2845039

the last king appears on the 48th card
https://math.stackexchange.com/questions/2845039/a-standard-deck-of-cards-is-shuffled-and-dealt-find-the-probability-that-the-la A standard deck of cards is shuffled and dealt. Find the probability that the last king appears on the 48th card?

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

then the species for the good/all configurations (with a king in position 48) is

$$\frac{good} {all} = \frac {E_{47}(X+Y)}{E_{47}(X+Y)} \frac {X}{X+Y} \frac {E_4(Y)} {E_4(X+Y)} $$

The e.g.f's are

$$good = \frac {(x+y)^{47}}{47!} x \frac {y^4}{4!}  $$

$$all ={ (x+y)^{47} \over 47!} (x+y) { (x+y)^4 \over 4!} $$

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

$$\frac {good} {all} = \frac {3243}{54145} \approx 0.06 $$

Working with GF :

$$(X+Y)^{47}\cdot X \cdot Y^4$$ The e.g.f. for all configurations is:

$$(X+Y)^{47}\cdot(X+Y)\cdot(X+Y)^4 $$