Holdem - Flop problem

[MarkD]

Given that you have AJo (let’s say AsJc) and the flop comes Kxx, how do I calculate the probability that my opponent(s) has a king? (I realize that this will be weighted by my read, but I don’t want to worry about that atm).

I am looking for the method, not just the result. In fact I care less about the result than I do about the method used to solve the problem.

Case 1: 1 opponent
Case 2: 2 opponents
Case 3: 3 opponents

I assume the answers are:
Case 1:
1-C(44,2)/C(47,2)

Case 2:
1-C(44,4)/C(47,4)

Case 3:
1-C(44,6)/C(47,6)

But this seems almost too simple so I think it’s wrong.

[BruceZ]

[ QUOTE ]
Given that you have AJo (let’s say AsJc) and the flop comes Kxx, how do I calculate the probability that my opponent(s) has a king? (I realize that this will be weighted by my read, but I don’t want to worry about that atm).

I am looking for the method, not just the result. In fact I care less about the result than I do about the method used to solve the problem.

Case 1: 1 opponent
Case 2: 2 opponents
Case 3: 3 opponents

I assume the answers are:
Case 1:
1-C(44,2)/C(47,2)

Case 2:
1-C(44,4)/C(47,4)

Case 3:
1-C(44,6)/C(47,6)

But this seems almost too simple so I think it’s wrong.

[/ QUOTE ]

You're wrong it's right. [img]/images/graemlins/tongue.gif[/img]

[LetYouDown]

At first glance, it looks right to me.

Edit: And Bruce beat me to it. He tends to be right a significant percentage of the time [img]/images/graemlins/grin.gif[/img].

[Supern]

Can you really calculate the risk like this?

The players you meet on the flop has decided to play his hand (if he's not in BB and unraised).

An Ace or a King are more likely to be played than other cards. So the real risk must be higher than this calculation suggests.

10-player table: 18 cards that you don't know:
1 - C(44/18)/C(47/18) = 77% risk that someone has been dealt a K.

At a tight table maybe top 1/3 of the king hands are played in unraised pots.
77% / 3 = 25% that a K is in play on the flop.

At a loose table maybe 1/2 of the king hands are played in unraised pots.
77% / 2 ~= 40 % that a K is in play on the flop.

Don't really know how I am going to include the number of opponents on the flop from here.
Or am I thinking about this the wrong way?

[BruceZ]

[ QUOTE ]
Can you really calculate the risk like this?

The players you meet on the flop has decided to play his hand (if he's not in BB and unraised).

An Ace or a King are more likely to be played than other cards. So the real risk must be higher than this calculation suggests.

10-player table: 18 cards that you don't know:
1 - C(44/18)/C(47/18) = 77% risk that someone has been dealt a K.

At a tight table maybe top 1/3 of the king hands are played in unraised pots.
77% / 3 = 25% that a K is in play on the flop.

At a loose table maybe 1/2 of the king hands are played in unraised pots.
77% / 2 ~= 40 % that a K is in play on the flop.

Don't really know how I am going to include the number of opponents on the flop from here.
Or am I thinking about this the wrong way?

[/ QUOTE ]

He wanted to ignore "reads", which I take to mean he wanted to assume random hands. To be realistic, of course we would have to consider the hands the players could actually hold. But your calculation caused me to think of something which I think is important, or maybe not, I don't know I just woke up, but I think it's important. That is, instead of first considering the set of all hands an opponent can play, and then figuring out what fraction of those hands have a king, we can instead use the percentage of random hands that have a king which we computed, and multiply that by the fraction of king hands that the opponent will play. This is a neater calculation which won't change fundamentally when we consider players with different sets of playable hands.

[Supern]

How is that different from my calculation?
Please explain more and calculate.

[BruceZ]

[ QUOTE ]
How is that different from my calculation?
Please explain more and calculate.


[/ QUOTE ]

It is your calculation.

[TaintedRogue]

But please......let's keep this one goin!
Let's look at the 3 problems realistically and determine what the odds are of the players having a king based upon what hands we think they would play with a K.
Let's say, whether it's 1,2 or 3 opponents, they would only play a K with:
AK, KQ,KJ,KT,K9 suited or unsuited.
How do we compute that one Bruce?

Thanks,
Ken