"Blockers" Mathematical Proof?

Think in ranges then it will make more sense ;).

Originally posted by DannyJQ
Think in ranges then it will make more sense ;).

This.

Imagine a situation where you squeeze KJo for example, villain has less combinations of AK, KQ,KK,JJ,AJ etc to defend with against the 3b.

Maybe not the best example, but might help a bit

Originally posted by metza

Before anyone jumps and says I don't understand the concept, I do.
eg A 5 means opponent cannot have A A :club:, A A :heart:, A A
Therefore in theory, opponent is 50% less likely to have AA

Did you guys not read this? You just repeated what I already said I know, which is frustrating and makes me think you only read the title

It's not like I don't get the concept. I'm not asking or debating about what it theoretically does to the opponents range. I just presented some ideas about preflop deck combinations and probabilities of any cards being in any position in the deck as being fixed and even, which seems to be in contradiction to this theory of blockers. What I want to know is how to rationalize these two things which both appear to be mathematically true but contradict each other.

You are looking at the probability of two specific hands (with suits known), and not taking into account the number of ways to make these hands. The probability of you having any two specific cards is the same, which I think is confusing you.

P(7 2 :heart:) = P(A 5 :club:) = P(A A :spade:)

This is still true with dead cards. As you mentioned, if villain has A A :heart:, we are actually equally likely to have A 5 or 7 2 :heart:. So you are right to notice that.

But we are ignoring the number of ways we can make these hands. You are correct that the two card strings have equal probability. When we have A 5 though, there are fewer total number of card strings that give him AA (there are 3)*, as opposed to KK (there are 6)*.

When we compute the probability of villain having AA, we sum the probability of each of the possible events (each card string is an event) that give him AA, and each of them are equally likely. This is where the blockers come in. We limit the total number of card strings, hence we lower the probability of them having AA.

* Ignoring order and the other player's cards. If you look at it another way, it will still be a factor of two different.

Originally posted by Optroot
You are looking at the probability of two specific hands (with suits known), and not taking into account the number of ways to make these hands. The probability of you having any two specific cards is the same, which I think is confusing you.

P(7 2 :heart:) = P(A 5 :club:) = P(A A :spade:)

This is still true with dead cards. As you mentioned, if villain has A A :heart:, we are actually equally likely to have A 5 or 7 2 :heart:. So you are right to notice that.

But we are ignoring the number of ways we can make these hands. You are correct that the two card strings have equal probability. When we have A 5 though, there are fewer total number of card strings that give him AA (there are 3)*, as opposed to KK (there are 6)*.

When we compute the probability of villain having AA, we sum the probability of each of the possible events (each card string is an event) that give him AA, and each of them are equally likely. This is where the blockers come in. We limit the total number of card strings, hence we lower the probability of them having AA.

* Ignoring order and the other player's cards. If you look at it another way, it will still be a factor of two different.

This is more the answer I was looking for, thanks. So it is a matter of conditional probability that negates the fact that all card strings are equally possible? Once we have this incomplete information we can then discern information about the card string after the fact?

I wouldn't exactly say I was ignoring the different ways to make these hands, I have acknowledged this in the first post, but more the fact that yes we do "limit" the total number of card strings, but by the time we already have A5, we're already within in one specific fixed card string of 52! possible so it's an after-the-fact limitation.

The thing then that I am dicey on is that its very hard to be certain that the overall strings will deal less often (again just an example hand) any variant of A5 vs any variant of AA less often than any variant of Q6 vs any variant of AA, simply because there are 52! (which is a shitload) iterations of card strings, therefore its very hard to know/solve if this does in fact occur less often, so I guess I will just have to trust that it does.

http://en.wikipedia.org/wiki/Bayes%27_theorem

This is what you're looking for. When you understand Bayes theorem you will understand this as well. Remember that you are looking for event one: ace was dealt, event two: somebody got two aces, you want to count that possibility that someone have two aces when ace was dealt. Then you should compare your outcome with posibilitty of event two without event one.

Btw there are difference betwen two random cards and two sepcific cards. There are different exemplar (?) of dealing with that.

I'm not gonna go into the maths even though I could. I will just use simple logic--->

You have a bag of 5 red stones and 5 blue stones. You take 2 red stones out of the bag.

What is the chance your opponent takes a red stone. Is it 1/2?

Of course not, some of the reds are gone - the chance is now only 3/8.

The card strings are always equally possible. The blockers only change the number of possible strings there are.

Be careful with saying that we are already in a specific card string. We actually are technically not when we try to calculate the probability of an unknown hand, because according to us, we don't know which one we are in. Given that we have A 5 :club:, there are a certain number of card strings possible that give the opponent AA, and a certain number that give them KK. The probability of each of these is proportional to the number of possible card strings we could be in.

I'll try to make this more clear with an example. Let's say we are playing a two player game with a simplified deck. There are only three cards with only ranks A, K, Q all of spades.

I think you would agree that there are 3! ways to order this deck, each of them equally likely. Suppose I get the first card, and you get the second.

All possible orderings are

P(AKQ) = 1/6
P(AQK) = 1/6
P(KAQ) = 1/6
P(KQA) = 1/6
P(QAK) = 1/6
P(QKA) = 1/6

Now suppose I am dealt an ace. Which of these strings could we still be in? There is only AKQ and AQK. So how many of these result in you getting a Q? We know that each of these are equally likely, so that means that each has probability 1/N = 1/2, where N is the number of possible card strings. Since only one of them results in you getting a queen (given I have an ace), then the probability of you having a queen given I have an ace is 1/2, as expected.

If we do this for an actual deck, the way that we calculate the probability is by 'counting' the number card strings that give us the result we want, and dividing it by the total number of card strings (52!).

Go read about conditional probability.. I think this is a case of : Given you have Probability A, what is the probability of B happening?

w34z3l gave a very good example..

Yeah in overall play. But sometimes luck plays a factor.

Then again I've shove A9
On Ax99

Called by AA.

Say you have A 5 Now we know there's eleven :diamond:s left in the deck. Of which it is impossible for Villian to have A :diamond:. Your post reads, it isn't possible for A :diamond:A to exist. This is true. You can always assume you have the nut flush, except in those rare straight flush draw boards. Villian simply cannot have a card that you have, is all you said.