If you draw 5 cards from a standard deck of 52 cards, then the probability of your hand having the Ace of Spades is: $$\frac{51\choose 4}{52\choose 5} = \frac{51!5!47!}{4!47!52!} = \frac{5}{52}$$ If, however, you choose one card a time until you've drawn 5 cards, the probability of having the Ace of Spades is: $$\frac{1}{52}+\frac{1}{51}+\frac{1}{50}+\frac{1}{49}+\frac{1}{48}=\frac{433507}{4331600}\approx\frac{1}{10}>\frac{5}{52}$$ Why does choosing one card a time increase the probability of finding the Ace of Spades, when the resulting hands are equivalently drawn?

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    $\begingroup$ Hint: instead of drawing 5 cards one at a time, let's say you draw 34. Would you then say your probability of drawing the ace of spades is $1/52 + 1/51 + \ldots + 1/19 \approx 1.04 > 1$? $\endgroup$
    – Hong Ooi
    Jul 4, 2015 at 13:31
  • $\begingroup$ Another hint : What you wanted to do is compute P(X=1)+P(X=2)+..+P(X=5)...with P(x=k) representing the probability of the Ace of Spade drawn exactly at the k th draw.This is one way to find the answer (not the most simple) but one legit way as these are disjunctive cases. The problem comes from how you compute these probabilities. For P(X=2), for example, If you had to answer the question "what is the probability that the ace of space is drawn exactly at the second draw, would you find 1/51 ? $\endgroup$
    – brumar
    Jul 4, 2015 at 13:39
  • $\begingroup$ It's not actually homework, just a genuine question. But I think I know what you're saying: I'm conflating the idea of the geometric distribution and dependent draws. So, it'd be better to say $P(X=1)+P(X=2|X\neq 1)+\dots+P(X=5|X\neq 1,2,3,4)$? $\endgroup$
    – AJS
    Jul 4, 2015 at 14:26
  • $\begingroup$ One thing I can't figure out is, why does it cancel out to $5/52$? $P(X=1) = 1/52$, $P(X=2\mid X\neq 1)=P(X=2\cap X\neq 1)/P(X\neq 1)=((51/52)*(1/51)/(51/52))$, giving me the original (incorrect) sum of fractions? $\endgroup$
    – AJS
    Jul 4, 2015 at 14:47
  • $\begingroup$ Focus on P(X=2∩X≠1) = P(X=2 |X≠1) * P(X≠1). What is that? How does the answer to this relate to probability of getting an ace of spaces for the first time on the 2nd card drawn? $\endgroup$ Jul 4, 2015 at 15:23

1 Answer 1


As @Glen_b suggested, it would be a good idea to summarize the comments part in an answer. I'll do that and also give an alternative for the formula related to the probabilistic point of view at the end of the answer. The apparent contradiction between the two computations came from this line :

the probability of having the Ace of Spades is: $$\frac{1}{52}+\frac{1}{51}+\frac{1}{50}+\frac{1}{49}+\frac{1}{48}=\frac{433507}{4331600}\approx\frac{1}{10}>\frac{5}{52}$$

The idea behind this computation was good but the logic was flawed. If we sum the probability of the Ace of Spade drawn exactly at the k th draw with k going from 1 to 5, we have the probability we want. But $\frac{1}{51}$, for example, does not represent the probability that the ace of spade is exactly drawn at the second attempt but the probability that the ace of space is drawn at the second attempt given that it has not been drawn at the first one.

AJS finally found the right formula

I got it! $1/52+((51/52)∗(1/51))+⋯+((51/52)∗(50/51)∗(49/50)∗(48/49)∗(1/48))=5/52$

With the idea that the probability to exactly draw the ace of spade at the $k$ th trial is the probability to not draw the ace of spade during previous attempts.... $$(51/52)*(50/51)...(52-k+1)/(52-k+2)$$...multiplied by the probability to draw the card at the $k$ th attempt $$1/(52-k+1)$$

A general rule of thumb that I have to avoid this kind of error is to be cautious when it comes to adding probabilities. If you can turn your problem around to avoid additions to the profit of multiplications, this is less prone to error.
A more classic approach giving a shorter path would have been to consider that the probability of having the Ace of Spades after 5 draw is 1-(the probability to not draw it with 5 draws) which gives, as you know : $$1-(51/52)*(50/51)*(49/50)*(48/49)*(47/48)=1-47/52=5/52$$

  • $\begingroup$ Thank you, brumar. I appreciate you taking the time to write it out. $\endgroup$
    – AJS
    Jul 4, 2015 at 23:22

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