Question

In Texas hold 'em poker, players are dealt 2 cards each. Among the undealt cards, three are turned over( called the "flop"), then a fourth card is turned over( called the "turn"), then a fifth card is turned over(called the "river"). The order of the three cards within the flop do not matter. If there are 9 players, then how many possible distributions of cards among the players, flop, turn, and river are possible?

Answer 1

Solution

Back-up Theory

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}…..............................................…(1)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN(Number, Number_chosen) [Number is n, Number_chosen is r]…………………………….........................................................................…………. (1a)

Now to work out the solution,

The first player can be given any 2 out of 52 cards in ⁵²C₂ ways [vide (1)].

Now, there are only 50 cards and hence the second player can be given any 2 out of 50 cards in ⁵⁰C₂ ways [vide (1)].

And so on, the last (9th) player can be given any 2 out of 36 cards in ³⁶C₂ ways [vide (1), also noting that 16 cards have been already dealt to the preceding 8 players].

After completing dealing to 9 players, only 34 cards are remaining out of which three can be the ‘flop’ cards in ³⁴C₃ ways and proceeding identically, ‘turn’ and ‘river’ cards be selected in ³¹C₁ and ³⁰C₁ ways respectively. Thus, total number of possible distributions of cards among the players, flop, turn, and river is:

(⁵²C₂)(⁵⁰C₂)(⁴⁸C₂)(⁴⁶C₂)(⁴⁴C₂)(⁴²C₂)(⁴⁰C₂)(³⁸C₂)(³⁶C₂)( ³⁴C₃)( ³¹C₁)( ³⁰C₁)

= 2.96953E+33 [vide (1a)] Answer

DONE