Factoring Unavailable Cards

[Dovberman]

The probability that cards you need have been dealt to other players can be factored into the odds calculations. Dealt cards are not available to be drawn. This greatly reduces your chance of making or improving your hand.

Suppose 10 players start the hand. 18 cards are in the hands of other players and are not available to be drawn. 32 cards remain available for the draw.

Example: Calculate the odds of flopping three of a kind when you have unpaired hole cards and 18 cards are not available.

Let’s say that you have A,K.

There are less than 3 aces available to be drawn. The probable number of aces in dealt hands is 18/52 * 4 or 1.4. The number of available aces is (3-1.4 ) or 1.6.

Consider all the possible flops where ‘a’ and ‘b’ are your unpaired hole cards.
‘x’ is any card not matching a hole card.

Hole Flop Probability Calc Odds

a,b x,a,a (30/32 * 1.6/31 * .6/30) .001
a,b a,a,x (1.6/32 * .6/31 * 30/30 ) .001
a,b a,x,a (1.6/32 * 30/31 * .6/30) .001
a,b x,b,b (30/32 * 1.6/31 * .6/30) .001
a,b b,b,x (1.6/32 * .6/31 * 30/30 ) .001
a,b b,x,b (1.6/32 * 30/31 * .6/30) .001

Your odds are now .001 * 6 = .006 or 0.6 percent. Compared to 1.47 if no aces are held by other players.

Doyle Brunson hints at the reduced odds in a table on Page 578 of "Super System".

[Wake up CALL]

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The probability that cards you need have been dealt to other players can be factored into the odds calculations.

[/ QUOTE ]

This first sentence in your post shows you are already confused.

Interestingly enough the last sentence in your post shows the only thing you and Doyle have in common: confusion! /images/graemlins/smile.gif

[ QUOTE ]
Doyle Brunson hints at the reduced odds in a table on Page 578 of "Super System".

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[Morbo]

You are certainly a bit confused here. If you don't know what cards an opponent has folded, the card might aswell be the next card on the board. When you talk about odds and probability of for instance flopping a set with a pocket pair, you take all the cards you don't see and calculate the odds that one of them is the next card to fall.

Think like this. If I held a deck of cards to you and asked you to pull one randomly, what are the change that the card you pull is an ace?

Naturally, it is 4/52, which equals about 7.6%

Now, if I remove half of the deck, does the chance alter itself? We can correctly estimate that 2 aces are removed, and so the chance now is 2/26 which also equals 7.6%

So you see, we might just calculate the probability using the whole deck, as it just makes things a whole lot easier.

[jpym84]

FWIW all of the probability tables in Super System are by Mike Caro, not Doyle Brunson.

[PairTheBoard]

This is a common misconception held by many novices trying to figure probabilities. If you stubbornly hold to the misconception despite the explanations given here you should probably stay out of the probability business.

Here's another way to look at it. Do the probabilties for the flop change if the flop is dealt from the bottom of the deck rather than the top? For that matter, should you worry about the value of the burn card when figuring the flop odds? As it is, in a 10 player game the flop is dealt from the 22nd card down from the top of the 52 card deck. 2 of those cards go in your hand which you see. The flop odds don't change whether the 3 flop cards come from the bottom of the deck, the 3,4,5 cards in the deck, or the 22,23,24 cards in the deck. The remaining 47 cards can be all burned, dealt out to people playing blackjack, served for desert, or partially dealt to others at your table. It's irrelvant what's done with them. All that matters for computing your flop odds is the two known in your hand and the 50 remaining unkown cards.

Now if you can do some Baysian calculations to estimate the probability distribution of cards players were dealt according to their preflop action, then you might theoretically improve on this. But that is far more complicated than what you are thinking of. And I've never seen anybody actually try to do it. It's akin to the theory of Bunching which is generally considered not worth worrying about.

PairTheBoard

[Dovberman]

I appreciate all the replies. It is intuitive that if you remove half of a randomly shuffled deck then only 2 aces remain to be drawn. I guess that intuition is not compatible with probability. I knew that this proposal would stir controvesy.

I have won a significant share of pots when I had a pocket pair and flopped 3 of a kind. I have seen pots lost by players with pocket pairs drawing for trips when another player had a card matching their hole card.

I did note the confusion of the replier who had a need to bash me. This person does not understand the intent of this forum is to enlighten in a positive way.

Consider this subject closed.

[SossMan]

[ QUOTE ]
It is intuitive that if you remove half of a randomly shuffled deck then only 2 aces remain to be drawn. I guess that intuition is not compatible with probability.

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Reread Morbo's post.....4/52 = 2/26

[jpym84]

[ QUOTE ]
Now if you can do some Baysian calculations to estimate the probability distribution of cards players were dealt according to their preflop action, then you might theoretically improve on this. But that is far more complicated than what you are thinking of. And I've never seen anybody actually try to do it. It's akin to the theory of Bunching which is generally considered not worth worrying about.

[/ QUOTE ]

Maybe I'm misunderstanding the question but I dont think its necessarily that complicated. Lets say your opponent looks like hes made a flush after a third flush hits on the turn. If you have the ace of that suit, your odds against making the nut flush are now 5.57:1, not 4.11:1, since you only have 7 outs, not 9.

[Dovberman]

I agree that it is not complicated to reduce the number of outs based on the probability of other players having a card that you need. You used similar logic in the example of seeing 2 flush cards on the board.

Of course, you do not know what other players are holding. However you should be able to estimate the odds of them holding hearts.

The idea of factoring unavailable cards is different than all current thinking about odds calculations. That does not mean that the earth must be flat. I need someone to disprove the calculations that I presented.

[PairTheBoard]

That really is an example where you are essentially using Baysian calculations according to the action. In this case it's an unusual action of a tell. If you have a relaiable teil which allows you to conculding something that specific about your opponent's holdings then of course you can use that to adjust probabilties for your draws.

But that is not what the original poster is doing. He's saying something like, 'Drawing to say AA, well with the cards dealt to the other players, someone probably had an ace so let's assume someone had an Ace. Look what that does to my chance of flopping a set of Aces.' He's got no REASON to conclude that someone has an Ace other than the fact that it is somewhat likely. He doesn't understand how to compute total probabilities by way of conditional probabilities.

PairTheBoard

[PairTheBoard]

Ok. Let me show you the correct way to compute something like this using the concept you are talking about. Suppose you are at a 10 player table and are dealt AA. You are wondering what chances you have of flopping a set of Aces. You think, well out of the 18 cards dealt to the other players somebody probably has an Ace. Then you compute the probabilty of flopping a set of Aces based on that assumption. What you are trying to compute is the conditional probability that you will flop a set GIVEN an Ace was dealt to another player. The correct way to compute the total probability of flopping a set is a follows:

P(1 Ace was dealt)*P(flop a set Given that 1 Ace was dealt)
+
P(0 Aces were dealt)*P(flop a set Given 0 Aces were dealt)
+
P(2 Aces were dealt)*P(flop a set Given 2 Aces were dealt)

If you do those calculations correctly you will come up with exactly the same answer as if you just completely ignore the cards being dealt to the other players. That is, unless you can do some Baysian adjustments to the 3 leading standard probabilities above by way of Tells and/or other action at the table. If you have a tell that someone else was also dealt AA then the first two terms above are zero and you're left with a zero for the last term. But barring such Baysian adjustments to the terms like P(exactly 1 Ace was dealt), the total probability of flopping a set will not change because of the complicated calculation. It's up to you if you want to study enough probility theory to understand this. Otherwise don't expect a lot of expert attention to be given to stubbornly promoted crackpot ideas. btw I'm not bashing you. I'm just telling you how it is.

PairTheBoard

[SossMan]

[ QUOTE ]
Ok. Let me show you the correct way to compute something like this using the concept you are talking about. Suppose you are at a 10 player table and are dealt AA. You are wondering what chances you have of flopping a set of Aces. You think, well out of the 18 cards dealt to the other players somebody probably has an Ace. Then you compute the probabilty of flopping a set of Aces based on that assumption. What you are trying to compute is the conditional probability that you will flop a set GIVEN an Ace was dealt to another player. The correct way to compute the total probability of flopping a set is a follows:

P(1 Ace was dealt)*P(flop a set Given that 1 Ace was dealt)
+
P(0 Aces were dealt)*P(flop a set Given 0 Aces were dealt)
+
P(2 Aces were dealt)*P(flop a set Given 2 Aces were dealt)

If you do those calculations correctly you will come up with exactly the same answer as if you just completely ignore the cards being dealt to the other players. That is, unless you can do some Baysian adjustments to the 3 leading standard probabilities above by way of Tells and/or other action at the table. If you have a tell that someone else was also dealt AA then the first two terms above are zero and you're left with a zero for the last term. But barring such Baysian adjustments to the terms like P(exactly 1 Ace was dealt), the total probability of flopping a set will not change because of the complicated calculation. It's up to you if you want to study enough probility theory to understand this. Otherwise don't expect a lot of expert attention to be given to stubbornly promoted crackpot ideas. btw I'm not bashing you. I'm just telling you how it is.

PairTheBoard

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Exquisite, PTB.

[pudley4]

[ QUOTE ]
I agree that it is not complicated to reduce the number of outs based on the probability of other players having a card that you need. You used similar logic in the example of seeing 2 flush cards on the board.

[/ QUOTE ]

Except that we can see the hearts on board, and reasonably deduce that our previously passive opponent does in fact have the flush when he raises when the 3rd heart comes on the turn.

Your "analysis" is nothing but guesswork.

[ QUOTE ]


Of course, you do not know what other players are holding. However you should be able to estimate the odds of them holding hearts.

[/ QUOTE ]

Before the flop, the odds of them holding hearts are exactly the same as the odds of any two cards still remaining in the deck being hearts.

[ QUOTE ]


The idea of factoring unavailable cards is different than all current thinking about odds calculations. That does not mean that the earth must be flat. I need someone to disprove the calculations that I presented.

[/ QUOTE ]

Here you go:

[ QUOTE ]
Let’s say that you have A,K.

There are less than 3 aces available to be drawn. The probable number of aces in dealt hands is 18/52 * 4 or 1.4. The number of available aces is (3-1.4 ) or 1.6.

[/ QUOTE ]

1 - The number of remaining cards is not 52, it's 50
2 - The number of remaining aces is not 4, it's 3

[Morbo]

I never read your calculations since I knew they were irrelevant since I knew you were wrong. Sorry to be such an [censored] about it but I don't know how to put it. Anyway, I will use your calculations and show you:

You fail to realize that since you are already holding AK, there are 50 cards left and 3 aces remaining. This makes the probability of further aces being dealt in this hand 18/50 * 3 = 1.08

The number of available aces is (3 - 1.08) or 1.96

Now you have calculated for the 20 cards dealt out to players, so we have 32 remaining cards. 1.96 / 32 = 0.06 = 6%. This figure represents the odds of the first card on the flop being an ace. I use this number for simplification reasons, if you want the odds of an ace flopping, use (1.96 / 32) + (1.96 / 31) + (1.96 / 30), which equals to about 18%

If we would instead calculate using all 3 remaining aces we also use the whole remaining deck. 3 / 50 = 0.06 = 6%, the same number. You get the same results, the method of just calculating with every unseen card does nothing but makes the whole calculation easier, and easy enough to use in action.

[Dovberman]

Now that makes sense. Thanks for correcting my logic.

[Dovberman]

Thanks for taking the time to explain this.

[Dovberman]

Thanks,

I understand but ignored conditional probabilities.

[BruceZ]

You might want to look at this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=566 041&Forum=,,,,,,,,All_Forums,,,,,,,,&Words=&Search page=3&Limit=25&Main=565745&Search=true&where=&Nam e=197&daterange=&newerval=&newertype=&olderval=&ol dertype=&bodyprev=#Post566041).