What is the probability that a five-card poker hand contains a straight flush, that is, five cards of the same suit of consecutive kinds?

My Approach /Attempt -:

Number of Straight Flush Possible=$\binom{10}{1}*\binom{4}{1}*\binom{13}{5}$

$\binom{10}{1} \Rightarrow$ a straight can start with any of the $\left(A,2,3,4,5,6,7,8,9 ,10\right)$

$\binom{4}{1} \Rightarrow$ pick one of the 4 suits

$\binom{13}{5} \Rightarrow$ select 5 cards from the selected suit containing 13 cards.

P(Straight Flush)=$\left ( \binom{10}{1}*\binom{4}{1}*\binom{13}{5} \right )/\binom{52}{5}$

Am i correct?

  • $\begingroup$ Assume only low Straight allowed,i mean only (A,2,3,4,5,6,7,8,9,10) allowed ,No (10,9,8,7,6,5,4,3,2,A) allowed neither (10,J,Q,K,A) $\endgroup$ – virat Sep 28 '16 at 6:32
  • $\begingroup$ Please choose informative and precise titles instead of vague ones. I showed you how to do so in your previous question. $\endgroup$ – Did Sep 28 '16 at 13:25

Two errors:

  • The ace is also the 1, so there's only 9 start cards;
  • Once you've picked your start card, you have no choice whatsoever in what the remaining cards are: if your straight starts with a 3, it must go 3 4 5 6 7.
  • $\begingroup$ There are 10 start cards because the ace can be high or low: A,2,3,4,5 is a straight but so is 10,J,Q,K,A. So there are 10*4=40 possible straight flushes. $\endgroup$ – bof Sep 28 '16 at 6:27
  • $\begingroup$ This is assuming we count the royal flush -- 10 J Q K A all of the same suit -- separately. If we don't, then there are once again 10 start cards. $\endgroup$ – Dan Uznanski Sep 28 '16 at 6:27
  • $\begingroup$ actually 1 will not be in the set ! edited! assume that it is compulsary low ..only (A,2,3,4,5,6,7,8,9) allowed! $\endgroup$ – virat Sep 28 '16 at 6:29
  • $\begingroup$ You mean 9 start cards if we define straight flush to exclude royal flush. $\endgroup$ – bof Sep 28 '16 at 6:29
  • $\begingroup$ 10 cards if we don't count the royal flush as separate, and 9 if we do. $\endgroup$ – Dan Uznanski Sep 28 '16 at 6:30

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