Poker Hands

by Polya
$$Rank = t_0 + 4t_1 + 6t_2 + 4t_3+ t_4 $$

A deck is the product of 13 ranks

$$Deck = Rank^{13} $$

624 + 3744 + 54912 + 123552 + 1098240 + 1317888	 = 2598960

Choose one : 52
$$One(X) = X \cdot \mathit 3 X $$

$$None(X) = \mathit 4 X $$

$$ Selection (X_1, X_2,...) = \prod_{i=1}^{13} s.One(X_i) + n.None(X_i) $$

The type 13 = 1 + 12 is described by :

$$ e.g.f. =  (s.y\frac{y^3}{3!} + n.\frac{y^4}{4!})^{13} $$

We are interested in the coefficient of $$s.n^{12}.(\frac{y^4}{4!})^{13}$$

Four of a kind 624
$$Four(X) = \mathit 4 X $$

$$One(X) = X \cdot \mathit 3 X $$

$$None(X) = \mathit 4 X $$

$$ Selection (X_1, X_2,...) = \prod_{i=1}^{13} (f. Four(X_i) + s.One(X_i) + n.None(X_i)) $$

Spread : 1317888
The type 13 = 5 + 8 is described by :

$$    \mathit 5 [ \underbrace {X}_{ Y · \mathit 3 (Y)} ] · \mathit 8 [ \underbrace {X}_{\mathit 4(Y)} ] $$

$$ e.g.f. =

\frac 1 {5!} \left( xy\frac{y^3}{5!} \right)^3

· \frac 1 {8!} \left(x\frac{y^4}{4!} \right)^8

= 1317888 \frac 1 {13!} \left(   x { y^4 \over 4!}      \right)^{13}$$

One pair : 1098240
The type 13 = 1 + 3 + 9 type is described by :

$$    [ \underbrace {X}_{\mathit 2(Y)  · \mathit 2(Y)}] · \mathit 3 [ \underbrace {X}_{ Y · \mathit 3 (Y)} ] · \mathit 9 [ \underbrace {X}_{\mathit 4(Y)} ] $$

$$ e.g.f. =

\frac 1 {1!} x \frac {y^2}{2!} \frac{y^2}{2!}

· \frac 1 {3!} \left( xy\frac{y^3}{3!} \right)^3

· \frac 1 {9!} \left(x\frac{y^4}{4!} \right)^{9}

= 1098240 \frac 1 {13!} \left(   x { y^4 \over 4!}      \right)^{13}$$

Two pairs : 123552
The type 13 = 2 + 1 + 10 type is described by :

$$   \mathit 2 [ \underbrace {X}_{\mathit 2(Y)  · \mathit 2(Y)}] · [ \underbrace {X}_{ Y · \mathit 3 (Y)} ] · \mathit {10} [ \underbrace {X}_{\mathit 4(Y)} ] $$

$$ e.g.f. = \frac {1}{2!} \left(x\frac{y^2}{2!}\frac{y^2}{2!} \right)^2 · xy\frac{y^3}{3!} · \frac {1}{10!} \left(x\frac{y^4}{4!} \right)^{10}  = 123552.\frac{x^{13}}{13!} \left( \frac{y^{4}} {4!} \right) ^{13}$$

Three of a kind : 54912
The type 13 = 1 + 2 + 10 type is described by :

$$              \underbrace {X}_{\mathit 3(Y)  · Y} · \mathit 2    [ \underbrace {X}_{ Y · \mathit 3 (Y)} ] · \mathit {10} [ \underbrace {X}_{\mathit 4(Y)} ] $$

$$ e.g.f. = x \frac{y^3}{3!} y · \frac 1 {2!} \left (x y\frac{y^3}{3!} \right )^2 · \frac {1}{10!} \left(x\frac{y^4}{4!} \right)^{10} = 54912.\frac{x^{13}}{13!} \left( \frac{y^{4}} {4!} \right) ^{13}$$

Full house : 3744
The type 13 = 1 + 1 + 11 type is described by :

$$        \underbrace {X}_{\mathit 3(Y)  · Y} · \underbrace {X}_{ \mathit 2 (Y) · \mathit 2 (Y) } · \mathit {11} [ \underbrace {X}_{\mathit 4(Y)} ] $$

$$ e.g.f. = x \frac{y^3}{3!} y · x \frac{y^2}{2!}\frac{y^2}{2!} · \frac {1}{11!} \left(x\frac{y^4}{4!} \right)^{11} = 3744.\frac{x^{13}}{13!} \left( \frac{y^{4}} {4!} \right) ^{13}$$