability that it wasn't the first, it wasn't the second, and so on.

Probability that the first card wasn't the Ace of Spaces is 51/52. The probability that the ace wasn't in the first two cards is that is wasn't in the first card and it wasn't in the second. That is $51/52*50/51$. The probabily that it wasn't in the first 19 cards is the probability that it wasn't the first AND it wasn't the second AND it wasn't the third.... Or in other words$51/52*50/51*49/50*48/49*......*34/35*33/34 = 33/52$

So the odds of drawing on the $20$ draw is = Probability that is wasn't drawn in the first 19 * Probability it wasn't drawn from the remaining = $33/52 * 1/33$ = $ 1/52$.

To further your calculations: Probability of drawing on 20 through 30 = P on 20 + P on 21 + ... + P on 30 = P on 20 but not first 19 + P on 21 but not first 20 +...= $33/52*1/33 + 32/52*1/32 + .... + 19/52*1/19 = 1/52 + .... + 1/52 = 11/52$.


Here's another thought. Total ways to draw 52 cards = 52!

Number of ways to draw 52 cards so that none of the first 19 cards were the Ace of Spades:A= 51*50*.......34*33*33*32*...*1 = 33*51!.

Number of ways to draw 52 cards so that none of the first 30 cards were the Ace of Sades:B = 51*50*..........23*22*22*...*1 = 22*\51!.

Number of ways to draw 52 cards so that the the ace of spades is in the 20 -30th: A - B = (33 -22)51! = 11* 51!

So Prob = 11*51!/52! = 11/52.


It really does work out but you are going to use conditional probability you must do it properly.


I think it can be a lot simpler.

The probability that it is drawn in the set of cards 20th to 30th is just 11/52.

Just divide your set in three blocks: cards from 1th 19th, 20th to 30th and 31th to 52th. Probability of the ace of spades to be in every block is 19/52, 11/52 and 22/52 respectively.


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