Cards dealt to other players when calculating odds

Hi,

I know this is a basic question but it is really bothering me. :)
When we calculate odds we don't take into account the odds that have been dealt to the other players at the table, but just the cards dealt to us and the community cards. Why not? I've heard the explanation that you work with what you know, but even then it can really make a difference. If there are ten players at the the table then after the flop there not 47 cards left but only 29.
I realize that we can't know whether or not the cards we need to make out hand are in the pack or held by other players but isn't there some way of taking that into consideration and why don't we?

When a big part of villain's range includes the outs you need, you can sometimes discount those outs to weigh your decision in the hand. Otherwise, there's no way of really knowing who folded what or what's still out there.

Say you have QJ and are drawing to the straight. The board is KT27 and villain has raised pre from UTG and fired both turn and river. His range could be heavily weighted towards AK here so you hypothetically could remove an Ace from your outs. So best case scenario is that you'd split if the Ace falls on the river, but you'd scoop with a 9. I'm sure there are much better examples, but you get the point.

There's also times where you and villain are drawing to the same hand but, villain is drawing to the nuts. Example - you're drawing to the straight, where villain has a straight draw and the nut flush draw.

There's no difference between a card sitting somewhere in the deck and it sitting in front of an opponent as long as you don't know what it is.

Yeah, as long as we don't know what the card is our chances of hitting it our real odds with the cards face up could be better or worse than what we calculate, but it should even out. I think there is a theory about card removal where like if the first 7 people fold at a 9 person table it's more likely that they had *bad* cards, so the next person has a better chance of being dealt *good* cards, but you shouldn't really take this into much consideration.

Okay, thanks. I get the point although it still seems like something is missing. I suppose this is one of those case where you have to trust the maths and not your intuition.

You feel like something is missing because you think that there must be a way to calculate the the effect of the opponents cards to the total equation. Actually there is. But it makes no difference.

To clear your mind lets work on a simple situation;

You have two cards and none of them are Ace of Spades, opponent has two cards and we don't know what they are. We deal one card. What is the probability of that card to be As.

Common calculation:
There are 50 unseen cards left and only one of them is As so the probability of As is 1/50. Easy...

If we want to take opponents cards into account:
What is the probability of the opponent not to be dealt As?
- (49/50) * (48/49) = (48/50) = 96%


96% of the time, opponent doesn't have As (now we have 48 cards left in the deck and one of them is As), What is the probability of As to be dealt?
- 1/48

4% of the time, opponent does have As (now we have 48 cards left in the deck and none of them is As), What is the probability of As to be dealt?
- 0/48 = 0


And in total, the probability of As to be dealt is:
- [(1/48) * 96%] + [0 * 4%] = [(1/48) * (48/50)] + 0 = 1/50


So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it? :)

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Originally Posted by Belt

You feel like something is missing because you think that there must be a way to calculate the the effect of the opponents cards to the total equation. Actually there is. But it makes no difference.

To clear your mind lets work on a simple situation;

You have two cards and none of them are Ace of Spades, opponent has two cards and we don't know what they are. We deal one card. What is the probability of that card to be As.

Common calculation:
There are 50 unseen cards left and only one of them is As so the probability of As is 1/50. Easy...

If we want to take opponents cards into account:
What is the probability of the opponent not to be dealt As?
- (49/50) * (48/49) = (48/50) = 96%


96% of the time, opponent doesn't have As (now we have 48 cards left in the deck and one of them is As), What is the probability of As to be dealt?
- 1/48

4% of the time, opponent does have As (now we have 48 cards left in the deck and none of them is As), What is the probability of As to be dealt?
- 0/48 = 0


And in total, the probability of As to be dealt is:
- [(1/48) * 96%] + [0 * 4%] = [(1/48) * (48/50)] + 0 = 1/50


So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it? :)

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LOL... schrödinger's cat.. GG sir

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Originally Posted by Belt

So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it? :)

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Never open the Box.

It's well known that that Cat is double dealing and not double dealing both at the same time and skewing the statistical data and not skewing the statistical data at the same time.


Remember the cat can be alive and dead at the same time as long as you never open the box.

Just Never Open the Box.

For those of you who never, ever even thought of studying quantum mechanics, and have noooo f'n idea about the 'inside joke' going on itt, I present the inevitable link to wikipedia:

Schrödinger's cat - Wikipedia, the free encyclopedia

The gist of it -- and how it applies to the above -- is as follows: in the 'schrodinger's cat' thought experiment, the cat could either be alive or dead; however, due to some obscure equation in quantum mechanics, the cat must be considered both alive and dead at the same time (i.e. it exists in a 'superposition' of both states). However, we could, at any time, "open the box" the cat is in and 'observe' whether it is indeed still alive.

The idea is analogous to our situation with whether or not the 'next card dealt' is the ace of spades. In probability theory, the card has a probability assigned to it, based on known information, that it is indeed the ace of spades. However until we actually observe it is that exact card, we can merely assign a probability.

(At least that's my rudimentary understanding of it. I looked it up in wikipedia, for god's sake. Go me.)

http://upload.wikimedia.org/wikipedi...ingers_cat.png

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The gist of it -- and how it applies to the above -- is as follows: in the 'schrodinger's cat' thought experiment, the cat could either be alive or dead; however, due to some obscure equation in quantum mechanics, the cat must be considered both alive and dead at the same time (i.e. it exists in a 'superposition' of both states). However, we could, at any time, "open the box" the cat is in and 'observe' whether it is indeed still alive.

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Well, yes and no. Without knowing much on quantum theory you got the good bits but its not really from some obscure equation but more of from the interpretation of quantam particles to have a wave-like probability distribution (think of the electron cloud if you remember elementary chemistry). Because there is a large number of possible states a particle can exist in, and the states are energetically accessable and indestinguishable, you can only say it has a probability of being in any particular state at a given time. Until you observe that state, you must account for them all.

I feel a better analogy would be when you put villain on a range of hands, until you see what his hand is you must account for his entire range.

Despite this, Belt's comment has a very close resemblance and is quite clever. :D

And I thought there'd be no way to learn something from this thread.Props to you sir for looking that up and explaining it to the rest of us mere mortals.

???Who makes the Quantum and do their mechanics make as much as say a Ferrari mechanic?

Why does everything always have to be about pussy? ;-)

Thanks all for the various explanations. And people say that poker is waste of time.

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Originally Posted by Belt

So, until you observe the opponents cards, they have no effect on the equation...

It's like Heisenberg and Schrödinger playing heads-up and the cat is dealing, isn't it? :)

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Awesome, awesome quote. Love it!

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Originally Posted by Tasha

Why does everything always have to be about pussy? ;-)

Thanks all for the various explanations. And people say that poker is waste of time.

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lol n1

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Originally Posted by fatguy'06

......... until you see what his hand is you must account for his entire range.........

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:clap: :clap: :clap:

Putting the Jam in this doughnut of a thread.

High-lighting the impurrrrrtance of putting ranges on opponents unknown cards.

:)

I think I can answer this pretty well. If we knew what cards were folded it'd be great.. sometimes they will hold/fold cards we need and sometimes they will still be live in the deck. It doesn't matter because over a lifetime of hands on average we will always have access to the right amount of cards because the times it works in our favor will be balanced by the times it will not.

Hmm, I'm not sure that works, but I suppose there isn't a point of view that says you should take them into consideration and that is enough.

Imagine this situation:

You deal two cards to yourself and two cards facedown to an empty seat. You can't see the cards in the empty seat.

What are the odds that the next card to be dealt off the deck is an ace if you hold:

1) an ace? 3 out of 50
2) two aces? 2 out of 50
3) no aces? 4 out of 50

Do the two cards in the empty seat matter? No -- you don't know what they are, so the probability remains the same from your point of view. Sure, the empty seat could have one or two aces -- but you can't possibly know that when you're figuring out your probability.

The same holds true when you're calculating probabilities at the table -- you don't know what the other players have in their hand, or what they folded. But their hands don't matter because to your point of view, there are 50 remaining unknown cards (or 47 on the flop, etc.).

I guess a simpler way to think of it is: you aren't calculating the odds based on how many cards are in the deck. You are calculating the odds based on how many cards are left that are unknown to you.

I preferred Magic's more mathematical explanation, but I get the point. It's hard to 'see' why they aren't counted but if that's they way probability works then that is fine by me. :cool: