Odds on 2 people hitting a flush.

Trampas

Can some tell the odds of this happening

Axs vs Kxs

Flopped is a flush and no more suited cards come down.

You would feel very strong if you have the Kxs

Crumbs

(9/48)*(8/47)*(7/46)*(39/45)*(38/44) = 0.003635

Approximately one-third of one percent.

karlh

my brain is sore.

can you explain that please Crumbs?

Dave

That's head's up though. What are the odd's at a 9 handed table.

Crumbs

Bah, I just do the maths, I'm not good at the whole explaining part.

We have two suited cards and we're assuming that someone else also has two of that suit. That leaves 9 of the suit elsewhere in the deck, of which there are 48 unknown cards.

So, to flop a flush, the probability is (9/48)*(8/47)*(7/46).

Then we want the turn to be some other suit. There are still 6 of our suit amongst the 45 unknowns, so, not our suit = 39/45.

Similarily for the river, 38/44.

Smack them all together, and hey presto.

It's not really a relevant calculation in practice.

karlh

would you then have to factor in the odds of the players specifically have Ax and Kx?

i'm just thinking its a lot more unlikely than 0.003% considering flopping it with Ax and Kx in two players hands. i haven't go the maths to back it up tho

Imposter

It's the odds of you having Axs which is:
1/13 * 12/51
times the chance of 1 in 9 (assuming a 10-handed table) of the other players having Kxs:
9 * 1/50 * 10/49 - I think this is right. Do I have to consider the other players?
and then finally times the chance of a flush on the flop:
9/48 * 8/47 * 7/46 - last 2 cards are irrelevant.

Giving:
.018 * .037 * .005 = .0000033
which in % terms is .00033%

Dave

I'm not even going to try and calculate, but do you not have to factor into account the odds of nothing hitting aswell?

Imposter

Just reread the first origianal post. Take crumbs' valus as the last to be multiplied.
So that's .018 * .037 * 0.0036 = .00000239
So: .00239% approx.

Dave wrote:

I'm not even going to try and calculate, but do you not have to factor into account the odds of nothing hitting aswell?

What do you mean by nothing hitting?

Dave

As in if a rainbow flop comes or if only 1 or 2 cards of the suit come on the flop?

bmc

Imposter wrote:

What do you mean by nothing hitting?

Yeah??

Flush draws always hit, don't they?

Hitman Actual

Dave wrote:

As in if a rainbow flop comes or if only 1 or 2 cards of the suit come on the flop?

Why do you need to work that out? The question is about flopping the flush, not missing. Or am I missing something?

Dave

I could be totally wrong, but just thinking back to Leaving Cert probability, but for some reason I thought you might have to take the odds of it not happening i.e. [1 - (odds of it not happening)] + (odds of it happening), now if someone could put me in my place.

Hitman Actual

Hmm... the probabilty of missing is equal to 1-probability of hitting, but I dont think you need to use that here. All you need is the probability of hitting.

Imposter

One other correction to my calculation for the prob of having Ax given that we know opponent has Kx. So the whole thing looks like:

It's the odds of you having Axs which is:
1/13 * 11/51 - you know the other person is going to have the K!
times the chance of 1 in 9 (assuming a 10-handed table) of the other players having Kxs:
9 * 1/50 * 10/49
and then finally times the chance of a flush on the flop:
9/48 * 8/47 * 7/46 * 39/45 * 38/44 - last 2 cards are irrelevant.

Giving:
.0166 * .0367 * .00364 = .000002217
which in % terms is .00022% approx

roryc

It happened to me in the fitz freeroll last thursday. I got a flopped flush beaten by a higher one

Crumbs

karlh wrote:

would you then have to factor in the odds of the players specifically have Ax and Kx?

Yeh, if you like.
I took the original question as taking the Axs and Kxs hands as a given and then finding the odds from there of flopping a flush, etc.

Imposter wrote:

It's the odds of you having Axs which is:
1/13 * 11/51

We can get the hand in two different orders, so it would be
(1/13 * 11/51) * 2 = 0.03318

Imposter wrote:

times the chance of 1 in 9 (assuming a 10-handed table) of the other players having Kxs:
9 * 1/50 * 10/49 - I think this is right. Do I have to consider the other players?

Again, two different ways to get the hand, so I'd make it
(1/50 * 10/49) * 2 * 9 = 0.07347

A flushy flop with blanks on the turn and river, as before = 0.003635

Final probability = 0.03318 * 0.07347 * 0.003635 = 0.00000886

Which is odds of around 112,851/1.

Maths threads are so much more fun than "my QQ got cracked by KT in a turbo MTT, what did I do wrong, the earth is rigged" threads!

Imposter

I'm pretty sure that the order you get the cards doesn't come into it.

As an example:
Total number of 2 card combinations = 52*51 = 2652
Total number of combinations of Axs = 12 (not A but suited with A)* 4 (suits) = 48
48/2652 = .018

Calculation of Axs the other way (by my theory):
1/13 * 12/51 = .018

So the order isn't important. The odds of one doesn't rely on the odds of the other. So they seem like they are independent events and therefore the order makes no difference.

Hectorjelly

I get a totally different figure. Tell me where Im going wrong.

52 cards in a deck
13 hearts
2h in one hand
2h in one hand
9 hearts left

To flop the flush all three cards have to be hearts. Each event is not independant.

Card 1: 48 cards left, 9 hearts: 18.75%
Card 2: 47 cards left, 8 hearts: 17.02%
Card 3: 46 cards left 7 hearrs: 15.22%

Chance of all three is card 1*card2*card3, which is .37%

Then the chance of the turn/river being a non heart is

Card 4: 45 cards left, 39 non hearts. 86.67%.
Card 5: 44 cards left, 39 non hearts. 88.64%

So the chance of all three events happening is .37% * 86.67% * 88.64%, which is .29%. This is about 1/343.

The reason I looked into it is that 112,851/1 couldnt be right. Would the fact that two cards if your suit are not in the deck really reduce your chance of flopping a flush by that much? Becuase I flop a lot of flushes.

Hitman Actual

Using combs, this is the odds I get for flopping the flush:

48 unknown cards, giving 48c3 flops = (48x47x46)/(3x2x1) = 17296 flops.

9 hearts left, giving 9c3 combinations = 84.

84/17296 = .00485 = 204/1 chance of the flush flop.

Then again, I'm hungover, so that could be wrong.

Edit: Odds of no more hearts is:

45 unknown cards left, giving 45c2 turn+river combs = 990.

39 non-heart cards left, giving 39c2 non-heart combs = 741.

741/990 = 0.74848 = ~3/1 on.

Edit-Edit: These are the same as the results that Crumbs got, so considering that two different methods have been used, they're reliable.

Crumbs

Imposter wrote:

Total number of 2 card combinations = 52*51 = 2652
Total number of combinations of Axs = 12 (not A but suited with A)* 4 (suits) = 48
48/2652 = .018

Total number of 2 card permutations = 52*51 = 2652
Total number of 2 card combinations = (52*51)/2 = 1326
48/1326 = 0.036

I may be on the wrong track here, not totally sure. (I'm still in a post-lunch semi-vegetative state.)

Hectorjelly wrote:

So the chance of all three events happening is .37% * 86.67% * 88.64%, which is .29%. This is about 1/343.

That seems about right. Up in post #2, I worked it out as 0.36%.

The reason I looked into it is that 112,851/1 couldnt be right. Would the fact that two cards if your suit are not in the deck really reduce your chance of flopping a flush by that much? Becuase I flop a lot of flushes.

The 112,851/1 figure came about when you take into account that you specifically hold Axs and one of your opponents holds Kxs, or vice-versa. (Although this result might be a bit off depending on the above stuff about whether order is relevant or not.)

Hitman Actual

Imposter wrote:

As an example:
Total number of 2 card combinations = 52*51 = 2652

Dont forget to divide 2652 by 2! Bloody Mullingar maths.

Hectorjelly

I assumed the question was, assuming that you hold Ax and your opponent holds Kx, what is the chance of flopping a flush. This is actually an interesting question.

I hate to see all this maths wasted, but as crumbs alluded to these questions are pretty irrelevant. The odds of any poker hand happening are close to astronomical, but somehow those flops just keep on coming.

mewso

So we are agreed. It's unlikely.