What are the odds in texas hold em of getting all 5 cards on the table of the same suit?

Question

What are the odds in texas hold em of getting all 5 cards on the table of the same suit?

Answer 1

1:1x 13/51x12/50x11/49x10/48 (if the cards come from a single deck)

1:349.5 for a single deck of cards

Answer 2

Jim is correct.  The correct formula is

(12/51 * 11/50 * 10/49 * 9/48)

which equals 0.001980792

Answer 3

1 in 510

Answer 4

Jim's answer is correct. The shortest way to framing this problem is by using combinations:

Denote:

Event A: the player is dealt two suited cards

Event B: the community cards are of the same suit.

The searched probability is P(A) x P(B).

P(A) = 4C(13,2)/C(52,2) (there are 13 cards for each symbol and 4 symbols; we have 4C(13,2) favorable combinations from a total of C(52,2)).

P(B) = C(11,3)/C(50,3) (for a specific symbol, there are still 11 cards in play  if the player was dealt a suit of that symbol; we have C(11,3) favorable combinations from a total of C(50,3)).

Then P(A and B) = (4C(13,2)/C(52,2))x(C(11,3)/C(50,3) = 0.00198079.

Answer 5

The probability is 0.198%.  You'd expect to see this happen, on average, once for every 505 hands (to the nearest whole number), so the odds are 504 to 1 against.

Mathematically, you can treat this as five consecutive random draws.  That the dealer deals the first three cards in a group, and burns a card before each deal, can be ignored for this purpose.

The first card is certain to be some suit.  Call that suit Wands for convenience.  Now, the remaining deck has 12 Wands in 51 cards, so the probability of getting a second Wand is 12/51.

After that happens, there are 11 Wands in 50 cards, so the probability that the third card is a Wand (given that the first two were) is 11/50 -- and similarly for the fourth and fifth cards (10/49 and 9/48, respectively).

The probability that all of these events will occur is found by multiplying the separate probabilities:

1 x 12/51 x 11/50 x 10/49 x 9/48 = 0.00198079

If the notion of "Wands" bothers you, just because modern decks have no such suit, then you can separately calculate the odds for each real suit.  The chance that the first card is a Spade is 1/4, so the probability of five Spades is:

1/4 x 12/51 x 11/50 x 10/49 x 9/48 = 0.000495

That's also the probability for five of each of the other suits.  To find the probability that some suit will occur five times, add these four equal probabilities, which means multiplying the expression above by 4, which means getting back to the same expression I gave originally.

The answer by sandynlily follows the correct mathematical procedure but gets the fractions slightly wrong.  After you've dealt the first card, there are only 12 cards of that suit left to be dealt.  Therefore, the probability that the second card will be suited is only 12/51, not the 13/51 that sandynlily used, and similarly for the other fractions (each numerator is too high by one).

The answer by Puzzled? is wrong because the table he or she linked to -- part of the page at http://www.pokerall-in.com/pokeroddsstar... -- is giving the probabilities of certain starting hands in five-card draw poker.  The probability of being dealt five suited cards is 1 in 505, but a few of those hands are in sequence as well as suited.  The ones that are in sequence have a higher poker rank.  Accordingly, that table includes them in other lines (straight flush or royal flush).  Here, however, mark.james18 asked simply about all five being of the same suit, so the correct answer must include straight flushes and royal flushes as well as simple flushes.