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In the game of poker, five cards from a standard deck of 52 cards are dealt to each player.

Assume there are four players and the cards are dealt five at a time around the table until all four players have received five cards.

a. What is the probability of the first player receiving a royal flush (the ace, king, queen, jack, and 10 of the same suit).

b. What is the probability of the second player receiving a royal flush?

c. If the cards are dealt one at a time to each player in turn, what is the probability of the first player receiving a royal flush?

I know that the probability of a royal flush is $\dfrac{1}{649740}$, because of ${_{52}}C_5=2598960$, and $\dfrac{4}{2598960} = \dfrac{1}{649740}$.

But I am struggling to understand how to determine the probability of the first player getting it, etc...

Any help would be greatly appreciated.

SlipEternal
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Krish Ahluwalia
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    For (a), I understand. For (b), are you looking for the probability that both the first player and the second player both get royal flushes? Or are you just looking for the probability that the second player gets a royal flush and the first player can get anything? Because it does not make a difference what order the players are in. Every player has the same chances to get a royal flush. The chances are much lower that more that one person gets a royal flush, though. – SlipEternal Feb 04 '20 at 20:37
  • For b, I am asking that just the second player gets a royal flush and the first player gets anything – Krish Ahluwalia Feb 04 '20 at 20:39
  • It further does not matter in the slightest the method in which the cards are dealt so long as it is a fair way of dealing the cards with no bias. (Of course, if you deal by very explicitly searching for a card you like and giving it to yourself, then giving a bad card to someone else, then searching for another card you like and giving to yourself, etc... then that throws the probabilities off). Giving the cards one at a time, or giving the cards five at a time, or however else... the probabilities will all remain the same. – JMoravitz Feb 04 '20 at 20:40
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    If you have a difficult time grasping that concept, then I suggest you read this related question of If you draw two cards, what is the probability that the second card is a queen?. The only difference really here is that rather than talking about a single queen being the second card, we are talking instead about a straight flush being the second hand. The same logic applies however to explain why the probabilities are the same. – JMoravitz Feb 04 '20 at 20:44
  • Here is some extra information: Assume there are four players and the cards are dealt five at a time around the table until all four players have received five cards. – Krish Ahluwalia Feb 04 '20 at 20:46
  • Totally irrelevant. The answer whether there was a single player, four players, or even ten players, whether the cards were dealt one a time, five at a time, three cards here and two cards there followed by two cards here and three cards there, or any other dealing pattern, whether we were talking about the first player or if we were talking about a different player... they will still be the same $\frac{4}{\binom{52}{5}}$ – JMoravitz Feb 04 '20 at 20:48
  • Ok so, do both the first and second player have the same odds of getting a royal flush? And what about part c? – Krish Ahluwalia Feb 04 '20 at 20:50
  • Read my previous comments again... slower this time. – JMoravitz Feb 04 '20 at 20:51
  • Ok, so you said that regardless of the fashion of which the cards are dealt, the probabilities remain the same. Furthermore, could you explain the matrix (I assume) that you included? – Krish Ahluwalia Feb 04 '20 at 20:57
  • $$\dbinom{52}{5} = {_{52}}C_5 = \dfrac{52!}{5!47!} = 2,598,960$$ They are just different notations for the same number. – SlipEternal Feb 04 '20 at 20:58
  • Ok, so the odds of a) and b) are 1/649740, then what about c? – Krish Ahluwalia Feb 04 '20 at 21:00
  • @KrishAhluwalia for (c) will the first player wind up with five cards out of the original 52? If so, then it does not matter how the cards are dealt. The probability is exactly the same. Note: If the first player will only receive one card, then there is a 0% chance of winding up with a royal flush that contains five cards. – SlipEternal Feb 04 '20 at 21:01
  • I understand, but to be honest, I don't think my professor would assign a problem that has the same answer for everything, y'know? – Krish Ahluwalia Feb 04 '20 at 21:03
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    That is precisely why the professor would assign this problem. It is precisely because while not immediately apparent, every answer is the same. That is what makes this such as interesting question. – SlipEternal Feb 04 '20 at 21:04

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