Am I calculating these odds right?

[Key West]

I was in a hold'em game a few minutes ago, and I flopped a flush with my 87 diamonds, and another player flopped a flush with his K3 diamonds. After he finished separating me from my chips, I realized I can't remember the last time I saw two people flop a flush. I've seen flush vs. flush by the river many times, of course, but not often on the flop, if ever.

So I started thinking about the odds of that happening, and it occurred to me that it's basically the same odds as dealing out 7 diamonds in a row, right? We don't know what any of the down cards were, so for all intents and purposes, it's as if they're dealt consecutively.

Therefore, the odds would be:

(13/52)*(12/51)*(11/50)*(10/49)*(9/48)*(8/47)*(7/46)

or

(0.25)*(0.23529)*(0.22000)*(0.20408)*(0.18750)*(0. 17021)*(0.15217)

or

0.00001

or

1 in 100,000.

Do I have that right or am I smoking crack? If I'm smoking crack, what are the true odds against and how do you calculate them?

And then the secondary question is, since I read him for a flush based on a pretty accurate tell, how confident to I have to be in my read versus the possibility of a 100,000-to-1 event occurring?

[RiverDood]

Hmmn.

Your calculation is correct only if you're playing heads up. But my guess is that this was a 6, 9 or 10-seated table. Then it's not such a total longshot. You should be calculating the odds that any two players have made flushes, not two particular opponents.

As a rough rule of thumb, knock off somewhere between one and two zeroes. It's still a very rare occurrence, but not quite as astonishing.

[mtrubo]

knock off the first (13/52) part of the calculation. Having it means that your calculating for the odds of a particular suit instead of a flush of any suit.

[Key West]

I think we can all agree that it doesn't matter how many players started out, what matters is that two players get all in on the flop, and that they both have two cards of that same suit, and that the flop is exactly three cards of the same suit. I don't see how that's any different than dealing out 7 consecutive cards of the same suit. We never saw any of the other players' cards because they were all mucked. Saying that the number of mucked players hands changes the odds is like saying that the burn cards somehow change the odds. That's just silliness. A down card is a down card, whether it's in a player's hand or in the deck.

[Key West]

Ah, that makes sense. Then it's actually:

(12/51)*(11/50)*(10/49)*(9/48)*(8/47)*(7/46)=0.00513% or 1/19,490.

My intial thought at the table was that it would be (1/125)*(1/125), each player's individual odds, multiplied. That comes to 1/15,625 and is surprisingly close, considering the range we're dealing with.

But I still think the odds of this event occurring are exactly the same as the odds of dealing out 7 consecutive cards of the same suit, and I'm sure I've calculated that correctly.

[K C]

Let's figure this out. First we know you have 2 diamonds. We need to know what the chances that someone else will have two, and that the next 3 cards will be diamonds where you both flop a flush. This is really what you want to know here for practical purposes since you wouldn't really be contemplating this unless you had the diamonds [img]/images/graemlins/smile.gif[/img]

We're going to take a fairly simple way to calculate this so it's not too drawn out [img]/images/graemlins/smile.gif[/img] We're also going to assume that any opponent with 2 diamonds will see the flop with them.

Your 9 oppontents all have a 0.22 chance of getting a diamond on the first card. There are 9 of them so that's 1.98 opponents or roughly 2 that have a diamond on the first card. Of these two, there is a 10/49 chance the second one will be a diamond as well, or 0.204. Multiplying this by 2 we now have a .408 chance that someone else will have 2 diamonds as well (good to keep in mind when you have a low flush [img]/images/graemlins/smile.gif[/img])

Now, for each of the three cards to hit consecutively on the flop, we have 9/48 * 8/47 * 7/46 = 0.004856. We multiply this by the probability an opponent has it and we get roughly 0.002 or 1 in 2000.

The main diffence here is that we're assuming that you have the two diamonds and don't need to calculate the probability of that into it. Of course if we don't assume this we have 13/52 * 12/51 or ~0.059, and when added in this will serve to reduce the probability to ~0.0001 or 1 in 10,000. So the rule of thumb suggested of subtracting a zero serves pretty true here [img]/images/graemlins/smile.gif[/img]

KC

[AaronBrown]

I agree with RiverDood and K C. If you're playing heads up, two players flopping a flush has the same probability of dealing 7 cards of the same suit. I get your answer, although I'd do the math a little differently. There are C(13,7) = 1,716 ways of getting 7 cards out of 13, times 4 because there are four suits, equals 6,864. There are C(52,7) = 133,784,560 ways of getting 7 cards out of 52. The ratio is 19,491 to 1.

But the number of opponents increases the odds because any pair of them could have the flush. With 10 players, there are 45 distinct pairs, so to a first approximation you have to multiply the chances by 45. If you watched a table of 10 play 433 hands, you would expect this to happen once (however, you might not notice, because a lot of those paired hole cards would have been folded).

At the extreme, if you dealt out three suited flop cards, then dealt 24 hands with the rest of the deck, you would be more likely than not to get at least two flushes. So the chance of this happening is greater than half the probability of dealing three cards of the same suit.

[Key West]

I still don't understand what the other opponents have to do with it. I mean, I think I get what you're saying, except that with your method you'd have to set aside a certain number of suited hands that certain players would play in certain conditions. Mucked cards are equal to undealt cards, because they're face-down and therefore have no distinguishing characteristics. I'm simply talking about the probability of two hands that are suited both catching the flush on the flop. Those chances are unaffected by other players, because the player's own p.o.v. is the only one that matters in the scenario.

[RiverDood]

AaronBrown's right -- but let's try another way of expressing it. Supposing you're playing six handed. All of the following outcomes would qualify as two diamond flushes that both were made on the flop

Player ----Scenarios---

YOU DD DD DD DD DD
Opp1 DD -- -- -- --
Opp2 -- DD -- -- --
Opp3 -- -- DD -- --
Opp4 -- -- -- DD --
Opp5 -- -- -- -- DD

It doesn't matter which of your opponents shows up with the second suited diamond hand. You're astonished and furious, regardless of who's got it.

As you add more and more opponents, the chances of bumping into a second suited hand start increasing significantly.

[Key West]

Whoa boy, did I screw up. I ran a simulation. Heads up, my opponent had a flush when I had a flush 2,609 times in 100,000 hands. But three handed, one or more of my opponents flopped a flush 5,200 times. Five handed, 10,183. And ten-handed, it was a whopping 23,267 times that someone else had a flush.

So, for the six-handed game I was in, there was a 12,913 in 100,000 chance of someone else flopping a flush at the same time I did. That's 13% of the time, or 1 in 8. Not exactly extremely rare.

Okay, I get it now. Basically, each opponent has the same odds of flopping a flush with his hand being suited, and the odds of another opponent having the same two suits as you clearly increase when there are more players in.

Once again, very simple concept but I went about it WAY the wrong way.