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Probability Of Poker Hands

  • Thread starter Bashyboy
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Homework Statement


In five-card poker, a straight consists of five cards with adja-cent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that
aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight?
What is the probability that it will be a straight flush (all cards in the same suit)?


Homework Equations





The Attempt at a Solution


First, I calculated the number of possible 5-card hands that can be dealt out: [itex]{{52}\choose{5}}=2598960[/itex]. To answer the first question, I imagined how the cards could be dealt out to generate a straight, since order doesn't matter. I know that, to make a straight, Jacks , Queens, Kings, Aces, 1s, 2s ,3s 4s, and 5s are out of the question. So, if I was dealt a 10, there would be 4 choices (two black and two red); and since either a red or a black will be chosen, then there are only two possibilities for the 9. If the 10 happened to be hearts or diamonds, then the the 9 would have to be a spades or clubs. Using this reasoning for the rest of them, I calculated that there would [itex]4 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 64 [/itex] different straight hands with 10 as the highest card. Thus, the probability would be [itex]\frac{64}{2598960}=.000024625[/itex]. However, the answer is .000394. What did I do wrong?
 

Answers and Replies

  • #2
D H
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Emphasis mine:
So, if I was dealt a 10, there would be 4 choices (two black and two red); and since either a red or a black will be chosen, then there are only two possibilities for the 9.
What makes you think this? There's nothing that says that a straight has to have alternating red and black cards. 6♣, 7♠, 8♣, 9♣, 10♣ -- that's a straight with the highest card a ten. The only tricky issue is a hand such as 6♣, 7♣, 8♣, 9♣, 10♣ -- that's a straight flush. Is this also a straight? In the real game, it isn't. It's much, much better. However, you can also look at a straight flush as just being a special kind of straight, and this makes the math a bit easier.
 
  • #3
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Yes, I just realized I read the problem incorrectly, and so I didn't properly know what a straight was. 6♣, 7♣, 8♣, 9♣, 10♣ would only be a straight flush. Wouldn't there be 4 different straight flushes that could be formed with 6,7,8,9 and 10? And so if I calculated 4^5, I would have to subtract out 4? Because 4^5 would be counting both the number of straights and straight flushes?
 
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  • #4
D H
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Do that and you'll come up with a value that is slightly smaller than the given "correct" answer of 0.000394. It appears that the authors consider a straight flush to be a straight.
 
  • #5
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Oh, I see. Thank you!
 
  • #6
haruspex
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Do that and you'll come up with a value that is slightly smaller than the given "correct" answer of 0.000394. It appears that the authors consider a straight flush to be a straight.
To be fair, the statement of the problem defines straights to include straight flushes.
 

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