straight flush odds

What are the odds of straigh flush when u have 56 diamons in your hand

I think the math works out the following way
There are 4 cases

56 789
56 234
56 478
56 347

Each of the case has the following odds

For 7 to hit it is 1/50

For 8 to hit it is 1/49
For 9 to hit it is 1/48

There are 5! Combinations for 5 cards on the board (5! = 1*2*3*4*5) = 120
And there are 4 cases (above)

So we have
(1/50 * 1/49*1/48 )* 120 * 4 = 0.4% = 1/245

Meaning 1 out of 245 hand will hit straight flush ?

Not that, because some of the time, opponents will hold the 7,8,9 of diamonds as well, meaning that it is impossible for you to hit your straight flush.

I don't know how to work it out, but I know that it is higher than 1 in 245 for a straight flush.

How did you come up with this equation? I don't understand how it is less likely for a 9 or 8 to come up than a 7 since each deck should have the same number of cards in each suit?

because the population changes (decreases) every time you draw.
lets assume there are 10 disctint cards and you draw one card
let' assume you are looking for A of spades
the first time the odds are 1/10 - > you draw the card and put it on the table
the secend time ..well u have 9 cards left in the deck ..so it is 1/9
let's assume you drew 8 times and A did not come up ..you are left with 2 cards ...what are the odds now .. 1/10 or 1/2 ?
that's why the odds change every time you draw a card (unless you return the card to the deck, which you shuffle)

cha4zz said:

Not that, because some of the time, opponents will hold the 7,8,9 of diamonds as well, meaning that it is impossible for you to hit your straight flush.

I don't know how to work it out, but I know that it is higher than 1 in 245 for a straight flush.

Click to expand...

pls note that what opponents hold in their hands and what is left in the deck is unknown. If you put your opponent on 78 and you pretty sure ur opponent got it .. then yes .. u should adjust the odds .. otherwise the only cards that are known to u is ur hand

I don't play by the numbers as far as a flush is concerned...all I know is if you have no higher than a K of any suite, you must assume that one opponent may be holding the A same suite & is waiting to go all-in against you! I had nearly 35,000 chips not long ago on Full Tilt And was confident that my K high flush would win...went all-in and LOST to an A high flush! That's what I would call a 'bad-beat'! Lesson learned!

...i would assume that's what's what u call an online poker when quads beat full house ...every 10th hand (to make it more entertaining)...or in tournaments, especially on pokerstars.com .. once you are a chip leader you can go all in with any junk and bust other players that had less money ... ever wondered why e.g. hubble free roll lasts 2hrs +-15 minutes ....? that's exactly why ...it is all rigged ..those funky odds do not happen in real world ..

chris_the_terrible said:

Meaning 1 out of 245 hand will hit straight flush ?

Click to expand...

That doesn't sound right.

I don't have time to do the calculations, but here's how you figure it out.

https://www.cardschat.com/forum/cash-games-11/flush-draw-odds-71038/post-424877.html

That doesn't sound right.

I don't have time to do the calculations, but here's how you figure it out.

https://www.cardschat.com/forum/cash-games-11/flush-draw-odds-71038/post-424877.html

that's what my friend says too ...and I agree the odds are kinda high ...but as the unabomber says .. if u have 56 of diamonds .. u have 40% of straight flush !!!! anyways .. the webpage u mentioned explains the basic of how to calculate odds .. plus it has one mistake too

(...)
P(1st spade)*P(2nd spade)*P(3rd spade)
= (11/50)*(10/49)*(9/48)
= 33/3920 = %0.84
(...)

this assumes we care about the order of the spades

in poker it does not matter what's the exact order of the cards hence the calculations needs to be multiplied by 3! = 1*2*3 = 6

(11/50*10/49*9/48) * 3!

3! takes into considerations all different combinations you can arrange 3 cards ...e.g.
ABCD can be arranged 4! ways ...1*2*3*4 = 24
and we do not want to double count those cases

I suck at math.

Obviously your odds will be better with 6/7s than they would be with A/Ks.

According to here, the odds of flopping a straight flush with two suited connectors is 4,899:1. So getting it by the river will be substantially less than that.

And the flop gives you 60% of your community cards.

I can't seem to find someone posting the odds.

(Getting quads by the river with a PP is 1 in 122 but that does us no good)


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This guy really is an absolute wizard: askthewizard

He'd probably say something like... the probabilty of hitting exactly the 7,8,9 of diamonds with 5 cards to draw equals:

(47 choose 2)*(3*2*1) / (50 choose 5) = 6486/2118760 = .31%

So with 4 different ways to hit it gives 4*.31 = 1.2% or 82-to-1.

Just a guess. -Ben

I got it.
the mistake I made was the following...
I assumed that I need to mix up the 5 cards (5!) ..but I never included them in the calculation...so knowing that per
Chances of Flopping...
http://www.pokerchipsupplystore.com/Texas_Hold_Em_Odds-c1353-wp5957.htm

Straight Flush holding suited connectors 4,899-1

this is how you should count your odds
1/50 * 1/49 * 1/48 * 4 * 3! = 0.02% or 1 to 4900


so 3! not 5! should be there ...3! = 1*2*3 = 6 - > this is how many ways u can mix the 3 cards on the board..the other 2 cards are irrelevant .. I do not include them in the calculation (1/50 * 1/49 * 1/48) so ...I do not need to mix them up

4 is for the 4 cases (that I described above)
the outcome is 0.02% or 1 to 4900
it all make sense now

And lets not forget that in most cases you'll be folding that straight flush as its less likely your going to call with 5/6 suited on a 9 table when people raised before the flop.

Unless your a loose player and willing to take more riscs.

chris_the_terrible said:

this is how you should count your odds
1/50 * 1/49 * 1/48 * 4 * 3! = 0.02% or 1 to 4900

Click to expand...

Ah, those are the odds to FLOP it. I did the odds to hit it by last card. (82-to-1) I wonder if that's right...

>He'd probably say something like... the probabilty of hitting exactly the >7,8,9 of diamonds with 5 cards to draw equals:

>(47 choose 2)*(3*2*1) / (50 choose 5) = 6486/2118760 = .31%

>So with 4 different ways to hit it gives 4*.31 = 1.2% or 82-to-1.

that's much too high .. if we know that odds of flapping sf are 1:4900 then odds of having the sf by the river would be around 1:1500..maybe 1:1000...but not 1:82...

what do u think of :

4* [(1,1)*(1,1)*(1,1)]*[(47,2)] / (50,5) = 0.2% = 1: 490

where e.g. (47,2) is newton binomial not permutation (which counts the order in which cards are dealt)

if you apply (50,5) and get 2118760 it means you do not take order into consideration ...well then ..let;s forget about 3! at the top as well ..to be consistent

Yer right! I might have a newton binomial nightmare tonight. It's really 3 choose 3 and that works out nicely:

4*(47 choose 2)*(3 choose 3)/(50 choose 5) = .002

p.s I strive for maximum cocktails and minimal math at the table.