From Carol Ash's The Probability Tutoring Book,

Draw from a deck without replacement. Find the probability that

a) the $10^{th}$ draw is a king and the $11^{th}$ draw is a non-king

b) the first king occurs on the $10^{th}$ draw

c) it takes $10$ draws to get $3$ kings

My attempts:

a) $$P(10^{th}\text{ is king and }11^{th}\text { is non king}) = P(1^{st}\text{ is king and }2^{nd}\text { is non king}) \\ = \frac{4}{52} \frac{48}{51}$$

b) To get the probability that the first king is on the $10^{th}$ we want $9$ non-kings first. Here is where my confusion begins. I first try a combinations approach: Of $48$ non-kings, pick 9 of them and then of $4$ kings pick $1$ of them.

This results in: $$\frac{{48 \choose 9}{4 \choose 1}} {52 \choose 10}$$

Using a different approach: Draw 9 non-kings one at a time, and then draw a king $$\frac{(48)(47)(46)...(40)}{(52)(51)(51)...(44)} \frac{4}{43}$$

I feel as though these two should produce the same numerical answer, yet the first method produces an answer(0.424) that's a factor of 10 away from the latter method(.0424)

c)$$P(10 \text{ draws to get } 3 \text{ Kings}) = P(10^{th} = \text{King}|2 \text{Kings before})P(2 \text{ Kings before}) = \frac{2}{43} \frac{{4 \choose 2}{48 \choose 7}}{52 \choose 10}$$

  • 1
    $\begingroup$ Part b, first approach is wrong since you overcounted ten times. Notice that even if the king is chosen on the 7th card, this scenario is counted in the expression. Everything else is correct $\endgroup$ Jun 30, 2020 at 2:29
  • $\begingroup$ @BenjaminWang But by that logic, shouldn't we need to divide by $10!$? Since we can permute the cards in 10! different orderings $\endgroup$
    – BigBear
    Jun 30, 2020 at 2:32
  • 1
    $\begingroup$ Well, by your logic, we should divide by (10!/9!) because we don't care about the order of the 9 non-king cards. Okay maybe i'll explain this better in an answer later $\endgroup$ Jun 30, 2020 at 2:34
  • 1
    $\begingroup$ Part (a) also seems wrong.. it's without replacement, hence at the 10th draw you have already lost 9 cards, which may or may not be kings $\endgroup$ Jun 30, 2020 at 4:29
  • 1
    $\begingroup$ I agree with Dhanvi, (a) should be considered as $\sum_{i=0}^4$P(i kings left from 10th) $\endgroup$
    – Harry Lew
    Jun 30, 2020 at 4:49

2 Answers 2


a) It is intuitively clear that you answer is correct as there are altogether $52\cdot51$ ways to choose the 10th and 11th cards and $4\cdot 48$ ways to draw a king and a non-king cards. This can be confirmed by the direct count: $$ \frac{\binom{50}{9} 9!\binom{4}{1}\binom{48}{1}}{\binom{52}{11}11!}=\frac{4\cdot 48}{52\cdot51}. $$

b) The probability is: $$ \frac{\binom{48}{9} 9!\binom{4}{1}}{\binom{52}{10}10!}=\frac{\binom{48}{9}}{\binom{52}{9}}\frac4{43}. $$ Of course the r.h.s. can be obtained directly as $\frac4{43}$ is the probability to draw a king card from 43 remaining cards. The missing permutation factors explain the deviation by factor $10$ in your results.

c) It is not quite clear what is here meant. If it is (as you assumed) that the 10th card is a king and there were two king cards before the probability is: $$ \frac{\binom{48}{7}\binom{4}2}{\binom{52}{9}}\frac2{43}, $$ which is except for one important point similar to your answer. The important point is the number $\binom{52}{\color{red}{9}}$ in the denominator. Here you made the same mistake as in the first approach to part b).


I may give my opinion on the part b first. The approach you’ve tried seems right, it is a hypergeometric case. We could consider it as the probability of 9 cards randomly drawn out from 52 cards, but 0 king. According to the hypergeometric distribution:

$P(0)=\left(\frac {{4\choose 0}{48\choose 9}}{{52\choose 9}}\right)$

After that, you can use $P(0)•\left(\frac{4} {48}\right)$ to get the probability that first king be drawn at the 10th.

In terms of part c, the concept of hypergeometric distribution can be applied again. It is nothing more than: $P(3)=\left(\frac {{4\choose 3}{48\choose 7}}{{52\choose 10}}\right)$


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