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Disagreement with common "probability of royal flush" explanation?
http://www.indepthinfo.com/probability-poker/royal... explains the common thinking on this question.
But...
To get a royal flush, you must first be dealt any one of 20 cards - the 10, J, Q, K, or A in any suit. The probability of that is 20/52.
You now have to be dealt cards in the suit that matches the card you have. So for your second card, there are only 4 out of 51 remaining cards in the deck that will help you. For your third card, there are only 3 out of the remaining 50 cards that will help you. For your fourth card, the probability drops to (2/49), and then 1/48. So the probability should be:
(20/52)(4/51)(3/50)(2/49)(1/48) = 1/649740
Multiply the number of outcomes by 4 and you have 1/2,598,960, which is the common answer, as well as what you would get if you used (5/52) in place of (20/52) when you multiply. But why can't you use 20/52 to calculate P(R.F.), since it doesn't matter what suit the first card is; that one just defines what the others need to be.
That's what I get for drinking and doing math. Thanks.
2 Answers
- nudnik0Lv 61 decade agoFavourite answer
Your reasoning is correct. The probability of a royal flush is 1/649,740. The probability of a royal flush in any given suit is 1/4 of this, or 1/2,598,960. The link you give is also in agreement with this.
- lithiumdeuterideLv 71 decade ago
Probability in a discrete case like this simply means the number of desired outcomes divided by the total number of outcomes.
If we consider a hand of five cards (in any order) to be an outcome, then there are only four outcomes that qualify as a royal flush. TJQKA in spades, diamonds, clubs, or hearts are the four.
To calculate the total number of possible outcomes, we simply compute the combination:
Total = C(52,5)
Total = 2598960
So, the probability of getting a royal flush when drawing five cards from a normal deck is 4 / 2,598,960 = 1 / 649,740
This method seems easier to me than thinking about drawing the cards one at a time and computing the probability of one of the remaining cards being in the deck.
