What is the odd of hitting flush when you have nothing

Question

What is the odd of filling 4 suited cards by river to make a flush? This would be starting from the "flop" in hold 'em, and starting from the fourth card, all of which are suited, in seven card stud.

Answer 1

Your question can be re-written like this:

What are the odds of having 4 cards with the same suit, from the 5 available community cards?

The answer can be found by using basic probabilities and computing probabilities of composed events.

You can have 5 boards in which 4 cards are of the same suit:

1. XXXXY
2. XXXYX
3. XXYXX
4. XYXXX
5. YXXXX

Here, X means the suit you're asking for, Y means any other suit. Of course, each of the 5 scenarios have the same likelihood of occurrence.

There are 13 cards of a suit. We'll take out the two preflop cards, which means that we have 12 cards out of 50 in the deck (suppose you're drawing for a flush, otherwise you wouldn't have asked this question). So, one of the 5 scenarios described above occurs with the probability of:

(12/50) * (11/49) * (10/48) * (9/47) * (38/46) =0.00177556684098246143971001906776

Because we have 5 scenarios, each having the same probability of occurrence and each one being independent of the other 4 ones, the answer you are seeking for is:

P=P(1 scenario)*5=0.88%,

So, you will have exactly 4 cards of a given suit, on the river, in 0.88% of the cases.

If you consider all community cards coming up hearts, this is 50*49*48*47*46/12*11*10*9*8, which is about 0.03%.

Answer 2

The odds of hitting a five flush, given four suited cards on the "flop" in hold'em are about 35%. They are 47% starting from four cards, all to a flush, in seven card stud.

Your chances of eventually making the flush go down with each new card that does not "hit" the flush. If it "hits," you chances go up to 100%.

Actually four to a flush (or straight) is GOOD "nothing." It's actually rarer than a pair, and is "nothing" just by the current rules of the game. In the early days of poker (19th century and pre hold 'em), a four-flush or straight actually beat one pair, with a four flush beating a four straight (reflecting their relative "rarity").

Answer 3

The first answer above is wrong as it ignores the odds of the fifth card not being of the same suit.

To get the odds of getting four cards of a particular suit and one of another suit, use the hypergeometric distribution with 4, 5, 12 and 50 as parameters. The answer is about 0.88%. We assumed your two cards are off-suit so you can double this to get about 1.76%.

Note this doesn't include the possibility of all five community cards coming up the same (specific) suit. This is about 0.0374%

Answer 4

If I understood your question correctly, you have 4 cards of the same suit at the flop and you want to know what are the odds of hitting the flush by the river.

Here is the scenario: out of the 52 total cards in the deck, 5 have already been drawn (3 community cards + 2 cards in your hand). We won't take into account other players' hands because we have no information whatsoever about what they're holding. This leaves us with 52 - 5 = 47 potential cards, of which 13 - 4 = 9 of the suit we're looking for.

You want to (1) draw at least one card of the suit of your choice; this is complementary to asking (2) not to draw any card of that suit. If P(x) are the odds that x will happen, we have P(1) = 1 - P(2). We are doing this because calculating P(2) and then deducing P(1) is easier than going straight for P(1).

Here is how to calculate P(2): there are 47 possible cards to be extracted at the turn and 47 - 9 = 38 of them are the ones we're looking for (remember: we're asking not to draw that particular suit). So we have 38/47 for the turn.

Assuming we did not hit that suit at the turn, there are now 46 cards left in the deck and 37 of them are good for us; this time we have 37/46. Now you just have to multiply them: (38/47) * (37/46) = 0,650323774283071. Remember, this is P(2), i.e. the opposite of what you asked! To obtain P(1) you just have to subtract P(2) from 1: P(1) = 1 - P(2) = 1 - 0,650323774283071 = 0,349676225716929 or about 34,97%.

In the seven card stud case the procedure is exactly the same; your final odds in this case are 1 - ((39/48) * (38/47) * (37/46)) = 0,471611933395005 or about 47,16%.

Answer 5

using combination
will use NcK

first just random flush
13c5*39c2 is a 5 card flush - the 39 is the other cards (52 - 13) to fill in the other two cards of the 7 also need to include 6 and 7 flush
4c1 ( 13c5*39c2 + 13c6*39c1 + 13c7) / 52c5 = 3.06% = 31.7 : 1
that includes straight-flush and is the same calc as wiki
for exactly a 5 card flush = 2.85% = 34.1 : 1

so now to the question - you do not hold suited cards - get 4 or 5 to match
2c1 ( 12c4*38c1 + 12c5 ) / 50c5 = 1.85% = 53 : 1
a random hand has a 65% better chance as a random hand can be suited
any higher suited card would have a higher flush
you should only be playing flush draws with connectors or Ax