View Full Version : Flopped flush

08-15-2002, 07:41 PM
In Texas Hold'em, what is the probability that one of your opponents has flopped a flush assuming that you do not have one of the suited cards on the flop?

With one opponent seeing the flop?

With two?

With 3?

With 4?


08-15-2002, 07:47 PM
This question doesn't answer your specific question but should help.

When you hold two suited cards, the odds of you flopping a flush are 118:1.

08-16-2002, 04:04 AM
Actually, the way you've worded the problem is a little ambiguous and needs more information. The important information here is not just how many players SEE the flop, but also how many players there are at the table to begin with. The reason for this is that the odds are affected (in a sense...) by the identity of the mucked cards of the players no longer in the hand.

To make the point very clear, suppose you were playing 3-handed, and one player folded. There's a certain probability your opponent has flopped a flush, given the board is a 3-flush. (I'm assuming that you're assuming the flop is given, otherwise, the problem doesn't make sense. So, you're basically asking for a conditional probability.) Now, however, suppose you have a table with 20 players, and 18 of them fold. You're still heads-up, but clearly the odds are greater now, because it's more likely that someone with high (or not so high) suited cards stayed around to see the flop.

The information I would need to give you an accurate answer is:

1. How many players at the table.

2. How many players see the flop.

3. The probability that a player will stay around to see the flop, given that the player has suited cards.

4. The probability that a player will stay around to see the flop, given that the player has unsuited cards.

Obviously, 3 and 4 will require some thought for input. BUT, the more accurate your rough estimates for 3 and 4 are, the more accurate the final calculation will be.

This is a very interesting problem. Let me think about it, using variables, and I should be able to come up with a general formula that gives the probability as a function of 4 variables.

08-16-2002, 07:32 AM
Assume a 10-handed game.

I would like to know the odds for different numbers of players. So what are the odds my opponent flops a flush if I am only against one player. What if I am against 2 players? 3 players? Etcetera. I would think that once you get to 5 or 6 opponents seeing the flop that the odds are pretty good that one of the opponents has a flush.

I am not too sure I understand 3 and 4. I think it is best to assume that the opponents could have any two random cards.

Thanks and good luck!

08-16-2002, 02:25 PM
To briefly try to explain why 3 and 4 are important, one way to put it is this. The cards in the hands of those who see the flop are NOT random. Their identity is influenced by the fact that they stayed around to see the flop. It's kind of hard to see why this matters at first, because it looks like there's no influence. But just because you can't see the influence, doesn't mean it's not there.

Let me see if I can give another example to show why 3 and 4 are important. Suppose you're sitting at a table, call it Table 1, where, for whatever bizarre reason, if one of the players holds suited cards, there is a 99% chance they stay to see the flop, while if they have unsuited cards, only a 1% chance.

Now, consider another table, Table 2. At Table 2, another bizarre phenomenon happens. Here, it's just reversed. If someone holds UNsuited cards, there's a 99% chance they stay to see the flop. If they hold suited cards, for some reason, there's only a 1% chance they stay to see the flop.

Now, imagine that both tables are dealt a hand simultaneously, and 5 players stay to see the flop. A 3-flush flops. At which table do you think it's more likely that someone has flopped a flush? It's no contest.

(Mathematically, what's going on here is just Bayes' Theorem. Sklansky mentions this result in some of his books, I think. It's a theorem that relates conditional probabilities.)

08-17-2002, 01:19 AM
Yes, I understand what you are saying. I was just trying to make the problem a little easier for you /images/smile.gif.

But now that I think about it, the problem is that many players play pairs. At first I thought suited hands would be like 75%, but since many players play pairs it probably drops down. So I think the best assumption is probably 50% of the time a caller on the flop would have suited cards.

If you would like to make another assumption that is fine.

I unfortunately am going out of town tomorrow on vacation and will be gone a week. If I can find an internet on the beaches of Hawaii I'll check the postings, /images/smile.gif. Otherwise I'll check back in a week!


08-17-2002, 01:41 AM
OK, after I posted above I thought about it a little more. To be more scientific, I counted the total number of hands in Groups 1 through 7 according to Sklansky's hand rankings. There are 70 hands of which 44 are suited so let's go with 63%.

So the assumption is that each caller has a 63% probability that he holds suited cards.

08-20-2002, 08:23 PM
Assuming a completely random holding (say 1 opponent in big blind getting free flop) 3 of the suit flopped, you don't have any of the suit, so 47 cards left with 10 of the suit. Percentage of the one person having two of the suit would be (10/47)*(9/46) or about 4.16% which is about 23:1 against. Using the 4.16% against multiple random hands (less realistic) calculate the odds that noone has flopped it and then subtract that from 1. So, for two people use 1-(.9584)^2 = 8.15 % or about 11.3:1 against,

for 3 people 1-(.9584)^3 = 11.98% or 7.35:1 against and for 4 people about 15.6% or 5.4:1 against. This is against completely random hands, so in real play you would probably want to adjust the odds down according to the texture of the game.