A standard deck of cards is shuffled and dealt. Find the probability that the last king appears on the 48th card?

I would like to verify if my reasoning for the solution is correct:

I attempted to use hypergeometric distribution to solve this problem. The first half of the setup deals with drawing 47 cards of which exactly 3 are Kings and 44 are non-kings. The second half of setup completes the missing conditional probability of drawing the last king on the 48th card which is $\frac {1}{5}$

$\frac {\binom {4}{3}* \binom{48}{44}}{\binom {52}{47}}* \frac {1}{5}$

Is my reasoning correct?

Thank you!


Yes, that is correct.

One way to check yourself would be to start from the other end of the deck. Then you're trying to find the probability that the first king you see appears as the $5$th card from the bottom. This probability is $$ \frac{\binom{48}{4}}{\binom{52}{4}} * \frac{4}{48} $$ (where the first factor is the probability that the bottom four cards are not kings, and the second factor is the conditional probability that the $5$th card from the bottom is a king).

And in fact, these are equal.

  • 2
    $\begingroup$ Or using multinonial coefficients: $\dfrac{\binom 43\binom{48}{44}}{\binom {52}{47}\binom 51}=\dfrac{\binom 4{3,1,0}\binom{48}{44,0,4}}{\binom{52}{47,1,4}}=\dfrac{\binom {48} 4\binom 41}{\binom {52}4\binom{48}1}$ Where the middle formula is the probability that the four kings will be placed three among the first fourty-seven positions, one in the middle one position, and none among the last four positions. $\endgroup$ – Graham Kemp Jul 9 '18 at 0:35
  • $\begingroup$ wow I like these alternative solutions! $\endgroup$ – pino231 Jul 9 '18 at 1:50
  • $\begingroup$ I think the first factor is better interpreted as $48/52 * 47/51 * 46/50 * 45/49$, which is the product of the probabilities the $i$'th card from the bottom is a king, conditional on $i-1,..,1$ not being kings. $\endgroup$ – user217285 Jul 9 '18 at 3:46

Let X be the sort of kings and Y the sort of non-kings.

then the G.F. for the good configurations (with a king in position 48) is

$(X+Y)^{47}.X.Y^4$ The G.F. for all configurations is:

$(X+Y)^{52} $

We now compare the coefficients of $X^4Y^{48} $ in the above expressions - since there are 4 kings and 48 non-kings.

$\frac {good} {all} = \frac {16215}{270725} \approx 0.06 $

By inspecting the G.F. one may see that the conditions in the problem could be altered in many ways; for example one may ask a non-king in the positions 10, 20, 30, and 40 and a king in the position 50, with the very same result.


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