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I guess the answer is pretty intuitive that its 52 C 4 to most people, but no one has been able to explain to me why it's not 52*51*50*49*48... My reasoning is the first time you choose you to have 52 cards to choose from, then 51, and so on and so on. Can someone please explain it to me?

cupcakelover1016
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  • The answer is $\binom {52}5$. Your method would be correct if the order mattered, but it does not. – lulu Feb 28 '18 at 01:13
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    Unless it's 5-card stud. – ncmathsadist Feb 28 '18 at 01:14
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    Simple example. You have exactly 5 cards in the deck and your hand needs 5 cards. Using your logic you can get a total of 54321 = 120 hands. You need to divide by the permutations 5!. – TyCobb Feb 28 '18 at 01:14
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    Most people would consider $A\spadesuit 2\spadesuit 3\heartsuit 4\clubsuit 5\clubsuit$ to be the same "hand" as $3\heartsuit 2\spadesuit 5\clubsuit A\spadesuit 4\clubsuit$. They both consist of the same cards, just written in a different order. Using $52\cdot 51\cdots 48$ you have overcounted each hand $5!$ times. Also, note it is $\binom{52}{5}$, not $52~C~4$ (the first notation is preferred first off, and the bottom number is a five, not a four) – JMoravitz Feb 28 '18 at 01:15
  • As an aside, although the standard interpretation is that a "hand of five cards" in poker is considered to be a subset of size five of the deck of cards where order doesn't matter, it really doesn't hurt anything to use the interpretation that order is important when it comes to probability calculations. E.g. the number of one-pair hands are $\frac{13\cdot\binom{12}{3}\binom{4}{2}4^3}{\binom{52}{5}}$ using order irrelevant and $\frac{13\cdot\binom{5}{2}\cdot4\cdot 3\cdot 48\cdot 44\cdot 40}{52\cdot 51\cdot 50\cdot 49\cdot 48}$ using order relevant. Both answers are equal despite their looks – JMoravitz Feb 28 '18 at 01:25
  • @JMoravitz Maybe it's because I'm not so sure of myself with counting, and maybe it's not ideal, but many times I prefer the permutations approach. – Ovi Feb 28 '18 at 01:40

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In this way you are overcounting the hands since for example the hand

  • A♦️K♦️Q♠️J♠️10 ♥️

  • K♦️Q♠️J♠️10 ♥️ A♦️

  • ...

is counted 5*4*3*2*1=5! and since this is true fir each hand we need to divide by 5! to obtain $\binom{52}{5}$ hands.

user
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