I have the following question:

"We draw one card at a time without replacement from the top of a shuffled standard poker deck and stop when we draw an ace. Let $X$ be the number of cards we have drawn, then calculate the probabilities for $X=10$, $X=50$, and $X < 10$."

I have a theory as to how to solve the first one: calculate the probability of the drawing 10 cards WITHOUT getting an ace ($\binom{48}{10} / \binom{52}{10}$) and later subtracting this from 1. For the other two, I am a little bit lost.

Regarding the second one, I think that if I draw 50 cards from the deck, the probability of drawing an ace should be 1, shouldnt it?

And unfortunately I'm lost on the last part. I would love a clue on this matter.


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    $\begingroup$ Since there are four aces, the first ace must appear by the $49$th card. Hence, the probability that the $50$th card is the first ace is $0$. $\endgroup$ – N. F. Taussig Dec 10 '17 at 22:28
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    $\begingroup$ Hint for the last part: compute the probability that there is no ace among the first $10$ and subtract. $\endgroup$ – lulu Dec 10 '17 at 22:29
  • $\begingroup$ @N.F.Taussig maybe its a problem with the phrasing of the question but from what I understand is that they are asking the probability of getting an Ace on the nth draw, rather than asking what is the probability of getting the first Ace (to me they are not the same question). So wouldn't it make sense that for the 50th draw it would be 100% probable that the draw is an Ace? $\endgroup$ – Frank Pinto Dec 10 '17 at 22:39
  • $\begingroup$ My understanding is that we draw cards until we obtain the first ace. Hence, $P(X = 10)$ means that the first ace appears in the $10$th position. $\endgroup$ – N. F. Taussig Dec 10 '17 at 22:41
  • $\begingroup$ @lulu wouldnt that just yield the same result as the first question of the problem? Could you elaborate a bit more on the subject? $\endgroup$ – Frank Pinto Dec 10 '17 at 22:41

$P(X = k)$ is the probability that the first ace appears in the $k$th position.

What is $P(X = 10)$?

For the first ace to appear in the tenth position, we must first choose nine non-aces, then choose an ace with the tenth draw. The probability that the first nine cards are not aces is $$\frac{\dbinom{48}{9}}{\dbinom{52}{9}}$$ The probability that one of the four aces is chosen as the tenth card from the $43$ remaining cards is $$\frac{4}{43}$$ Hence, the desired probability is $$\frac{\dbinom{48}{9}}{\dbinom{52}{9}} \cdot \frac{4}{43}$$

What is $P(X = 50)$?

Since there are four aces in the $52$-card deck, the first ace must appear no later than the $49$th position. Hence, $P(X = 50) = 0$.

What is $P(X < 10)$?

The probability that $P(X < 10)$ is found by subtracting the probability that the first nine cards are not aces from $1$.

$$P(X < 10) = 1 - \frac{\dbinom{48}{9}}{\dbinom{52}{9}}$$

  • $\begingroup$ This seems to be computing the chance that there is exactly one ace among the 10 first cards drawn. But here we want that ace to come last among those 10 cards, so you need to divide the probability by 10. $\endgroup$ – hmakholm left over Monica Dec 10 '17 at 22:51
  • $\begingroup$ @HenningMakholm Have I fixed the error? $\endgroup$ – N. F. Taussig Dec 10 '17 at 23:01
  • $\begingroup$ Yes, I think so. $\endgroup$ – hmakholm left over Monica Dec 10 '17 at 23:07

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