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Poker is a game that plays with a single deck of cards. However, for the purpose of learning, I want to know the probability of poker hands in multiple decks. To simplify the problem, let's take a popular three-card poker game, and calculate probabilities of a pair and three-of-a-kind ranks for a single deck, and 8 decks.

Quasi answered my question and provided formulas for a pair and a three-of-a-kind probabilities. I then followed Quasi's approach, and calculated 3-card, single deck combinations for other ranks (see below). But when I applied an 8 deck factor to the single deck equations the results doesn't look right. Are the equations below correct for single deck, and what modifications are needed for 8-deck?

royal flush equation: COMBIN(4,1)

straight flush equation: COMBIN(12,1)*COMBIN(4,1) - COMBIN(4,1)

straight equation: COMBIN(12,1) * (4 ^ 3 - 4)

flush equation : COMBIN(4,1)*(COMBIN(13,3) - 12)

user469843
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  • You might just compute the probabilities assuming you draw with replacement, which corresponds to infinite decks. The only change comes because you deplete the useful cards, which is a larger effect with fewer decks. – Ross Millikan Aug 06 '17 at 00:26

1 Answers1

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Assume $k$ standard decks.

Probability of a pair: $$\large{\frac { \binom{13}{1}\binom{4k}{2}\binom{48k}{1} } { \binom{52k}{3} }} $$ Explanation:

  • Choose the rank for the pair: $\binom{13}{1}$ choices.
  • Choose the two cards for that rank: $\binom{4k}{2}$ choices.
  • Choose the non-pair card: $\binom{48k}{1}$ choices.

Probability of three-of-a-kind: $$\large{\frac { \binom{13}{1}\binom{4k}{3} } { \binom{52k}{3} }} $$ Explanation:

  • Choose the rank for the triple: $\binom{13}{1}$ choices.
  • Choose the cards for that rank: $\binom{4k}{3}$ choices.

Update:

Answering the additional questions in the OP's edit . . .

Probability of a royal flush: $$\large{\frac {\binom{4}{1}\binom{k}{1}^3} {\binom{52k}{3}} } $$ Explanation:

  • Choose the suit: $\binom{4}{1}$ choices.
  • Choose the cards for that suit: $\binom{k}{1}^3$ choices.

Probability of a straight flush (but not royal): $$\large{\frac {\binom{4}{1}\binom{11}{1}\binom{k}{1}^3} {\binom{52k}{3}} } $$ Explanation:

  • Choose the suit: $\binom{4}{1}$ choices.
  • Choose the rank for the high card: $\binom{11}{1}$ choices.
  • Choose $3$ cards, one for each rank: $\binom{k}{1}^3$ choices.

Probability of a straight (but not a flush): $$\large { \frac {\binom{12}{1}\left(\binom{4k}{1}^3-\binom{4}{1}\binom{k}{1}^3\right)} {\binom{52k}{3}} } $$ Explanation:

  • Choose the rank for the high card: $\binom{12}{1}$ choices.
  • Choose $3$ cards, one for each rank: $\binom{4k}{1}^3$ choices.
  • Subtract the count for the flushes: $\binom{4}{1}\binom{k}{1}^3$ choices.

Probability of a flush (but not a straight): $$\large { \frac {\binom{4}{1}\left(\binom{13k}{3}-\binom{12}{1}\binom{k}{1}^3\right)} {\binom{52k}{3}} } $$ Explanation:

  • Choose the suit: $\binom{4}{1}$ choices.
  • Choose $3$ cards from that suit: $\binom{13k}{3}$ choices.
  • Subtract the count for the straights: $\binom{12}{1}\binom{k}{1}^3$ choices.
quasi
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  • royal flush COMBIN(4,1) straight flush COMBIN(12,1)COMBIN(4,1) - COMBIN(4,1) straight COMBIN(12,1) (4 ^ 3 - 4) flush COMBIN(4,1)*(COMBIN(13,3) - 12) I calculated the 3-card, single deck combinations for other ranks are shown above. But when I applied an 8 deck factor to the single deck equations the results doesn't look right. Are my equations correct for single-deck, and what modifications are needed for 8-deck? – user469843 Aug 09 '17 at 22:10